ShunyaSaarthi in Action | JEE Advanced Physics

ShunyaSaarthi · Chapter 01 Torque and pivot choice

Rohan and Seema illustrated chapter scene

01

Three forces on a hinged rod

Foundation · Force diagrams and conservation laws

A uniform rod of length L and finite moment of inertia rotates in a horizontal plane about a frictionless vertical hinge at O. All stated forces lie in that plane; gravity has no torque about the rotation axis. Two forces of magnitude F act at x=L/3 and x=L: the first makes 60° anticlockwise with the rod, the second 30° clockwise. A third force P acts along the rod at x=L/2. Find the signed net torque about O, then identify all P for which the rod has zero angular acceleration.

Illustrated setup: Three forces on a hinged rod — **Read the diagram**The hinge is the pivot: only components perpendicular to the rod turn it. The axial push acts along a line through the hinge, so it adds no moment.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\tau_O=\frac{FL}{6}(\sqrt3-3)<0

Rohan and Seema discuss the setup

Rohan: If P is huge, can it stop this rod turning?

Seema: First mark each force's line of action about O.

Rohan: P points along the rod, so its lever arm is zero?

Seema: Exactly. Add only the other two signed torques.

Step-by-step solution

Set anticlockwise positive: τ₁=(L/3)F sin60°=√3 FL/6; τ₂=−LF sin30°=−FL/2.
The line of action of P passes through O, so τ_P=0 for every P.
Thus τ_net=FL(√3−3)/6<0. No value of P makes angular acceleration zero for F,L>0; an axial push cannot supply the missing torque. The rod's inertia is irrelevant to the zero-torque test.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for three forces on a hinged rod show?

Draw O at the left, two force arrows at L/3 and L, and P directly along the rod. Mark the perpendicular force components.

2

METHOD

Which law or relation earns the method marks?

Sum signed moments about the hinge; the axial force has no lever arm.

3

RESULT

What should you box in the final answer?

The net torque is clockwise and cannot be cancelled by changing the axial force.

4

CHECK

What assumption, limit or direction must you check?

The conclusion assumes nonzero F and L and the stated force directions.

5

AVOID

What tempting mistake loses marks?

Do not multiply the axial force by L as if it were perpendicular.

02

Square plate and a diagonal force

Foundation · Force diagrams and conservation laws

A square plate of side a is pivoted at its centre. Three in-plane forces act: F upward at the right-upper corner, 2F leftward at the right-lower corner, and an unknown force of magnitude Q along the diagonal from the left-lower to the right-upper corner. Find the torque and determine whether Q can balance the first two forces.

Illustrated setup: Square plate and a diagonal force — **Read the diagram**The centre pivot gives the diagonal force a zero moment arm. The upward and leftward corner forces have opposite signed moments; add those before considering the adjustable force.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\tau_C=\frac{aF}{2}-aF=-\frac{aF}{2}

Rohan and Seema discuss the setup

Rohan: Could a stronger diagonal force balance the plate?

Seema: Extend that diagonal through the central pivot.

Rohan: It passes through the pivot, whatever Q is!

Seema: Then calculate the moments of the other two forces.

Step-by-step solution

At (+a/2,+a/2), upward F gives τ₁=+aF/2.
At (+a/2,−a/2), leftward 2F gives τ₂=−aF.
The diagonal force has zero moment about the centre whatever Q. Result τ=−aF/2; no Q can balance it. Do not assume every adjustable force provides adjustable torque.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for square plate and a diagonal force show?

Centre at origin, corners (±a/2,±a/2). Extend the unknown diagonal force's line through the centre.

2

METHOD

Which law or relation earns the method marks?

Use each force line and the centre as the moment axis.

3

RESULT

What should you box in the final answer?

The specified two forces leave a clockwise moment; changing the diagonal force cannot balance it.

4

CHECK

What assumption, limit or direction must you check?

Extend the diagonal line to verify it passes through the centre.

5

AVOID

What tempting mistake loses marks?

Do not assume a large force always creates a large torque.

03

Loaded beam with a cable

Foundation · Force diagrams and conservation laws

A nonuniform horizontal beam of length L has weight W acting at x=0.7L. It is hinged at x=0 and held by a cable attached at x=L and anchored above and to the left. The cable makes an acute angle θ with the leftward horizontal. A downward point load P is at x=0.4L. With +x rightward and +y upward, find tension and both signed hinge-reaction components; state the condition for a taut cable.

Illustrated setup: Loaded beam with a cable — **Read the diagram**The cable pulls up-left. With +x rightward, its horizontal component is negative, so the hinge pushes right. Weight W acts at 0.7L and load P at 0.4L.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

T=\frac{0.7W+0.4P}{\sin\theta}

Rohan and Seema discuss the setup

Rohan: The beam has two loads. Which force should we solve first?

Seema: Take moments about the hinge to remove its reactions.

Rohan: Then tension is fixed by the vertical force component?

Seema: Yes. Use both force balances to find the hinge reaction.

Step-by-step solution

Torque about the hinge: TL sinθ=0.7WL+0.4PL; hence T=(0.7W+0.4P)/sinθ.
The cable components are (−T cosθ,+T sinθ). Horizontal balance gives Hₓ=+T cosθ; vertical balance Hᵧ=W+P−T sinθ=0.3W+0.6P.
A taut upward-supporting cable requires sinθ>0 for positive W,P. A cable angle near zero demands unbounded tension in this ideal model.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for loaded beam with a cable show?

Put W at 0.7L, P at 0.4L, tension T at L, and hinge reactions Hₓ,Hᵧ at 0.

2

METHOD

Which law or relation earns the method marks?

Take moments about the hinge, then resolve horizontal and vertical forces.

3

RESULT

What should you box in the final answer?

Box the cable tension and both signed hinge-reaction components.

4

CHECK

What assumption, limit or direction must you check?

Require an upward cable component and nonzero sine of its angle.

5

AVOID

What tempting mistake loses marks?

Do not put the load at the beam midpoint; its position is 0.4L.

04

All zero-torque reference points

Foundation · Force diagrams and conservation laws

A force F=(F_x,F_y) acts at (a,b) on a lamina in the xy plane. Find every point (x_0,y_0) about which its torque vanishes. Describe geometrically why choosing a point on that set simplifies an equation of motion.

Illustrated setup: All zero-torque reference points — **Read the diagram**A force causes zero torque about any point on its own straight line of action. Move the trial pivot along that line; the perpendicular lever arm stays zero.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

(a-x_0)F_y-(b-y_0)F_x=0

Rohan and Seema discuss the setup

Rohan: Where can I move the pivot so this force makes no torque?

Seema: Put it anywhere on the force's line of action.

Rohan: So a zero moment doesn't mean zero force?

Seema: Correct. Translation can still change.

Step-by-step solution

About (x₀,y₀), τ_z=(a−x₀)F_y−(b−y₀)F_x.
Set τ_z=0: (a−x₀)F_y=(b−y₀)F_x. For a nonzero force this is precisely its entire line of action.
This pivot removes F's torque from an equation, although the force still affects translation. If F=0, every point works.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for all zero-torque reference points show?

Draw the line of action of F through (a,b), then place O anywhere on that same line.

2

METHOD

Which law or relation earns the method marks?

Set the two-dimensional cross product of position and force to zero.

3

RESULT

What should you box in the final answer?

Every allowed zero-moment pivot lies on the entire force line of action.

4

CHECK

What assumption, limit or direction must you check?

For zero force, every point is an exception.

5

AVOID

What tempting mistake loses marks?

Do not confuse zero torque about a pivot with zero effect on translation.

05

A fixed horizontal force on a turning rod

Foundation · Force diagrams and conservation laws

A rigid rod of length L rotates in a horizontal plane about a frictionless vertical pivot at its left end. A constant in-plane force F directed along +x acts at its right end. At rod angle φ measured anticlockwise from +x, find the applied torque τ(φ), its sign and all positions where it vanishes. Explain why a constant force need not give constant angular acceleration. For a motion experiment, release it from rest at 0<φ₀<π; gravity has no torque about the vertical axis.

Illustrated setup: A fixed horizontal force on a turning rod — **Read the diagram**As the rod turns, its perpendicular distance from the horizontal force line changes. Read that distance from the dashed segment before assigning the torque sign.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\tau_z=-FL\sin\phi

Rohan and Seema discuss the setup

Rohan: The push stays horizontal. Why does the torque change?

Seema: The perpendicular distance changes as the rod turns.

Rohan: It becomes zero when the rod lies along the push?

Seema: Yes. Write the position cross the force with a sign.

Step-by-step solution

r=(L cosφ,L sinφ), F=(F,0).
τ_z=rₓF_y−r_yFₓ=−FL sinφ. It vanishes at φ=nπ and changes sign across these positions.
For a rigid rod with fixed-axis inertia I, α(φ)=−(FL/I)sinφ: constant force is not constant torque when its lever arm changes.
If instead φ₀=0 and the rod is exactly at rest, the axial force gives zero torque and it remains at rest in this ideal model.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for a fixed horizontal force on a turning rod show?

At rod angle φ, resolve its endpoint position perpendicular to rightward F.

2

METHOD

Which law or relation earns the method marks?

Write the endpoint position at angle φ before taking its cross product with the horizontal force.

3

RESULT

What should you box in the final answer?

The torque varies with the sine of the rod angle and changes sign across a straight alignment.

4

CHECK

What assumption, limit or direction must you check?

Check it vanishes when the rod and force are collinear.

5

AVOID

What tempting mistake loses marks?

A force fixed in direction does not give a constant torque on a rotating rod.

ShunyaSaarthi · Chapter 02 Angular language and motion

06

A changing angular acceleration

Intermediate · Vectors and calculus

A wheel starts from rest with angular acceleration α(t)=at−bt² for 0≤t≤2a/b, where a,b>0. Find every instant of maximum angular speed, total angular displacement, and the intervals in which its angular acceleration opposes its angular velocity.

Illustrated setup: A changing angular acceleration — **Read the diagram**These scaled graphs show why maximum signed angular velocity and maximum angular speed differ. Compare the positive peak, the zero crossing and the final negative endpoint.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\omega(t)=\frac{at^2}{2}-\frac{bt^3}{3}

Rohan and Seema discuss the setup

Rohan: Where is angular speed greatest on this interval?

Seema: Integrate acceleration, then inspect the whole interval.

Rohan: The positive peak might not be the largest speed?

Seema: Right. Compare absolute values, including the endpoint.

Step-by-step solution

With ω(0)=0, integrate: ω=at²/2−bt³/3=t²(a/2−bt/3).
ω rises to a local maximum a³/(6b²) at t=a/b, crosses zero at 3a/(2b), then falls to −2a³/(3b²) at 2a/b. The largest angular speed |ω| on the stated closed interval occurs at the final endpoint; distinguish this from maximum signed velocity.
Integrate again: θ−θ₀=at³/6−bt⁴/12. At t=2a/b, net angular displacement is zero, despite a nonzero traveled angle.
α opposes nonzero ω only for a/b<t<3a/(2b).

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for a changing angular acceleration show?

Sketch α(t)=t(a−bt), then ω(t); mark t=a/b and t=3a/(2b).

2

METHOD

Which law or relation earns the method marks?

Integrate angular acceleration twice, keeping its sign on each interval.

3

RESULT

What should you box in the final answer?

Identify maximum angular speed by comparing the turning point with the interval endpoint.

4

CHECK

What assumption, limit or direction must you check?

Distinguish net angular displacement from total angle travelled.

5

AVOID

What tempting mistake loses marks?

Do not report the maximum signed angular velocity as the maximum speed.

07

Points at R and 2R

Foundation · Force diagrams and conservation laws

Two particles are fixed on the same radial ray (the same spoke) at radii R and 2R on a rigid wheel with a fixed axis. At an instant the wheel has ω and α of opposite signs. Compare their tangential and centripetal acceleration vectors and determine when their total accelerations are perpendicular to their velocities.

Illustrated setup: Points at R and 2R — **Read the diagram**The two points lie on one radial ray. In this example ω is anticlockwise and α clockwise; inward and tangential acceleration magnitudes both double at the outer point.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\vec a(r)=\alpha r\,\hat t-\omega^2r\,\hat r

Rohan and Seema discuss the setup

Rohan: Do the inner and outer marks have the same acceleration?

Seema: They share angular quantities, but each linear part scales with radius.

Rohan: Can acceleration be normal to velocity here?

Seema: Only if the tangential component vanishes.

Step-by-step solution

Tangential magnitudes are |α|R and 2|α|R; radial magnitudes ω²R and 2ω²R. Corresponding acceleration vectors are parallel and the outer vector is twice the inner.
Velocity is tangential. Total acceleration is perpendicular to velocity only if its tangential part vanishes, α=0 (or the point is at r=0). The given nonzero opposite-signed ω,α do not meet this condition.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for points at r and 2r show?

At each point draw inward a_c=ω²r and tangential a_t=αr; the farther pair is twice as long.

2

METHOD

Which law or relation earns the method marks?

Split acceleration into radial and tangential components at each radius.

3

RESULT

What should you box in the final answer?

At twice the radius, each component and hence the full acceleration vector doubles.

4

CHECK

What assumption, limit or direction must you check?

Vector doubling requires the same radial ray. At different azimuths only the corresponding acceleration magnitudes necessarily have the 2:1 ratio.

5

AVOID

What tempting mistake loses marks?

Opposite signs of angular speed and angular acceleration do not make the total acceleration radial.

08

Cubic angular displacement

Intermediate · Vectors and calculus

A disc of radius R rotates with θ(t)=At³−Bt², A,B>0. A point initially at the positive x axis is observed when its angular velocity first returns to zero after t=0. Find its position modulo 2π, total path length traveled by that point, and its instantaneous acceleration.

Illustrated setup: Cubic angular displacement — **Read the diagram**The scaled angle decreases to its minimum at t*. At that instant the point stops, so centripetal acceleration vanishes, while the positive tangential acceleration remains.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

t_*=\frac{2B}{3A},\quad \theta(t_*)=-\frac{4B^3}{27A^2}

Rohan and Seema discuss the setup

Rohan: The wheel stops again. Has the mark returned to its start?

Seema: Find the angle at that time; stopping is not returning.

Rohan: And its radial acceleration is then zero?

Seema: Yes. The tangential part can still remain.

Step-by-step solution

ω=3At²−2Bt=t(3At−2B); the first later zero is t*=2B/(3A).
θ(t*)=−4B³/(27A²), so the point is at polar angle −4B³/(27A²) modulo 2π and has traveled distance 4RB³/(27A²).
α(t)=6At−2B=2B. Since ω(t*)=0, centripetal acceleration vanishes and the instantaneous acceleration is purely tangential of magnitude 2BR, in the positive rotation direction.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for cubic angular displacement show?

Draw θ(t), decreasing from zero to its minimum before the marked instant.

2

METHOD

Which law or relation earns the method marks?

Differentiate θ to find the later zero of angular velocity, then substitute in θ.

3

RESULT

What should you box in the final answer?

Box the final polar angle modulo a full revolution, path length and tangential acceleration.

4

CHECK

What assumption, limit or direction must you check?

Count travelled angle using absolute angular motion, even if signed θ is negative.

5

AVOID

What tempting mistake loses marks?

Do not include centripetal acceleration at the instant angular speed is zero.

09

Two reversing wheels

Advanced · Vectors and calculus

Two wheels begin with angular velocities +ω and −2ω, and constant angular accelerations −α and +α/2 respectively, where ω,α>0. They carry painted marks that coincide at t=0. Find all later coincidence times and identify any repeated coincidences after either wheel reverses.

Illustrated setup: Two reversing wheels — **Read the diagram**Plot the unwrapped relative angle. Every crossing of a level 2πn is a coincidence; mark the reversal times independently of those crossings.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\theta_1(t)-\theta_2(t)=2\pi n

Rohan and Seema discuss the setup

Rohan: How do we count meetings after the wheels reverse?

Seema: Use unwrapped angles and set their difference to 2πn.

Rohan: Should we keep every quadratic root?

Seema: Keep only real, positive times, then mark the reversals.

Step-by-step solution

θ_A=ωt−αt²/2; θ_B=−2ωt+αt²/4. Thus Δθ=3ωt−3αt²/4.
For each integer n solve Δθ=2πn: 3αt²/4−3ωt+2πn=0; retain real roots t>0. Explicitly t=[3ω±√(9ω²−6παn)]/(3α/2). For n<0 only the plus branch is positive.
A reverses at ω/α, B at 4ω/α. Evaluate the retained roots against these times to classify coincidences. For n=0, besides t=0 they coincide again at t=4ω/α, exactly when B reverses; later n<0 solutions occur after that.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for two reversing wheels show?

Plot their angular positions unwrapped; coincidences occur when their difference crosses an integer multiple of 2π.

2

METHOD

Which law or relation earns the method marks?

Find the two unwrapped angular positions and set their difference equal to any integer number of full turns.

3

RESULT

What should you box in the final answer?

Keep every positive real coincidence time, including the later zero-difference event.

4

CHECK

What assumption, limit or direction must you check?

Test each root against the reversal times of both wheels.

5

AVOID

What tempting mistake loses marks?

Do not equate the unwrapped angles only at zero difference; marks can coincide after multiple turns.

10

Measured acceleration direction

Foundation · Force diagrams and conservation laws

A bead fixed to a turntable at radius r has measured acceleration magnitude A and its acceleration makes angle β with the inward radial direction, 0<β<π/2. Determine the possible instantaneous angular speed and magnitude of angular acceleration. Discuss what the measurement cannot reveal about their signs.

Illustrated setup: Measured acceleration direction — **Read the diagram**At the rim point, resolve the measured acceleration into inward A cosβ and tangential A sinβ. These give angular-speed and angular-acceleration magnitudes.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

|\omega|=\sqrt{\frac{A\cos\beta}{r}},\quad |\alpha|=\frac{A\sin\beta}{r}

Rohan and Seema discuss the setup

Rohan: The acceleration arrow is tilted. Can we find the spin?

Seema: Resolve it inward and tangentially.

Rohan: Will that tell us clockwise or anticlockwise?

Seema: Not from acceleration alone; note the sign ambiguity.

Step-by-step solution

rω²=A cosβ, so |ω|=√(A cosβ/r).
r|α|=A sinβ, so |α|=A sinβ/r.
Acceleration alone cannot distinguish clockwise from anticlockwise rotation. If the tangential side of the inward direction is known, it fixes α's direction, but not ω's sign.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for measured acceleration direction show?

Resolve the observed A into inward A cosβ and tangential A sinβ.

2

METHOD

Which law or relation earns the method marks?

Resolve measured acceleration into inward and tangential parts.

3

RESULT

What should you box in the final answer?

Box the magnitude of angular speed and magnitude of angular acceleration.

4

CHECK

What assumption, limit or direction must you check?

The acceleration direction alone does not reveal the sign of angular speed.

5

AVOID

What tempting mistake loses marks?

Do not take the square root of the tangential component to find angular speed.

ShunyaSaarthi · Chapter 03 Moment of inertia

11

Remove and relocate a ring arc

Advanced · Force diagrams and conservation laws

A thin circular ring of mass M, radius R, has a small segment of angular width 2φ centred at angle zero removed. The removed part is then attached as a point mass at the diametrically opposite rim point. Find the new centre of mass and moment of inertia about the original ring centre. Take the ring's original uniform linear density and use the removed arc's exact mass.

Illustrated setup: Remove and relocate a ring arc — **Read the diagram**Remove the right-hand arc and relocate its exact mass to the opposite rim. The centroid moves left, while every bit still lies a distance R from the original centre.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

I_O=MR^2

Rohan and Seema discuss the setup

Rohan: If mass stays on the rim, does the ring's inertia change?

Seema: About the original centre, every moved bit remains at R.

Rohan: But the centre of mass shifts left?

Seema: Yes. Average the removed arc before relocating its mass.

Step-by-step solution

Removed mass μ=Mφ/π. Its centre is at x=R sinφ/φ, since averaging R cosψ over −φ≤ψ≤φ gives R sinφ/φ.
Original ring has zero first moment. New x_CM=[−μR−μR sinφ/φ]/M=−R(φ+sinφ)/π; y_CM=0.
Every removed bit and the replacement point lie at distance R from O, so I_O remains MR². If an axis through the new CM is requested, subtract Mx_CM² by the parallel-axis theorem.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for remove and relocate a ring arc show?

Ring centre O, removed arc symmetric about +x, replacement point at (−R,0). The arc's centre is not on the rim.

2

METHOD

Which law or relation earns the method marks?

Use the exact arc mass and its average x coordinate; the replacement is a point on the opposite rim.

3

RESULT

What should you box in the final answer?

The original-centre inertia stays equal to that of the full ring; the centre of mass moves left.

4

CHECK

What assumption, limit or direction must you check?

A CM-axis inertia requires one more parallel-axis shift.

5

AVOID

What tempting mistake loses marks?

Do not put the removed arc’s own centre at radius R.

12

Off-centre hole in a disc

Intermediate · Force diagrams and conservation laws

A uniform disc of radius R has a circular hole of radius R/3 cut with its centre R/2 from the disc's centre. Determine the remaining object's centre of mass and I about an axis normal to the disc through that centre of mass, in terms of the original disc mass M.

Illustrated setup: Off-centre hole in a disc — **Read the diagram**Subtract the off-centre hole as a negative disc. Then shift from the original centre to the remaining body’s centroid using its remaining mass.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

I_{\rm remain,CM}=I_{\rm full,O}-I_{\rm hole,O}-M_{\rm remain}d^2

Rohan and Seema discuss the setup

Rohan: Can I subtract just the hole's central-disc inertia?

Seema: Include the hole's offset from the original axis too.

Rohan: Then shift the remaining plate to its new centre?

Seema: Exactly. Use the parallel-axis theorem twice.

Step-by-step solution

Remaining mass M'=8M/9. Its centre is x_CM=−[(M/9)(R/2)]/(8M/9)=−R/16.
About the original centre, I_O=(1/2)MR²−[(1/2)(M/9)(R/3)²+(M/9)(R/2)²]=(151/324)MR².
Shift from O to the remaining object's CM: I_CM=I_O−M'(R/16)²=(1199/2592)MR². Both the missing disc's intrinsic I and its offset are essential.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for off-centre hole in a disc show?

Treat the hole as a negative disc of mass M/9 at x=R/2.

2

METHOD

Which law or relation earns the method marks?

Subtract the removed disc as negative mass, including its own inertia and its offset.

3

RESULT

What should you box in the final answer?

Box the remaining centroid and inertia about that new centroid.

4

CHECK

What assumption, limit or direction must you check?

Use the remaining mass, not original mass, for the final axis shift.

5

AVOID

What tempting mistake loses marks?

Do not subtract only the hole’s centre-axis inertia.

13

Four masses and the least-inertia line

Advanced · Vectors and calculus

Four point masses m,2m,3m,4m occupy (a,0),(0,a),(−a,0),(0,−a). Find the centroid, I about the z axis through the centroid, and the in-plane direction of a line through the centroid that minimizes I about that line.

Illustrated setup: Four masses and the least-inertia line — **Read the diagram**The four unequal masses lie at equal radii. First find their weighted centroid C, then find the principal line through C that gives the smaller in-plane inertia.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

I_z=\sum_i m_i\left[(x_i-x_{\rm CM})^2+(y_i-y_{\rm CM})^2\right]

Rohan and Seema discuss the setup

Rohan: Which line through the masses has least inertia?

Seema: Find their weighted centre, then the principal axes.

Rohan: A coordinate axis may not be the best one?

Seema: Right. Diagonalize the in-plane inertia matrix.

Step-by-step solution

Total mass 10m and centroid C=(−a/5,−a/5).
By subtracting the centroid shift from Σmr², I_z,C=(Σm(x²+y²))−10m(2a²/25)=10ma²−(4/5)ma²=(46/5)ma².
In-plane axis inertia matrix through C is (ma²/5) [[28,2],[2,18]]. Its smaller eigenvalue is I_min=(23−√29)ma²/5.
A minimum-axis direction (uₓ,uᵧ) obeys uᵧ/uₓ=−(5+√29)/2. The perpendicular principal line gives the larger eigenvalue; their sum equals I_z,C.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for four masses and the least-inertia line show?

Plot the four masses on coordinate axes. The optimal in-plane axis passes through the weighted centroid, not the origin.

2

METHOD

Which law or relation earns the method marks?

Calculate the weighted centroid, then diagonalize the in-plane inertia matrix there.

3

RESULT

What should you box in the final answer?

Box the smaller principal inertia and its in-plane direction.

4

CHECK

What assumption, limit or direction must you check?

The two principal in-plane inertias add to the centroidal perpendicular-axis inertia.

5

AVOID

What tempting mistake loses marks?

Do not search axes through the original origin when the required axis passes through the centroid.

14

A bent uniform rod

Intermediate · Force diagrams and conservation laws

A uniform rod of mass M and length L is bent without stretching into two perpendicular arms of lengths L/3 and 2L/3 joined at one end. Find its centre of mass and I about an axis perpendicular to the plane through the centre of mass.

Illustrated setup: A bent uniform rod — **Read the diagram**The short arm lies along +x and the long arm along +y. Uniform density makes their masses proportional to their lengths. The marked centroid is off the rod itself.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

I_{\rm CM}=\sum_{j=1}^2\left(\frac{m_j\ell_j^2}{12}+m_jd_j^2\right)

Rohan and Seema discuss the setup

Rohan: The bent rod still has length L. Is I unchanged?

Seema: No. Mark the two arm masses and their centroids.

Rohan: Compute around the joint before shifting axes?

Seema: That's the cleanest route to I about the centre.

Step-by-step solution

The arms have masses M/3 and 2M/3, with centres (L/6,0) and (0,L/3). Therefore C=(L/18,2L/9).
About the joint, I_O=(M/L)[(L/3)³+(2L/3)³]/3=ML²/9.
Shift to C: I_C=I_O−M[(L/18)²+(2L/9)²]=(19/324)ML².

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for a bent uniform rod show?

Put the common joint at (0,0); shorter arm along +x and longer along +y.

2

METHOD

Which law or relation earns the method marks?

Treat the two perpendicular arms as rods with the same linear density.

3

RESULT

What should you box in the final answer?

Box the weighted centroid and the centroidal perpendicular inertia.

4

CHECK

What assumption, limit or direction must you check?

Their masses follow the one-third and two-thirds length fractions.

5

AVOID

What tempting mistake loses marks?

Do not use one straight-rod inertia after the rod is bent.

15

Two joined discs

Foundation · Force diagrams and conservation laws

Two identical thin solid discs, each of mass M and radius R, are rigidly joined in one plane with centres at (−R,0) and (+R,0). Find I about the z axis through their combined centre of mass and about the parallel z-directed axis through the outermost point (2R,0). Explain the parallel-axis shifts. Both axes are perpendicular to the disc plane.

Illustrated setup: Two joined discs — **Read the diagram**The two discs touch at their combined centre. Both marked axes are normal to the page; the outermost axis is at x=2R, not an in-plane tangent line.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

I_{\rm CM}=2\left(\frac12MR^2+MR^2\right)

Rohan and Seema discuss the setup

Rohan: Do I treat the joined discs as one big disc?

Seema: Add each disc's own I and its centre offset.

Rohan: Then move from their midpoint to the tangent axis?

Seema: Yes, using the combined mass for that shift.

Step-by-step solution

Each disc contributes I_own=(1/2)MR² and MR² from its centre's offset. Hence I_mid=2[(1/2)MR²+MR²]=3MR².
A parallel tangent axis 2R from the combined centre has I_tangent=I_mid+(2M)(2R)²=11MR². Use the total mass for the second shift.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for two joined discs show?

Place disc centres at x=−R and x=+R. Mark perpendicular-to-page axes at x=0 and x=2R; an in-plane tangent line is a different axis.

2

METHOD

Which law or relation earns the method marks?

Shift both disc inertias to their shared midpoint, then shift the combined body to the tangent axis.

3

RESULT

What should you box in the final answer?

Box midpoint inertia 3MR² and outer tangent-axis inertia 11MR².

4

CHECK

What assumption, limit or direction must you check?

Use total mass 2M in the final parallel-axis shift.

5

AVOID

What tempting mistake loses marks?

Do not shift just one disc to the outer tangent line.

ShunyaSaarthi · Chapter 04 Torque and angular acceleration

16

Rod falling from horizontal

Foundation · Force diagrams and conservation laws

A uniform rod of length L and mass M is hinged at one end and released from rest from horizontal. Find initial angular acceleration and hinge reaction components immediately after release. Then obtain its angular speed when vertical using energy.

Illustrated setup: Rod falling from horizontal — **Read the diagram**At release the rod has no angular speed, so its centre has tangent but no radial acceleration. The weight turns it about the hinge; at the later vertical position use the centre’s height drop.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\alpha_0=\frac{Mg(L/2)}{ML^2/3}=\frac{3g}{2L}

Rohan and Seema discuss the setup

Rohan: At release, what acceleration does the rod's centre have?

Seema: First get angular acceleration from weight's hinge torque.

Rohan: There is no centripetal part while ω is zero?

Seema: Correct. Then Newton's law gives the hinge force.

Step-by-step solution

I_hinge=ML²/3. Initial torque MgL/2 gives α₀=3g/(2L), rotating downward.
The centre initially accelerates downward by α₀L/2=3g/4. Newton's vertical equation H_y−Mg=−3Mg/4 gives H_y=Mg/4; H_x=0.
At the bottom the centre has fallen L/2. Energy: MgL/2=(1/2)(ML²/3)ω²; therefore ω=√(3g/L). The hinge force at the bottom is a different quantity from its initial value.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for rod falling from horizontal show?

Draw the centre at L/2, weight straight down and hinge at one end. At release ω=0, so no radial acceleration yet.

2

METHOD

Which law or relation earns the method marks?

Initially use hinge torque and centre acceleration; later use energy.

3

RESULT

What should you box in the final answer?

Box the initial angular acceleration, initial hinge components and speed at the bottom.

4

CHECK

What assumption, limit or direction must you check?

At release angular speed is zero, so centre radial acceleration is zero then.

5

AVOID

What tempting mistake loses marks?

Do not use the bottom hinge reaction as if it were the initial hinge reaction.

17

Speed-dependent driving torque

Intermediate · Vectors and calculus

A disc of moment of inertia I turns about a fixed axle under driving torque τ_0(1−ω/Ω) and resisting torque cω, with positive constants. Starting from rest, find terminal angular speed and the time needed to reach half that speed.

Illustrated setup: Speed-dependent driving torque — **Read the diagram**The decreasing drive-torque curve and increasing resisting-torque curve meet at terminal speed. Their difference determines angular acceleration about a fixed axle.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

I\dot\omega=\tau_0-\left(\frac{\tau_0}{\Omega}+c\right)\omega

Rohan and Seema discuss the setup

Rohan: The motor weakens as the wheel speeds up. Where does it settle?

Seema: Set drive torque equal to resisting torque.

Rohan: How do we find the half-speed time?

Seema: Solve the first-order speed equation from rest.

Step-by-step solution

I dω/dt=τ₀−(τ₀/Ω+c)ω. Define K=τ₀/Ω+c>0.
Set acceleration zero: ω_term=τ₀/K.
Solve the linear equation from rest: ω(t)=ω_term[1−exp(−Kt/I)]. Half-speed occurs at t=(I/K)ln2.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for speed-dependent driving torque show?

On a torque-versus-ω graph, the motor line slopes downward and resisting torque slopes upward; their crossing is terminal speed.

2

METHOD

Which law or relation earns the method marks?

Balance driving and resisting torque, then solve the first-order angular-speed equation.

3

RESULT

What should you box in the final answer?

Box the terminal speed and time to reach half of it.

4

CHECK

What assumption, limit or direction must you check?

The torque slope is positive in the decay constant and the speed tends to a finite limit.

5

AVOID

What tempting mistake loses marks?

Do not set torque equal to a constant throughout the run.

18

Hanging mass and a fixed-axle cylinder

Foundation · Force diagrams and conservation laws

A light string is wound round a solid cylinder of mass M and radius R on a fixed axle. A mass m hangs from the free end. Find acceleration, tension and cylinder angular acceleration; show the limit as M→0 and specify why it differs from a cylinder moving translationally.

Illustrated setup: Hanging mass and a fixed-axle cylinder — **Read the diagram**The hanging mass translates while the cylinder turns about a stationary axle. String tension creates the pulley torque, and no slip ties the mass acceleration to angular acceleration.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

mg-T=ma,\quad TR=I\frac{a}{R}

Rohan and Seema discuss the setup

Rohan: Does the hanging mass pull with its full weight?

Seema: Some force accelerates it; string tension spins the cylinder.

Rohan: Should I add cylinder translation too?

Seema: No. Its axle fixes the centre; use only its torque.

Step-by-step solution

mg−T=ma, TR=Iα and a=αR.
With I=MR²/2, T=(M/2)a. Thus a=mg/(m+M/2), T=mMg/[2m+M], α=a/R.
M→0 gives a→g and T→0. Do not add a cylinder centre-of-mass acceleration term: its axle holds its centre stationary.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for hanging mass and a fixed-axle cylinder show?

Two separate diagrams: falling mass with mg,T; cylinder with one string torque TR. The cylinder's centre is fixed.

2

METHOD

Which law or relation earns the method marks?

Use force balance on the mass, torque on the fixed-axle cylinder and no-slip string acceleration.

3

RESULT

What should you box in the final answer?

Box acceleration, string tension and pulley angular acceleration.

4

CHECK

What assumption, limit or direction must you check?

As pulley mass tends to zero, acceleration tends to g and tension tends to zero.

5

AVOID

What tempting mistake loses marks?

Do not include translational acceleration of the fixed cylinder centre.

19

Off-centre pin and near-end force

Intermediate · Force diagrams and conservation laws

A uniform rod of length L and mass M is free to rotate in a horizontal plane about a fixed vertical pin at a point L/4 from one end. A horizontal force F perpendicular to the rod is applied at its near end. Find the initial angular acceleration and the pin force immediately after release.

Illustrated setup: Off-centre pin and near-end force — **Read the diagram**The applied force sits on the short side of the pin, so its moment points clockwise. The centre of mass is on the other side and accelerates transversely; that fixes the signed pin reaction.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\alpha_0=-\frac{F(L/4)}{I_{\rm pin}}=-\frac{12F}{7ML}

Rohan and Seema discuss the setup

Rohan: Why can a near-end force make the rod turn opposite to it?

Seema: The pin is between the force and the centre.

Rohan: So the torque sign comes from the force's position?

Seema: Yes. Then find the centre acceleration and pin reaction.

Step-by-step solution

I_pin=ML²/12+M(L/4)²=(7/48)ML².
Signed torque τ=−FL/4 gives α=−12F/(7ML). Initially ω=0, so a_CM=α(L/4)=−3F/(7M) in the transverse direction.
Net transverse force is F+H_y=Ma_CM, hence H_y=−10F/7; H_x=0. The pin force can exceed the applied force in magnitude.
The diagram uses +x to the right, +y upward and anticlockwise angular acceleration positive; therefore the clockwise acceleration is negative.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for off-centre pin and near-end force show?

Pin at x=L/4, rod CM a further L/4 to the right, near-end force F upward at x=0. The force turns the rod clockwise.

2

METHOD

Which law or relation earns the method marks?

Find inertia about the off-centre pin and include the force’s signed moment.

3

RESULT

What should you box in the final answer?

Box clockwise angular acceleration and the signed pin reaction.

4

CHECK

What assumption, limit or direction must you check?

At release radial acceleration is zero; centre tangential acceleration sets the pin force.

5

AVOID

What tempting mistake loses marks?

Do not assume the pin reaction must be smaller than the applied force.

20

A brake with torque proportional to angle

Intermediate · Vectors and calculus

A wheel's moment of inertia about a fixed axis is I and it starts with angular speed ω_0. A brake produces resisting torque τ=kθ, where θ measures angular displacement since braking began. Find its stopping angle and discuss whether the braking time has an elementary expression.

Illustrated setup: A brake with torque proportional to angle — **Read the diagram**The resisting torque magnitude grows linearly with angle. Its area is the energy removed before the first stop; the corresponding time integral is elementary.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\frac12I\omega_0^2=\int_0^{\theta_s}k\theta\,d\theta

Rohan and Seema discuss the setup

Rohan: The brake torque grows with angle. Will stopping time be simple?

Seema: Integrate torque work first to find ω as a function of angle.

Rohan: The final time integral looks like a circular arc?

Seema: It is a quarter-period integral; evaluate it exactly.

Step-by-step solution

Work of the brake from 0 to θ is −∫kθ dθ=−kθ²/2.
At rest, Iω₀²/2=kθ_stop²/2, so θ_stop=ω₀√(I/k) for ω₀>0.
During forward motion, ω²=ω₀²−(k/I)θ². Integrate dt=dθ/ω from zero to θ_stop: t_stop=(π/2)√(I/k). It does have an elementary expression (quarter of the associated oscillator period).

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for a brake with torque proportional to angle show?

Plot rotational kinetic energy versus θ; it is a downward parabola and reaches zero at the first stopping angle.

2

METHOD

Which law or relation earns the method marks?

Integrate the angle-dependent braking torque as work, then integrate angle over angular speed for time.

3

RESULT

What should you box in the final answer?

Box the first stopping angle and its finite stopping time.

4

CHECK

What assumption, limit or direction must you check?

Choose the positive root for angle and check the speed vanishes there.

5

AVOID

What tempting mistake loses marks?

Do not replace changing braking torque by its final value; the time integral is elementary.

ShunyaSaarthi · Chapter 05 Work and energy

21

Rod swinging through the bottom

Intermediate · Force diagrams and conservation laws

A uniform rod of mass M and length L is hinged at one end and starts at rest at angle θ_0 from the downward vertical. Find its speed at the lower end at the bottom and the hinge force there; assume it reaches the bottom without obstruction.

Illustrated setup: Rod swinging through the bottom — **Read the diagram**The initial rod makes θ₀ with the downward vertical. At the bottom gravity gives zero torque, but the centre still has upward radial acceleration.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\frac12I_O\omega_b^2=Mg\frac L2(1-\cos\theta_0)

Rohan and Seema discuss the setup

Rohan: The rod reaches the bottom. Is its hinge force just Mg?

Seema: No. Find speed from the centre's drop first.

Rohan: At the bottom the centre also accelerates upward?

Seema: Exactly. Use radial acceleration in the force balance.

Step-by-step solution

Energy gives Mg(L/2)(1−cosθ₀)=(1/2)(ML²/3)ω_b²; so ω_b²=(3g/L)(1−cosθ₀) and lower-end speed v_b=√[3gL(1−cosθ₀)].
At the bottom α=0 because gravity's line passes through the hinge. The centre's acceleration is upward, a_c=ω_b²L/2.
H_y−Mg=Mω_b²L/2, so H_y=Mg[1+(3/2)(1−cosθ₀)], H_x=0. Zero torque at the bottom does not mean zero hinge force.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for rod swinging through the bottom show?

Draw the rod initially θ₀ from downward vertical and finally vertical; the centre drops (L/2)(1−cosθ₀).

2

METHOD

Which law or relation earns the method marks?

Use the drop of the rod centre for energy, then radial force balance at the bottom.

3

RESULT

What should you box in the final answer?

Box the lower-end speed and bottom hinge force.

4

CHECK

What assumption, limit or direction must you check?

At the bottom gravity has zero moment but the centre has upward radial acceleration.

5

AVOID

What tempting mistake loses marks?

Zero angular acceleration at the bottom does not imply zero hinge reaction.

22

Solid sphere versus spherical shell

Intermediate · Force diagrams and conservation laws

A solid sphere and thin spherical shell of equal M,R are released from rest and roll without slipping down identical tracks with vertical drop h. Find their final speeds, ratio of descent times on a straight incline of angle θ, and friction needed at each incline angle.

Illustrated setup: Solid sphere versus spherical shell — **Read the diagram**Both objects start from rest at the same height. The solid sphere and shell have different inertia factors. On the incline friction acts uphill at the contact point.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

v^2=\frac{2gh}{1+I/(MR^2)}

Rohan and Seema discuss the setup

Rohan: Which reaches the bottom first, the sphere or shell?

Seema: Compare how much energy each stores in spin.

Rohan: More rotational inertia means less centre speed?

Seema: Yes. Check the friction each needs to roll.

Step-by-step solution

Rolling energy: Mgh=(1/2)Mv²(1+k), hence v²=2gh/(1+k).
Solid sphere k=2/5 gives v_s²=10gh/7. Thin shell k=2/3 gives v_h²=6gh/5.
On the same straight incline, a=g sinθ/(1+k) and t=√(2s/a). Thus t_s/t_h=√[(7/5)/(5/3)]=√21/5.
Uphill static friction magnitudes are f_s=(2/7)Mg sinθ and f_h=(2/5)Mg sinθ. Check μ_s≥f/(Mg cosθ) separately for each.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for solid sphere versus spherical shell show?

Sketch a solid sphere and a thin spherical shell released from the same height. At the bottom their centre speeds differ; label inertia factors 2/5 and 2/3.

2

METHOD

Which law or relation earns the method marks?

For each body use translation plus spin energy with its own inertia factor.

3

RESULT

What should you box in the final answer?

The solid sphere reaches greater speed; box both speeds, time ratio and uphill friction values.

4

CHECK

What assumption, limit or direction must you check?

Static friction must be sufficient on both tracks.

5

AVOID

What tempting mistake loses marks?

Do not assign the shell the solid sphere’s moment of inertia.

23

Motor and opposing torque

Intermediate · Vectors and calculus

A motor exerts torque τ(θ)=τ_0(1−θ/Θ) on a rotor of I from θ=0 to θ=Θ. A constant opposing torque τ_f also acts. Starting from ω_0, find ω at Θ and the smallest ω_0 that guarantees the rotor can traverse the entire interval.

Illustrated setup: Motor and opposing torque — **Read the diagram**On a torque-angle plot, motor work is the sloping area and opposing work is a rectangle. Their net area changes rotational kinetic energy across the prescribed angle.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\omega_\Theta^2=\omega_0^2+\frac{(\tau_0-2\tau_f)\Theta}{I}

Rohan and Seema discuss the setup

Rohan: Can the rotor cross the whole angle Θ?

Seema: Compare motor work, braking work and initial kinetic energy.

Rohan: Is checking only the final energy enough?

Seema: Here the energy curve is concave; check its endpoints.

Step-by-step solution

Motor work to Θ is τ₀Θ/2. Resistive work is −τ_fΘ.
Energy yields ω_Θ²=ω₀²+(τ₀−2τ_f)Θ/I; choose the nonnegative square root while turning forward.
K(θ)=Iω₀²/2+(τ₀−τ_f)θ−τ₀θ²/(2Θ), a concave function. Its minimum over 0≤θ≤Θ lies at an endpoint.
Therefore the infimum initial speed for reaching Θ is √{max[0,(2τ_f−τ₀)Θ/I]}. If it arrives at Θ with zero speed, its ability to proceed beyond Θ depends on the torque law outside the stated interval.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for motor and opposing torque show?

Shade area under the motor torque-versus-angle triangle; subtract the opposing rectangle.

2

METHOD

Which law or relation earns the method marks?

Integrate motor torque versus angle and subtract the opposing torque work.

3

RESULT

What should you box in the final answer?

Box final speed and the threshold initial speed for traversing the interval.

4

CHECK

What assumption, limit or direction must you check?

Check the kinetic-energy curve stays nonnegative throughout, not just at a guessed interior point.

5

AVOID

What tempting mistake loses marks?

Do not equate motor torque at the endpoint with its average over the interval.

24

Cylinder rolling uphill

Intermediate · Force diagrams and conservation laws

A uniform disc of mass M and radius R rolls up a rough fixed incline that rises to the right with initial centre speed v_0 and no slip throughout. Find its maximum vertical rise and initial and final signs of static friction. Explain carefully whether the friction force must always be nonzero.

Illustrated setup: Cylinder rolling uphill — **Read the diagram**The incline rises rightward. The disc initially moves uphill and spins clockwise, while its acceleration is downhill. Uphill contact friction supplies an anticlockwise torque that slows the initial spin.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

a=-\frac{g\sin\theta}{1+I/(MR^2)}

Rohan and Seema discuss the setup

Rohan: A cylinder rolls uphill. Which way does friction act?

Seema: Sketch its spin and the torque needed to slow that spin.

Rohan: So friction can point uphill too?

Seema: Yes. Solve translation and rotation before trusting intuition.

Step-by-step solution

Let uphill translation and clockwise rotation be positive. Translation: −Mg sinθ+f=Ma. Torque about centre: −fR=Iα; no-slip a=Rα.
With k=I/(MR²)=1/2, solve a=−g sinθ/(1+k)=−(2/3)g sinθ and f=[k/(1+k)]Mg sinθ=(1/3)Mg sinθ, uphill.
Initial energy is (1/2)Mv₀²(1+k), so maximum vertical rise H=(1+k)v₀²/(2g)=3v₀²/(4g). The ideal rolling constraint gives the same uphill friction direction during ascent and immediately after reversal.
For θ>0 and k>0, friction is nonzero under this model. The momentary zero speed at the turning point does not switch off the needed torque.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for cylinder rolling uphill show?

Translation points uphill but acceleration downhill; friction points uphill and its torque reduces the initial clockwise spin.

2

METHOD

Which law or relation earns the method marks?

Choose uphill translation and clockwise rotation positive; link them by no slip.

3

RESULT

What should you box in the final answer?

Box the maximum rise; friction points uphill although acceleration points downhill.

4

CHECK

What assumption, limit or direction must you check?

Friction remains needed at the instant the cylinder stops and reverses.

5

AVOID

What tempting mistake loses marks?

Do not reverse friction merely because the direction of motion reverses.

25

Falling mass coupled to a spring and pulley

Advanced · Force diagrams and conservation laws

A mass m hangs from a string wound around a massive pulley of I and radius R; the other end is attached to a horizontal spring k, arranged so a downward mass displacement y increases its extension from x_0 to x_0+y. Assume x_0≥0. With mass released at rest, formulate energy as a function of y and find the first turning point. State the valid interval before the string could slacken.

Illustrated setup: Falling mass coupled to a spring and pulley — **Read the diagram**A tangent string connects spring and hanging mass. The common displacement controls spring extension and pulley rotation. Classify both downward and upward excursions before checking for a slack string.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\frac12\left(m+\frac I{R^2}\right)\dot y^2+\frac k2\left[(x_0+y)^2-x_0^2\right]-mgy=0

Rohan and Seema discuss the setup

Rohan: As the mass falls, what happens to the spring?

Seema: Its extension increases by the same displacement y.

Rohan: Energy gives the turning point, but is that all?

Seema: Check both string tensions stay nonnegative.

Step-by-step solution

Let y be positive downward and set x=x₀+y, the spring extension. With no string slip, K=(1/2)(m+I/R²)ẏ². Write M_eff=m+I/R².
Energy relative to release is K+(k/2)[(x₀+y)²−x₀²]−mgy=0. The nonzero formal turning displacement is y_turn=2(mg/k−x₀).
If 0≤x₀<mg/k, the first excursion is downward: 0≤y≤y_turn. If x₀=mg/k, the release point is equilibrium and there is no excursion.
If mg/k<x₀≤2mg/k, the first excursion is upward: y_turn≤y≤0. The turning-point extension is 2mg/k−x₀≥0. At x₀=2mg/k the spring-side tension reaches zero only at the turning point.
If x₀>2mg/k, the formal turning extension would be negative. A string cannot transmit the spring compression required by that continuation. The taut model ends first at x=0, y=−x₀, with ẏ²=kx₀(x₀−2mg/k)/M_eff>0. Subsequent slack-string motion needs a new model.
Throughout any stated taut interval, a=[mg−kx]/M_eff, T_spring=kx and T_mass=m(g−a). For x≥0 and I>0, T_spring≥0 and T_mass>0. Thus the spring-side tension determines the slack boundary. Also require the assumed string wrap to persist.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for falling mass coupled to a spring and pulley show?

Mark y downward for the mass and extension x=x₀+y. The pulley's angular speed is ẏ/R.

2

METHOD

Which law or relation earns the method marks?

Write the spring extension as initial extension plus downward mass displacement.

3

RESULT

What should you box in the final answer?

Box the energy equation and the first reachable turning point. For x₀>2mg/k, report the slack boundary y=−x₀ instead of an unreachable formal turning point.

4

CHECK

What assumption, limit or direction must you check?

Classify x₀ relative to mg/k and 2mg/k; require nonnegative spring extension and both string tensions.

5

AVOID

What tempting mistake loses marks?

Do not omit pulley rotational energy or assume tension equals spring force on both sides.

ShunyaSaarthi · Chapter 06 Rolling without slipping

26

Cylinder and general inertia factor

Foundation · Force diagrams and conservation laws

A uniform solid cylinder rolls without slipping down an incline θ. Find a, f, and the minimum coefficient μ_s. Then repeat for a body with I=kmR² and compare the k→0 and k→∞ limits, noting which idealized bodies are physically realizable.

Illustrated setup: Cylinder and general inertia factor — **Read the diagram**Friction acts exactly at the contact point, uphill along the tangent. Gravity and the normal pass through the cylinder centre; friction provides the rolling torque.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

a=\frac{g\sin\theta}{1+\kappa},\quad f=\frac{\kappa}{1+\kappa}mg\sin\theta

Rohan and Seema discuss the setup

Rohan: Why does a hollow body accelerate differently?

Seema: Write I as kmR²; k measures mass distribution.

Rohan: Does no slip automatically hold?

Seema: Only if available static friction meets the required value.

Step-by-step solution

Translation: mg sinθ−f=ma. Rotation: fR=Iα; constraint a=αR.
Set I=kmR². Then f=kma, so a=g sinθ/(1+k) and f=[k/(1+k)]mg sinθ.
Feasible static friction requires μ_s≥f/(mg cosθ)=[k/(1+k)]tanθ. For a solid cylinder k=1/2: a=(2/3)g sinθ, f=(1/3)mg sinθ and μ_s≥(1/3)tanθ.
k→0 gives pointlike/no-spin-inertia behavior a→g sinθ. Algebraically k→∞ gives a→0 and f→mg sinθ, but ordinary bodies confined within a radius R about their centre have k≤1; the large-k limit needs an altered mass model.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for cylinder and general inertia factor show?

On a fixed incline, draw mg sinθ downhill; contact friction uphill generates the required clockwise spin.

2

METHOD

Which law or relation earns the method marks?

Use downhill force balance, centre torque and the rolling constraint.

3

RESULT

What should you box in the final answer?

Box acceleration, uphill static friction and minimum static-friction coefficient.

4

CHECK

What assumption, limit or direction must you check?

The formal very-large-inertia limit may not describe an ordinary body contained within R.

5

AVOID

What tempting mistake loses marks?

Do not set static friction to zero just because the contact point is instantaneously at rest.

27

Spool pulled on upper or lower axle tangent

Advanced · Force diagrams and conservation laws

A spool has outer radius R, axle radius r<R, mass M and I about its centre. The spool is initially at rest. A string tangent to the inner axle is pulled at angle θ above the horizontal while the outer rim rolls without slipping on the floor. For 0≤θ<π/2, determine the initial direction of centre motion as a function of θ for both string-winding geometries; identify the critical angle(s).

Illustrated setup: Spool pulled on upper or lower axle tangent — **Read the diagram**The outer rim contacts the floor at C while the pull is tangent to the small inner axle. Its moment about C depends on whether the string leaves above or below; the lower case can reverse direction.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

a_{\rm upper}=\frac{FR(R\cos\theta+r)}{I+MR^2},\quad a_{\rm lower}=\frac{FR(R\cos\theta-r)}{I+MR^2}

Rohan and Seema discuss the setup

Rohan: Can pulling a spool to the right make it roll left?

Seema: Take torque about the floor contact for each string tangent.

Rohan: The lower string has a critical angle?

Seema: Yes. Compare R cosθ with the axle radius r.

Step-by-step solution

Take rightward centre acceleration positive and clockwise angular acceleration α=a/R. Around the floor contact, effective fixed-contact inertia for rolling is I+MR².
The force supplies clockwise driving moment F(R cosθ+r) for the upper tangent: a_upper=FR(R cosθ+r)/(I+MR²), always rightward for 0≤θ<π/2.
For the lower tangent, a_lower=FR(R cosθ−r)/(I+MR²). It moves right for cosθ>r/R, left for cosθ<r/R, and has zero initial acceleration at θ_c=arccos(r/R).
Verify with ΣFₓ=F cosθ+f=Ma and torque about the centre. The sign result assumes the string exits the specified upper/lower side and sufficient static friction.
The contact solution also requires N=Mg−F sinθ>0 and |f|=|Ma−F cosθ|≤μ_s N. At N=0 contact is marginal and friction capacity vanishes; if N<0 the spool lifts and the floor-rolling model fails. These results classify initial motion from rest, not the velocity of a spool already moving.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for spool pulled on upper or lower axle tangent show?

Use the floor contact as an instantaneous torque axis. Upper tangent force has a lever arm R cosθ+r; lower tangent has R cosθ−r.

2

METHOD

Which law or relation earns the method marks?

Take moments of the pull about the instantaneous floor contact for each axle tangent.

3

RESULT

What should you box in the final answer?

Upper pull drives right; lower pull changes direction at the critical pull angle.

4

CHECK

What assumption, limit or direction must you check?

Check release from rest, maintained contact N=Mg−F sinθ>0 and |Ma−F cosθ|≤μ_s N.

5

AVOID

What tempting mistake loses marks?

Do not apply the same lever arm to both tangent positions.

28

Sliding-to-rolling cylinder

Intermediate · Force diagrams and conservation laws

A uniform solid cylinder of radius R initially has centre speed v_0 to the right and angular speed ω_0 clockwise on a rough horizontal floor with kinetic friction coefficient μ>0. Find the time and centre displacement until pure rolling begins for all three regimes v_0−Rω_0 positive, zero, and negative, with signed friction explicitly shown.

Illustrated setup: Sliding-to-rolling cylinder — **Read the diagram**Read relative slip at the bottom before choosing friction. If the bottom slides right relative to the floor, friction points left; if it slides left, the force reverses.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

u_0=v_0-R\omega_0,\quad t_{\rm roll}=\frac{|u_0|}{3\mu g}

Rohan and Seema discuss the setup

Rohan: The cylinder slides and spins. When will it roll cleanly?

Seema: Track the bottom slip speed u=v−Rω.

Rohan: Does friction always point left?

Seema: It opposes the sign of u, which can be either way.

Step-by-step solution

For a solid cylinder, a=−μg sgn(u), α_clockwise=(2μg/R)sgn(u), so u̇=a−Rα=−3μg sgn(u).
If u₀=v₀−Rω₀≠0, rolling begins at t_r=|u₀|/(3μg). The centre displacement is x_r=v₀t_r−(μg/2)sgn(u₀)t_r².
At that instant v_r=v₀−u₀/3=(2v₀+Rω₀)/3 and ω_r=v_r/R. If u₀=0, t_r=0 and kinetic friction never acts in this ideal setup.
If the computed x_r is negative, it denotes displacement left of the starting point; do not silently replace it by a path length.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for sliding-to-rolling cylinder show?

At the bottom, draw relative slip u=v−Rω with clockwise ω positive. Friction points opposite the sign of u.

2

METHOD

Which law or relation earns the method marks?

Form the signed bottom-point slip and direct kinetic friction against it.

3

RESULT

What should you box in the final answer?

Box time to rolling, signed displacement and final centre speed for each slip sign.

4

CHECK

What assumption, limit or direction must you check?

If initial slip is zero, rolling begins immediately with no kinetic-friction phase.

5

AVOID

What tempting mistake loses marks?

Do not call a negative displacement a positive distance travelled.

29

Solid sphere on a moving belt

Intermediate · Force diagrams and conservation laws

A uniform solid sphere of mass m and radius R touches a conveyor belt moving right at maintained speed U. At t=0 the sphere has zero spin and centre velocity v₀ along +x. With kinetic friction coefficient μ>0 until no slip, find the final centre velocity, signed clockwise angular velocity and time to roll relative to the belt.

Illustrated setup: Solid sphere on a moving belt — **Read the diagram**Use the solid-sphere inertia. No slip relative to the belt means v−Rω=U, not zero. Friction opposes u=v−Rω−U until the contact velocities match.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

v_f=\frac{5v_0+2U}{7},\quad \omega_f=\frac{5(v_0-U)}{7R},\quad t_r=\frac{2|v_0-U|}{7\mu g}

Rohan and Seema discuss the setup

Rohan: The belt moves. Is rolling defined by v=Rω?

Seema: Not here: the bottom surface speed must equal U.

Rohan: Then the belt can change the body's energy?

Seema: Yes. Its motor supplies or removes work.

Step-by-step solution

Use I=2mR²/5 for a uniform solid sphere and clockwise angular velocity positive. The belt-relative slip is u=v−Rω−U; initially u₀=v₀−U.
While sliding, a=−μg sgn(u) and α_clockwise=(5μg/2R)sgn(u). Therefore u̇=−(7/2)μg sgn(u).
Slip reaches zero after t_r=2|v₀−U|/(7μg). Integrating gives v_f=v₀−2u₀/7=(5v₀+2U)/7 and ω_f=5u₀/(7R).
Check v_f−Rω_f=U. If v₀=U, t_r=0 and no kinetic-friction phase occurs. The belt motor can supply or remove energy; mechanical energy of the sphere alone is not conserved.
Example: v₀=0 and U>0 gives v_f=2U/7 and ω_f=−5U/(7R): the sphere translates right and spins anticlockwise, with its bottom moving right at U.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for solid sphere on a moving belt show?

Compare the bottom surface velocity v−Rω with belt speed U, not with zero.

2

METHOD

Which law or relation earns the method marks?

Use the sphere inertia 2mR²/5 and track u=v−Rω−U until relative slip vanishes.

3

RESULT

What should you box in the final answer?

Box v_f=(5v₀+2U)/7, clockwise-signed ω_f=5(v₀−U)/(7R), and t_r=2|v₀−U|/(7μg).

4

CHECK

What assumption, limit or direction must you check?

The final bottom-point speed must equal the maintained belt speed.

5

AVOID

What tempting mistake loses marks?

Do not impose conservation of the sphere’s mechanical energy on a driven belt.

30

Cylinder rolling on an accelerating plank

Advanced · Force diagrams and conservation laws

A uniform solid cylinder of mass M and radius R rolls without slipping on a horizontal plank of mass P which slides without friction on the ground. A horizontal force F is applied to the cylinder's centre. Find both translational accelerations, angular acceleration and contact friction, assuming sufficient static friction.

Illustrated setup: Cylinder rolling on an accelerating plank — **Read the diagram**F acts through the cylinder centre. Static friction points left on the cylinder and right on the plank. Their contact accelerations agree, giving a_c−Rα=a_p.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

a_c-R\alpha=a_p

Rohan and Seema discuss the setup

Rohan: The plank also moves. What is the no-slip condition?

Seema: Match the cylinder contact acceleration to the plank's.

Rohan: So friction acts oppositely on the two bodies?

Seema: Exactly. Draw separate diagrams before solving.

Step-by-step solution

Cylinder: Ma_c=F+f; clockwise angular acceleration α=−fR/I. Plank: Pa_p=−f.
No-slip at the interface: a_c−Rα=a_p. Substitute to find f=−F/[1+MR²/I+M/P].
For a solid cylinder I=MR²/2: f=−FP/(3P+M), a_p=F/(3P+M), a_c=F(2P+M)/[M(3P+M)], and α=2FP/[MR(3P+M)] clockwise.
Friction acts left on the cylinder, right on the plank. Require |f|≤μ_s N for the assumed no-slip solution.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for cylinder rolling on an accelerating plank show?

Separate cylinder and plank. Let rightward friction on the cylinder be signed f; the plank receives −f. No-slip applies to relative contact acceleration.

2

METHOD

Which law or relation earns the method marks?

Draw separate cylinder and plank forces and use relative no-slip acceleration.

3

RESULT

What should you box in the final answer?

Box both translational accelerations, angular acceleration and signed friction.

4

CHECK

What assumption, limit or direction must you check?

Check the friction required does not exceed available static friction.

5

AVOID

What tempting mistake loses marks?

Do not set the cylinder contact acceleration to zero in the laboratory frame.

ShunyaSaarthi · Chapter 07 Angular momentum and collisions

31

Particle sticks to a pinned rod

Intermediate · Force diagrams and conservation laws

A particle of mass m moving at speed v hits and sticks to the end of an initially stationary uniform rod of length L and mass M, pivoted at its other end. Its incident velocity makes angle φ with the rod. Find the angular speed just after impact and kinetic energy lost. Explain why angular momentum about the pivot can be conserved during impact even though linear momentum is not.

Illustrated setup: Particle sticks to a pinned rod — **Read the diagram**The incident path is oblique. Only v sinφ contributes to angular momentum about the left pin; the sign follows the approach direction. The initially stationary rod starts rotating after sticking.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\omega_+=\frac{mvL\sin\phi}{(M/3+m)L^2}

Rohan and Seema discuss the setup

Rohan: The particle sticks. Can I conserve energy at impact?

Seema: Use angular momentum about the pin during the short collision.

Rohan: Only the velocity across the rod contributes?

Seema: Yes. Compute the energy loss afterward.

Step-by-step solution

During short impact, the pin impulse has zero moment about its own location. Initial L_pivot=mvL sinφ, with sign set by the approach direction.
Final inertia is I=(M/3+m)L². Hence ω=mvL sinφ/[(M/3+m)L²]=mv sinφ/[L(M/3+m)].
The loss from the particle's initial kinetic energy is ΔE=(1/2)mv²−(1/2)Iω²=(1/2)mv²−m²v²sin²φ/[2(M/3+m)]. The axial motion and inelastic sticking both contribute to loss.
Linear momentum need not be conserved because the pin supplies an external impulse.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for particle sticks to a pinned rod show?

Draw the incident velocity component perpendicular to the rod; the component along the rod contributes no angular momentum about the pin.

2

METHOD

Which law or relation earns the method marks?

During impact take angular momentum about the fixed pin; include only the incident transverse velocity.

3

RESULT

What should you box in the final answer?

Box the immediate angular speed and the nonnegative energy loss.

4

CHECK

What assumption, limit or direction must you check?

The pin impulse has zero moment about the pin even though it changes total linear momentum.

5

AVOID

What tempting mistake loses marks?

Do not conserve kinetic energy in a sticking collision.

32

Skater pulls masses inward

Intermediate · Force diagrams and conservation laws

A skater modeled as two point masses m at radius r and a torso of inertia I_0 rotates at ω_0. The masses are pulled to r/2. Find final ω, work done by the skater, and the ratio of angular kinetic energies.

Illustrated setup: Skater pulls masses inward — **Read the diagram**Two equal masses move from r to r/2 about the same axis. Angular momentum remains constant while the skater’s work increases rotational kinetic energy.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

(I_0+2mr^2)\omega_0=\left(I_0+\frac{mr^2}{2}\right)\omega_f

Rohan and Seema discuss the setup

Rohan: The skater pulls weights inward. Why does spin speed up?

Seema: External torque is negligible, so angular momentum remains.

Rohan: Where does the extra kinetic energy come from?

Seema: From work done pulling the masses inward.

Step-by-step solution

I_i=I₀+2mr² and I_f=I₀+2m(r/2)²=I₀+mr²/2.
Angular momentum conservation gives ω_f=(I_i/I_f)ω₀.
K_f/K_i=I_i/I_f; work done by the skater is W=K_f−K_i=(1/2)I_iω₀²(I_i/I_f−1)>0. Conserved angular momentum does not mean conserved kinetic energy.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for skater pulls masses inward show?

Draw two masses at r and then r/2; external torque about the axle is zero, while the skater does internal work.

2

METHOD

Which law or relation earns the method marks?

Conserve axial angular momentum while the skater draws two masses inward.

3

RESULT

What should you box in the final answer?

Box the increased angular speed and positive work done by the skater.

4

CHECK

What assumption, limit or direction must you check?

The ratio of final to initial kinetic energy equals the inverse inertia ratio.

5

AVOID

What tempting mistake loses marks?

Conserved angular momentum does not mean kinetic energy is conserved.

33

Mass sticks gently to a rotating disc

Intermediate · Force diagrams and conservation laws

A uniform disc of moment of inertia I rotates at ω₀ about a frictionless fixed vertical axle. A small mass m, initially with zero horizontal velocity and zero angular momentum about that axle, is lowered gently and released onto the disc at radius a. Neglect its vertical impact energy and all external torque about the axle during sticking. Find the final angular velocity and rotational mechanical energy lost.

Illustrated setup: Mass sticks gently to a rotating disc — **Read the diagram**The added mass initially has zero angular momentum about the axle. Gentle vertical placement increases the final inertia and dissipates rotational kinetic energy.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\omega_f=\frac{I\omega_0}{I+ma^2}

Rohan and Seema discuss the setup

Rohan: A mass lands on the spinning disc. Which law survives?

Seema: Use angular momentum about the fixed vertical axle.

Rohan: Does the larger final I mean slower rotation?

Seema: Yes. Then subtract energies to find the loss.

Step-by-step solution

Initial L=Iω₀, final I'=I+ma². Thus ω_f=Iω₀/(I+ma²).
The rotational kinetic energy lost is (1/2)Iω₀²−(1/2)(I+ma²)ω_f²=[Ima²/(2(I+ma²))]ω₀².
The question's gentle placement avoids unspecified vertical impact energy. The axle may apply a force but negligible torque about its axis.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for mass sticks gently to a rotating disc show?

The small mass arrives with negligible angular momentum about the vertical axle; once attached it enlarges I.

2

METHOD

Which law or relation earns the method marks?

Conserve angular momentum about the axle before and after gentle attachment.

3

RESULT

What should you box in the final answer?

Box reduced common angular speed and rotational energy lost.

4

CHECK

What assumption, limit or direction must you check?

Assume negligible vertical impact and negligible external axial torque.

5

AVOID

What tempting mistake loses marks?

Do not give the slowly lowered mass an initial tangential angular momentum.

34

Central force and circular orbit

Advanced · Vectors and calculus

A particle moves under an attractive central force F(r)=−kr² along the radial direction. State what is conserved, derive its effective potential for nonzero angular momentum L, and determine whether a circular orbit is stable.

Illustrated setup: Central force and circular orbit — **Read the diagram**The force points radially inward, so it has no torque about O. At fixed angular momentum, combine the radial potential with the angular-momentum barrier to test the circular orbit.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

U_{\rm eff}(r)=\frac{L^2}{2mr^2}+\frac{kr^3}{3}

Rohan and Seema discuss the setup

Rohan: Can a strange central force hold a circular orbit?

Seema: Build the effective potential at fixed angular momentum.

Rohan: How do I tell if the orbit is stable?

Seema: Find a stationary radius and test the second derivative.

Step-by-step solution

A central force exerts zero torque about its centre, so L=mr²φ̇ is conserved. Since F_r=−kr², potential U(r)=kr³/3 (up to a constant).
U_eff(r)=L²/(2mr²)+kr³/3. A circular orbit has U_eff'=−L²/(mr³)+kr²=0, so r_c=[L²/(mk)]^(1/5).
U_eff''=3L²/(mr⁴)+2kr>0 at r_c: the circular orbit is radially stable for small perturbations.
Total mechanical energy E=(1/2)mṙ²+U_eff(r) is also conserved because the central force is time independent and derives from U(r). The mass m and k are positive; the circular-orbit analysis assumes L≠0.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for central force and circular orbit show?

Plot effective potential: centrifugal term rises at small r and kr³/3 rises at large r, making a well.

2

METHOD

Which law or relation earns the method marks?

Combine radial potential with the angular-momentum barrier to form effective potential.

3

RESULT

What should you box in the final answer?

Box the circular-orbit radius and conclude radial stability from a positive second derivative.

4

CHECK

What assumption, limit or direction must you check?

A central force has zero torque about the force centre.

5

AVOID

What tempting mistake loses marks?

Do not infer stability just from zero first derivative.

35

Simultaneous impacts on a free rod

Advanced · Force diagrams and conservation laws

An initially stationary uniform rod of mass M and length L lies free on a smooth horizontal plane, with no pin or fixed centre. Put its endpoints at x=±L/2 and take +y upward in the top view. Two equal masses m strike and stick simultaneously: the left particle has velocity +vŷ and the right particle −vŷ. Find final angular velocity and energy loss. Repeat for both incident velocities +vŷ.

Illustrated setup: Simultaneous impacts on a free rod — **Read the diagram**This rod has no fixed pin. The left particle moves along +y and the right along −y: their momenta cancel, but both give clockwise angular momentum about the centre.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

L_z=\pm mvL,\quad I_f=\frac{ML^2}{12}+\frac{mL^2}{2}

Rohan and Seema discuss the setup

Rohan: The two projectiles have opposite velocities. Does the rod move?

Seema: Their linear momenta cancel; their torques add.

Rohan: What if both velocities point the same way?

Seema: Then moment cancels and the joined body translates.

Step-by-step solution

Opposite velocities give initial L_z=−mvL about the rod's centre. Final inertia I=ML²/12+2m(L/2)²=L²(M/12+m/2), so ω_f=−mvL/I. The centre does not translate because total linear momentum is zero.
Initial energy is mv². Final rotational energy is L_z²/(2I); heat/deformation loss is mv²−m²v²L²/(2I).
If both incident velocities are +vŷ, initial L_z about the centre is zero and total momentum is 2mvŷ. The joined system translates at V=2mv/(M+2m) and has ω_f=0; kinetic-energy loss is mv²−(1/2)(M+2m)V².

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for simultaneous impacts on a free rod show?

Put the rod ends at x=±L/2. For opposite transverse velocities choose left +vŷ and right −vŷ; their linear momenta cancel but torques add.

2

METHOD

Which law or relation earns the method marks?

Use transverse momentum and torque about the rod centre for both impact directions.

3

RESULT

What should you box in the final answer?

Opposite incoming velocities give rotation with no CM motion; same velocities give CM motion with no rotation.

4

CHECK

What assumption, limit or direction must you check?

Compute energy loss separately for each sticking case.

5

AVOID

What tempting mistake loses marks?

Do not assume equal and opposite velocities cancel angular momentum as well as linear momentum.

ShunyaSaarthi · Chapter 08 Equilibrium and coupled bodies

36

Atwood machine with a massive pulley

Foundation · Force diagrams and conservation laws

Masses m_1 and m_2 hang on opposite sides of a pulley with mass M, radius R and I=MR²/2. String is light, inextensible and does not slip. Derive a, both tensions and pulley angular acceleration for m_2>m_1; determine the limit M→0.

Illustrated setup: Atwood machine with a massive pulley — **Read the diagram**The heavier right mass descends while the left rises. Its side tension and the left side tension must differ to turn a pulley with inertia; use one common acceleration magnitude.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

a=\frac{(m_2-m_1)g}{m_1+m_2+I/R^2},\quad \alpha=\frac aR

Rohan and Seema discuss the setup

Rohan: Can we call both pulley tensions T?

Seema: A massive pulley needs their difference to accelerate.

Rohan: So I write two block equations and one torque equation?

Seema: Yes. Tie all accelerations with no slip.

Step-by-step solution

Take m₂ downward: m₂g−T₂=m₂a, T₁−m₁g=m₁a.
Pulley: (T₂−T₁)R=I(a/R)=(M/2)Ra. Add equations: a=(m₂−m₁)g/(m₁+m₂+M/2).
T₁=m₁(g+a), T₂=m₂(g−a), α=a/R. As M→0, T₂−T₁→0 and the usual massless-pulley formula follows.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for atwood machine with a massive pulley show?

Draw T₁ and T₂ as separate arrows. Their difference, not either one alone, turns the pulley.

2

METHOD

Which law or relation earns the method marks?

Write separate mass force equations and a torque equation for the massive pulley.

3

RESULT

What should you box in the final answer?

Box acceleration, both distinct tensions and angular acceleration.

4

CHECK

What assumption, limit or direction must you check?

The massless-pulley limit makes the tension difference vanish.

5

AVOID

What tempting mistake loses marks?

Do not set the two tensions equal when pulley inertia is finite and acceleration is nonzero.

37

Ladder, wall, ground and person

Intermediate · Force diagrams and conservation laws

A uniform ladder of length L and mass M rests against a frictionless vertical wall on rough ground, making angle θ to the horizontal. A person of mass m stands x from its foot. Find normal forces and minimum ground friction coefficient for equilibrium; determine maximal allowed x for given μ_s.

Illustrated setup: Ladder, wall, ground and person — **Read the diagram**With the wall on the right, its normal force points left. The floor supplies upward normal force and rightward friction. Take moments about the foot to remove both ground forces.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

N_wL\sin\theta=g\left(\frac{ML}{2}+mx\right)\cos\theta

Rohan and Seema discuss the setup

Rohan: How far up the ladder can a person climb?

Seema: Take torque about its foot, then check ground friction.

Rohan: The wall contributes only a horizontal force?

Seema: Correct for a frictionless wall.

Step-by-step solution

Torque about the foot: N_w L sinθ=[Mg(L/2)+mgx]cosθ. Thus N_w=g cotθ(M/2+mx/L).
Horizontal and vertical balance: |f_g|=N_w and N_g=(M+m)g.
Minimum coefficient is μ_min=(M/2+mx/L)cotθ/(M+m).
For fixed μ_s, x≤(L/m)[μ_s(M+m)tanθ−M/2], additionally 0≤x≤L. If the right side is negative, not even x=0 is allowed; if above L, the whole ladder length is allowed.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for ladder, wall, ground and person show?

At the frictionless wall draw only a horizontal reaction N_w; ground supplies vertical N_g and horizontal f_g.

2

METHOD

Which law or relation earns the method marks?

Moment-balance about the ladder foot; wall reaction is horizontal only.

3

RESULT

What should you box in the final answer?

Box wall force, ground normal, minimum friction coefficient and allowed person position.

4

CHECK

What assumption, limit or direction must you check?

Intersect the computed position bound with the physical interval along the ladder.

5

AVOID

What tempting mistake loses marks?

Do not add friction at the explicitly smooth wall.

38

Cable-supported loaded beam

Foundation · Force diagrams and conservation laws

A horizontal uniform beam of length L and mass M is hinged at its left end. A cable attached at the right end is anchored above and to the left, making an acute angle θ with the leftward horizontal. A mass m hangs from position x along the beam. With +x rightward and +y upward, find tension and signed hinge reaction for 0≤x≤L. Determine whether the vertical reaction changes sign.

Illustrated setup: Cable-supported loaded beam — **Read the diagram**The cable pulls up-left, so the hinge’s horizontal reaction is rightward. Increasing the load position changes tension, but the vertical hinge reaction remains positive for 0≤x≤L.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

TL\sin\theta=Mg\frac L2+mgx

Rohan and Seema discuss the setup

Rohan: As the load moves right, can the hinge lift downward?

Seema: Find tension from torque, then calculate the hinge's vertical force.

Rohan: Does its sign ever change on this beam?

Seema: Evaluate the formula across 0≤x≤L.

Step-by-step solution

Moment balance at hinge: TL sinθ=MgL/2+mgx, so T=g(M/2+mx/L)/sinθ.
Hₓ=+T cosθ and Hᵧ=(M+m)g−T sinθ=g[M/2+m(1−x/L)].
For positive M and 0≤x≤L, Hᵧ≥Mg/2>0. There is no sign change within the beam, even though the cable's vertical share grows with x.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for cable-supported loaded beam show?

Tension acts at x=L; the hinge has two signed components; beam weight at L/2 and hanging mass at x.

2

METHOD

Which law or relation earns the method marks?

Balance moment about hinge, then resolve cable and hinge forces.

3

RESULT

What should you box in the final answer?

Box cable tension and both signed hinge reactions as load position varies.

4

CHECK

What assumption, limit or direction must you check?

The vertical hinge reaction stays positive on the physical interval for positive masses.

5

AVOID

What tempting mistake loses marks?

Do not announce a sign change at a position outside the beam.

39

Block, rough table and hanging mass

Intermediate · Force diagrams and conservation laws

Two blocks m₁,m₂ are connected by a light inextensible string over a fixed pulley of inertia I and radius R. Block m₁ is already sliding right on a horizontal table with coefficient μₖ; block m₂ is descending. With no string slip, find the signed acceleration and both tensions while these velocity directions persist. Classify when the descending mass speeds up, moves at constant speed or slows down. Explain why a start from rest needs static-friction data.

Illustrated setup: Block, rough table and hanging mass — **Read the diagram**Rightward sliding gives leftward kinetic friction. The hanging mass is descending, but its signed acceleration can be positive, zero or negative; negative means decelerating descent.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

a=\frac{g(m_2-\mu_km_1)}{m_1+m_2+I/R^2}

Rohan and Seema discuss the setup

Rohan: The table block has friction and the pulley has inertia. Start where?

Seema: Draw three bodies and use separate tensions.

Rohan: Does a negative answer mean the mass cannot be descending?

Seema: No. It means a descending mass slows down until it stops.

Step-by-step solution

Take rightward acceleration of m₁ and downward acceleration of m₂ as the same signed a. Friction on the sliding table block remains leftward while its velocity remains rightward: T₁−μₖm₁g=m₁a and m₂g−T₂=m₂a.
Pulley torque gives T₂−T₁=(I/R²)a. Thus a=g(m₂−μₖm₁)/(m₁+m₂+I/R²), T₁=m₁(a+μₖg), and T₂=m₂(g−a).
For m₂>μₖm₁, a>0 and the descending mass speeds up. For equality it descends at constant speed. For m₂<μₖm₁, a<0: it can still descend while decelerating, until its velocity reaches zero.
A negative acceleration does not invalidate the assumed downward velocity. At a stop, reassess static friction and possible reversal; do not continue the same kinetic-friction sign blindly. Starting from rest requires μ_s, not just μₖ.
Check: m₁=4 kg, m₂=1 kg, μₖ=1/2, g=10 m/s² and I/R²=1 kg give a=−5/3 m/s², T₁=40/3 N, T₂=35/3 N. An initial downward speed of 2 m/s lasts 1.2 s before stopping.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for block, rough table and hanging mass show?

Table block moves right; friction μ_k m₁g left; hanging block moves down. Separate tensions T₁ and T₂.

2

METHOD

Which law or relation earns the method marks?

Use separate tensions, leftward kinetic friction on the table block and pulley torque.

3

RESULT

What should you box in the final answer?

Box signed acceleration and both tensions; classify speeding up, constant-speed descent and decelerating descent.

4

CHECK

What assumption, limit or direction must you check?

The formulas apply while the stated sliding velocities persist. Reassess friction at a stop; a start from rest requires μ_s.

5

AVOID

What tempting mistake loses marks?

Do not infer velocity direction from acceleration sign, or set the two pulley tensions equal.

40

Beam between two frictionless surfaces

Foundation · Force diagrams and conservation laws

A uniform beam of length L and mass M has one end on a frictionless horizontal floor and the other against a frictionless vertical wall. Show that static equilibrium at an acute angle without any other support is impossible; identify which missing horizontal force makes the torque equation inconsistent.

Illustrated setup: Beam between two frictionless surfaces — **Read the diagram**Both surfaces are smooth. The floor supplies only an upward normal; no friction arrow belongs here. Horizontal equilibrium would force the wall normal to zero and leave gravity’s moment unbalanced.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\sum F_x=0\Rightarrow N_w=0,\qquad \sum\tau_{\rm foot}\ne0

Rohan and Seema discuss the setup

Rohan: This beam touches two smooth surfaces. Why can't it rest?

Seema: List the directions each surface can push.

Rohan: There is no horizontal force to balance the wall?

Seema: Exactly—and gravity's torque still needs a counter-moment.

Step-by-step solution

Horizontal equilibrium would require N_wall=0 because the floor is frictionless.
About the foot, gravity has nonzero moment Mg(L/2)cosθ for an acute angle; with N_wall=0 nothing opposes it.
Hence the stated static equilibrium is impossible. A horizontal floor-friction force (or another lateral support) is the missing force that permits a nonzero wall reaction.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for beam between two frictionless surfaces show?

Wall force horizontal; floor force vertical; weight downward at the beam centre. There is no horizontal counterforce.

2

METHOD

Which law or relation earns the method marks?

Use horizontal force balance before the moment equation.

3

RESULT

What should you box in the final answer?

Conclude that static balance is impossible with both contacts frictionless at an acute angle.

4

CHECK

What assumption, limit or direction must you check?

Zero wall reaction leaves gravity’s torque unbalanced about the foot.

5

AVOID

What tempting mistake loses marks?

Do not invent a horizontal floor-friction force on a smooth floor.

ShunyaSaarthi · Chapter 09 Advanced mixed regimes

41

Impact followed by an upright swing

Advanced · Force diagrams and conservation laws

A uniform rod pivoted at one end hangs vertically at rest. A horizontal particle of mass m and speed v strikes and sticks to its free end. Find the limiting energy threshold v* for the rod-particle system to approach the upright position; distinguish asymptotic approach at threshold from finite-time passage and the separate taut-string constraint, and state the pivot model.

Illustrated setup: Impact followed by an upright swing — **Read the diagram**Conserve angular momentum about the pin during impact, then energy during the swing. At the exact energy threshold the upright position is approached only asymptotically.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\frac{(mvL)^2}{2(M/3+m)L^2}=gL(M+2m)

Rohan and Seema discuss the setup

Rohan: How fast must the particle arrive to lift the rod upright?

Seema: First conserve angular momentum during impact about the pin.

Rohan: Then use energy for the upward swing?

Seema: Yes. A rigid pin needs no taut-string top-speed rule.

Step-by-step solution

Initial rod hangs down. Transverse impact gives pivot angular momentum mvL, and after sticking I=(M/3+m)L². Thus ω_+=mvL/I.
Rotational energy after impact is (mvL)²/(2I). To reach upright, the rod's centre rises L and the attached mass rises 2L: ΔU=gL(M+2m).
Equating at threshold gives v*=√[2gL(M+2m)(M/3+m)]/m. At equality the assembly approaches the unstable upright position with speed tending to zero as t→∞; it does not arrive in finite time. The expression is a threshold (infimum).
A rigid pin can sustain the needed radial force. For a nonzero-speed full revolution, require v>v*; do not import the separate taut-string condition used for a mass on a string.
Near the upright position let δ be the remaining angle. At threshold, ω²≈[gL(M/2+m)/I]δ², so dt is proportional to dδ/δ and the time diverges logarithmically. Finite-time crossing requires v>v*.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for impact followed by an upright swing show?

Draw two stages with different laws: short impact about the pin, then gravitational rise to the inverted vertical.

2

METHOD

Which law or relation earns the method marks?

Conserve angular momentum about pin only during impact, then use energy for the upward swing.

3

RESULT

What should you box in the final answer?

Box the energy threshold v*. At equality the top is approached asymptotically; finite-time passage requires a larger incident speed.

4

CHECK

What assumption, limit or direction must you check?

A rigid pin can push or pull. At v=v*, angular speed tends to zero at the unstable top only as time tends to infinity.

5

AVOID

What tempting mistake loses marks?

Do not import a taut-string condition when the system uses a rigid pin.

42

Rolling sphere enters a smooth loop

Advanced · Force diagrams and conservation laws

A uniform solid sphere is released from rest and rolls without slipping down a centre-of-mass height h and then enters a smooth vertical circular track of radius R measured along the sphere's centre path. Find the minimum h for maintaining contact at the top. Account for what happens to spin once the sphere enters the smooth segment.

Illustrated setup: Rolling sphere enters a smooth loop — **Read the diagram**The sphere gains translation and spin on the rough approach. On the smooth loop spin stays constant; translation pays for the rise. At the top contact threshold, normal force is zero.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

v_b^2=\frac{10gh}{7},\quad v_t^2=v_b^2-4gR\ge gR

Rohan and Seema discuss the setup

Rohan: The sphere rolls before the loop. Does it keep rolling inside?

Seema: The loop is smooth; there is no torque to change its spin.

Rohan: So only translation energy pays for the climb?

Seema: Exactly. At the top check the normal force is nonnegative.

Step-by-step solution

At loop entry after rolling down h, (1/2)mv_b²(1+2/5)=mgh, so v_b²=10gh/7.
The loop is smooth: no torque about sphere centre from N or mg; its angular speed remains fixed. Translational energy alone changes by 2mgR between bottom and top: v_t²=v_b²−4gR.
At the top, inward radial equation mg+N=mv_t²/R. Minimum contact has N=0, so v_t²=gR.
Hence v_b²≥5gR and h≥(7/2)R. Treat R as the centre-path radius. Applying rolling conservation through the smooth loop would give a wrong threshold.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for rolling sphere enters a smooth loop show?

On the rough approach, spin and translation share energy; on the smooth loop, there is no tangential contact force, so spin stays constant and cannot pay for climbing.

2

METHOD

Which law or relation earns the method marks?

Use rolling energy on the rough approach; in the smooth loop conserve translational energy separately from constant spin.

3

RESULT

What should you box in the final answer?

Box the minimum starting height needed for contact at the loop top.

4

CHECK

What assumption, limit or direction must you check?

At the top threshold the normal force reaches zero and centre-path radius is R.

5

AVOID

What tempting mistake loses marks?

Do not continue imposing rolling without slip on the smooth loop.

43

Coaxial discs stick by friction

Foundation · Force diagrams and conservation laws

A spinning disc of I_1,ω_0 is placed coaxially against a second disc of I_2 initially at rest; interfacial friction locks them together, axles exert negligible torque. Find common speed and heat generated. Then evaluate the special case I_2=3I_1.

Illustrated setup: Coaxial discs stick by friction — **Read the diagram**Two discs share a fixed common axis. Their internal friction transfers angular momentum until they have one angular velocity, while mechanical energy is dissipated.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\omega_f=\frac{I_1\omega_0}{I_1+I_2}

Rohan and Seema discuss the setup

Rohan: The two discs lock together. Is kinetic energy conserved?

Seema: Friction dissipates energy, but axial angular momentum remains.

Rohan: Their final speed is one common value?

Seema: Yes. Compute the missing energy as heat.

Step-by-step solution

I₁ω₀=(I₁+I₂)ω_f, giving ω_f=I₁ω₀/(I₁+I₂).
Dissipated energy Q=(1/2)I₁ω₀²−(1/2)(I₁+I₂)ω_f²=(I₁I₂/[2(I₁+I₂)])ω₀².
For I₂=3I₁, ω_f=ω₀/4 and Q=(3/8)I₁ω₀².

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for coaxial discs stick by friction show?

Draw one fast disc, one stationary; friction torque is internal to the pair, so total axial L persists while kinetic energy falls.

2

METHOD

Which law or relation earns the method marks?

Conserve total axial angular momentum across the frictional coupling.

3

RESULT

What should you box in the final answer?

Box common angular speed and positive heat; in the three-to-one case speed is one-quarter initial.

4

CHECK

What assumption, limit or direction must you check?

Friction is internal to the disc pair but converts kinetic energy into heat.

5

AVOID

What tempting mistake loses marks?

Do not conserve kinetic energy while the two discs lock together.

44

A moving spinning rod caught at one end

Advanced · Force diagrams and conservation laws

A uniform horizontal rod slides on a frictionless plane with centre speed v perpendicular to the rod toward +y and counterclockwise angular speed ω about its centre. Its LEFT end sticks instantaneously to a fixed pin when it reaches it. Determine angular speed immediately after capture and energy lost; explicitly choose the impact-time conservation axis.

Illustrated setup: A moving spinning rod caught at one end — **Read the diagram**The illustration separates pre-capture from post-capture. During the brief capture, take angular momentum about the future left pin and include both orbital and spin contributions.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

\omega_f=\frac\omega4+\frac{3v}{2L}

Rohan and Seema discuss the setup

Rohan: A moving spinning rod catches on a pin. What survives?

Seema: Add spin and orbital angular momentum about the new pin.

Rohan: Could the capture lose no energy at all?

Seema: Only if that end already has zero velocity.

Step-by-step solution

The pin's impact impulse has zero moment about itself. Initial L_pin=(ML²/12)ω+MvL/2.
After capture I_pin=ML²/3, so ω' = L_pin/I_pin=ω/4+3v/(2L), counterclockwise for v,ω>0.
Initial kinetic energy K_i=(1/2)Mv²+(ML²ω²/24). Final K_f=(ML²/6)(ω/4+3v/(2L))². Loss is K_i−K_f, which is nonnegative; expanding gives (M/8)(v−ωL/2)².
The expression vanishes when the left endpoint already has zero velocity before capture, v=ωL/2. This is a strong physical check.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for a moving spinning rod caught at one end show?

Pin at the left end; before capture add orbital angular momentum Mv(L/2) to spin angular momentum I_CMω about that future pin.

2

METHOD

Which law or relation earns the method marks?

During capture add initial spin and orbital angular momentum about the future pin.

3

RESULT

What should you box in the final answer?

Box the captured angular speed and nonnegative kinetic-energy loss.

4

CHECK

What assumption, limit or direction must you check?

Pin impulse has no moment about that pin; include the centre velocity contribution.

5

AVOID

What tempting mistake loses marks?

Do not conserve angular momentum about the moving centre instead of the fixed impact point.

45

Rough region followed by smooth floor

Advanced · Force diagrams and conservation laws

A solid cylinder with initial centre speed v_0>0 rightward and signed clockwise angular speed ω_0 encounters a horizontal rough floor with kinetic friction coefficient μ>0, followed by a coplanar smooth floor after a distance D to its right. Assume its centre stays right-moving until pure rolling begins. Determine the threshold D needed to attain pure rolling by the time it reaches the boundary, and describe the subsequent motion on the smooth section for each regime of initial slip.

Illustrated setup: Rough region followed by smooth floor — **Read the diagram**The rough and smooth floors are coplanar, with no obstacle at the boundary. Friction can remove slip only while the body remains on the rough section.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

u=v-R\omega,\quad t_r=\frac{|u_0|}{3\mu g},\quad D_{\min}=x_r

Rohan and Seema discuss the setup

Rohan: Will the cylinder be rolling before the rough strip ends?

Seema: Find when its bottom slip first reaches zero.

Rohan: What if the smooth region starts earlier?

Seema: Then its current translation and spin each remain unchanged.

Step-by-step solution

From Q28, t_r=|u₀|/(3μg) and x_r=v₀t_r−(μg/2)sgn(u₀)t_r². Under the stated right-moving assumption, the threshold is D*=x_r.
If D≥D*, the cylinder enters the smooth region in pure rolling with v_r=(2v₀+Rω₀)/3 and ω_r=v_r/R; both remain constant there.
If 0≤D<D*, it is still slipping at the boundary. Find t_D from D=v₀t_D−(μg/2)sgn(u₀)t_D², taking the first root 0≤t_D<t_r. Entry values are v_D=v₀−μg sgn(u₀)t_D and ω_D=ω₀+(2μg/R)sgn(u₀)t_D.
On the smooth section both v_D and ω_D remain constant separately, so its slip persists. The smooth surface cannot create the torque needed to finish synchronizing them.
D=D* means rolling begins exactly at the boundary; rolling strictly before it requires D>D* when u₀≠0. For u₀>0 the stated strictly right-moving condition requires v_r=(2v₀+Rω₀)/3>0. If u₀=0, D*=0 and the body rolls from the start.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for rough region followed by smooth floor show?

Draw initial slip u₀=v₀−Rω₀, rolling point x_r, and the boundary at x=D. Use clockwise ω positive.

2

METHOD

Which law or relation earns the method marks?

On rough floor solve signed slip evolution until rolling or the boundary, whichever comes first.

3

RESULT

What should you box in the final answer?

Box the boundary-distance threshold and the subsequent constant speed and spin on smooth floor.

4

CHECK

What assumption, limit or direction must you check?

After entering the smooth region, both translation and spin persist independently.

5

AVOID

What tempting mistake loses marks?

Do not assume a smooth floor can generate the torque needed to remove residual slip.

ShunyaSaarthi · Chapter 10 Exam laboratory

$Rohan and Seema illustrated chapter scene$

46

Multiple-correct rolling claims

Foundation · Force diagrams and conservation laws

Multiple correct: An axisymmetric wheel or round rigid body of radius R executes planar rolling without slipping on a stationary horizontal surface, with no additional spin about another axis. Evaluate: (A) contact point has zero instantaneous velocity in the ground frame; (B) contact point has zero acceleration; (C) static friction must be nonzero; (D) total kinetic energy can be written ½I_contact ω². Justify all options and state the axis theorem used.

Illustrated setup: Multiple-correct rolling claims — **Read the diagram**For a uniformly rolling wheel, the bottom point has zero instantaneous velocity but upward acceleration ω²R. No friction is needed for this constant-speed example.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

K=\frac12I_{\rm contact}\omega^2,\quad I_{\rm contact}=I_{\rm CM}+MR^2

Rohan and Seema discuss the setup

Rohan: The contact point is at rest. Is its acceleration zero too?

Seema: Instantaneous zero velocity does not imply zero acceleration.

Rohan: Do we always need friction to keep rolling?

Seema: No. Check each statement against a constant-speed wheel.

Step-by-step solution

A true: no slip on stationary ground gives v_contact=0 at that instant.
B false: zero velocity does not imply zero acceleration; e.g. a rim point on a uniformly rolling wheel has acceleration toward the centre.
C false: a wheel can roll at constant speed on level ground with no required friction force.
D true: K=(1/2)Mv_CM²+(1/2)I_CMω²=(1/2)(I_CM+MR²)ω²=(1/2)I_contactω² by the parallel-axis theorem. Answer: A, D. The contact is an instantaneous velocity axis, not automatically an axis usable for every torque derivative.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for multiple-correct rolling claims show?

Mark the bottom contact as instantaneously stationary, but draw its acceleration and the centre's acceleration separately.

2

METHOD

Which law or relation earns the method marks?

Evaluate contact velocity, contact acceleration, required friction and the parallel-axis kinetic energy independently.

3

RESULT

What should you box in the final answer?

Choose A and D only.

4

CHECK

What assumption, limit or direction must you check?

The floor contact is an instantaneous velocity centre, not necessarily a stationary acceleration point.

5

AVOID

What tempting mistake loses marks?

Do not infer zero acceleration or compulsory static friction from zero contact velocity.

47

Numerical cylinder check

Foundation · Force diagrams and conservation laws

Numerical answer: A uniform solid cylinder rolls down a 30° fixed rough incline through vertical height h=1.5 m, starting from rest. Use g=10 m/s². Find v² at bottom in m²/s², acceleration in m/s² and minimum μ_s; report exact expressions where needed.

Illustrated setup: Numerical cylinder check — **Read the diagram**At the contact, uphill friction provides clockwise angular acceleration. Gravity and normal pass through the centre. Check the force result independently with energy.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

v^2=\frac{4gh}{3},\quad a=\frac23g\sin30^\circ

Rohan and Seema discuss the setup

Rohan: Can we solve this cylinder with forces and with energy?

Seema: Use translation and torque for acceleration and friction.

Rohan: Then check v² from the vertical drop?

Seema: Yes. Two methods should agree on the final speed.

Step-by-step solution

For a solid cylinder k=1/2, a=g sin30°/(1+k)=10/3 m/s².
f=[k/(1+k)]mg sin30°=mg/6; with N=mg cos30°, μ_min=f/N=1/(3√3).
Energy gives v²=2gh/(1+k)=2(10)(1.5)/(1.5)=20 m²/s².
Geometric check: path length s=h/sin30°=3 m and 2as=2(10/3)(3)=20. Results: 20 m²/s², 10/3 m/s², 1/(3√3).

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for numerical cylinder check show?

Separate the downhill force balance from rotational torque, then cross-check by energy over the vertical drop.

2

METHOD

Which law or relation earns the method marks?

Cross-check cylinder dynamics with the energy gained over the vertical drop.

3

RESULT

What should you box in the final answer?

Box 20 square metres per square second, 10/3 metres per square second, and 1/(3√3).

4

CHECK

What assumption, limit or direction must you check?

Check both path-length kinematics and the minimum-friction inequality.

5

AVOID

What tempting mistake loses marks?

Do not confuse vertical drop with incline distance.

48

Conservation during a pinned-rod impact

Foundation · Force diagrams and conservation laws

Multi-select: A particle strikes a rod pivoted at one end and sticks. During the short impact, assess which quantities are generally conserved for the particle-plus-rod system: linear momentum, angular momentum about pivot, angular momentum about rod centre, mechanical energy. State the impulse assumptions underlying each choice.

Illustrated setup: Conservation during a pinned-rod impact — **Read the diagram**The pin can deliver a large impulse but its lever arm about itself is zero. That preserves angular momentum about the pivot during the brief impact, while sticking loses energy.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

L_{\rm pivot}^{\rm before}=L_{\rm pivot}^{\rm after}

Rohan and Seema discuss the setup

Rohan: Which quantity survives when a particle sticks to a pinned rod?

Seema: Inspect the pin impulse about each proposed axis.

Rohan: It has no moment only about the pin itself?

Seema: Correct. Sticking also destroys mechanical energy.

Step-by-step solution

Linear momentum of particle+rod is generally not conserved because the pin supplies an external impulse.
Angular momentum about the pivot is conserved during the short impact if external gravity's impulse and axle friction torque are negligible; the pin impulse has no lever arm there.
Angular momentum about the rod centre is not generally conserved because the pin impulse has a lever arm about that centre. Mechanical energy is not conserved for sticking.
The conclusion depends on the short-duration impulse model; a substantial external torque about the pivot would invalidate item 2.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for conservation during a pinned-rod impact show?

Draw the external pin impulse at the pivot; its moment is zero about the pivot but generally nonzero about the rod centre.

2

METHOD

Which law or relation earns the method marks?

Mark the external pin impulse and its lever arm about each proposed axis.

3

RESULT

What should you box in the final answer?

Angular momentum about pivot survives the short impact; the other three quantities generally do not.

4

CHECK

What assumption, limit or direction must you check?

This assumes gravity impulse and axial friction torque are negligible during impact.

5

AVOID

What tempting mistake loses marks?

Do not conserve angular momentum about the rod centre when the pin impulse has a moment there.

49

Upper versus lower inner-axle tangent

Advanced · Force diagrams and conservation laws

Linked comprehension: A spool with outer radius R and inner radius R/2 rolls without slipping on level ground. A horizontal force F pulls a string tangent to the top of its inner axle. Given I=MR²/2, find the translational acceleration and friction force; then move the rightward pull to a string tangent to the BOTTOM of the inner axle and re-evaluate. Include a signed diagram for each case.

Illustrated setup: Upper versus lower inner-axle tangent — **Read the diagram**Both rightward pulls are tangent to the inner axle, but the upper and lower lines sit at different heights above floor contact C. Their different moments produce different accelerations and friction.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

a_{\rm upper}=\frac{FR(R+r)}{I+MR^2},\quad a_{\rm lower}=\frac{FR(R-r)}{I+MR^2}

Rohan and Seema discuss the setup

Rohan: Why do the two string positions move the spool differently?

Seema: Measure each force's lever arm from the floor contact.

Rohan: Can the upper case need no friction at all?

Seema: For these particular I and r values, yes. Verify it.

Step-by-step solution

Effective rolling inertia about floor contact is I+MR²=(3/2)MR². For a horizontal rightward pull at the upper inner-axle tangent, a=FR(R+r)/(I+MR²)=F/M.
Cylinder translation F+f=Ma gives f=0 for this particular I and r. Angular acceleration is clockwise a/R=F/(MR).
Move the same rightward pull to the lower inner-axle tangent: a=FR(R−r)/(I+MR²)=F/(3M). Hence f=Ma−F=−2F/3, leftward on the spool, and clockwise α=F/(3MR).
The zero upper-tangent friction is a special parameter result, not a general feature of spools. Confirm sufficient static friction for the lower case: μ_s Mg≥2F/3.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for upper versus lower inner-axle tangent show?

Draw the outer floor contact and two horizontal force lines at heights R+r and R−r above it, with r=R/2.

2

METHOD

Which law or relation earns the method marks?

Draw both force lines tangent to the inner axle and take moments about outer floor contact.

3

RESULT

What should you box in the final answer?

Upper pull gives acceleration F/M and zero friction; lower pull gives F/(3M) and friction left of magnitude 2F/3.

4

CHECK

What assumption, limit or direction must you check?

Require enough static friction for the lower-tangent rolling case.

5

AVOID

What tempting mistake loses marks?

Do not swap the upper and lower force lever arms.

50

Two independent pulley methods

Foundation · Force diagrams and conservation laws

Open-ended verification: Design a two-method solution for a massive pulley and two hanging masses: first Newton-plus-torque, then energy-plus-constraint. Show their acceleration formulas agree, check the zero-pulley-inertia limit, and identify one plausible but wrong use of equal tensions that each method would expose.

Illustrated setup: Two independent pulley methods — **Read the diagram**The right mass falls and left mass rises through one shared string displacement. Separate tensions create pulley torque; the same result follows by giving both masses and the pulley kinetic energy.Navy: geometry · red: forces · blue: motion. Directions and constraints refer to the stated model.

(m_2-m_1)g=\left(m_1+m_2+\frac I{R^2}\right)a

Rohan and Seema discuss the setup

Rohan: Can one pulley problem have two independent solutions?

Seema: Use two tensions and torque, then check with energy.

Rohan: What mistake would equal tensions cause?

Seema: A massive pulley would have zero torque despite accelerating.

Step-by-step solution

Newton: m₂g−T₂=m₂a; T₁−m₁g=m₁a; (T₂−T₁)R=I a/R. Thus a=(m₂−m₁)g/(m₁+m₂+I/R²).
Energy over a small descent dx of m₂: net gravity work (m₂−m₁)g dx increases kinetic energy of both blocks and pulley. Differentiate [(1/2)(m₁+m₂+I/R²)v²] with respect to x, using a=v dv/dx, to get the same formula.
With I→0 both methods give a=(m₂−m₁)g/(m₁+m₂). Assuming T₁=T₂ prematurely makes pulley torque zero and would force α=0 for finite I, contradicting no slip when a≠0.

5 exam sketch notes · questions & answers

1

SKETCH

What should the first sketch for two independent pulley methods show?

Newton method uses two tensions and a pulley torque; energy method uses one shared displacement x and the rotational speed v/R.

2

METHOD

Which law or relation earns the method marks?

Derive acceleration once from two mass equations plus pulley torque and again from work and energy.

3

RESULT

What should you box in the final answer?

Both methods give the same inertia-adjusted acceleration; the zero-inertia limit is the ordinary Atwood result.

4

CHECK

What assumption, limit or direction must you check?

The common string displacement fixes pulley angular speed as mass speed divided by radius.

5

AVOID

What tempting mistake loses marks?

Equal tensions on a finite-inertia pulley would imply zero torque and contradict its angular acceleration.

Every problem starts with a picture you can reason from.

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ShunyaSaarthi · Chapter 01 Torque and pivot choice

Three forces on a hinged rod

Step-by-step solution

5 exam sketch notes · questions & answers

Square plate and a diagonal force

Step-by-step solution

5 exam sketch notes · questions & answers

Loaded beam with a cable

Step-by-step solution

5 exam sketch notes · questions & answers

All zero-torque reference points

Step-by-step solution

5 exam sketch notes · questions & answers

A fixed horizontal force on a turning rod

Step-by-step solution

5 exam sketch notes · questions & answers

ShunyaSaarthi · Chapter 02 Angular language and motion

A changing angular acceleration

Step-by-step solution

5 exam sketch notes · questions & answers

Points at R and 2R

Step-by-step solution

5 exam sketch notes · questions & answers

Cubic angular displacement

Step-by-step solution

5 exam sketch notes · questions & answers

Two reversing wheels

Step-by-step solution

5 exam sketch notes · questions & answers

Measured acceleration direction

Step-by-step solution

5 exam sketch notes · questions & answers

ShunyaSaarthi · Chapter 03 Moment of inertia

Remove and relocate a ring arc

Step-by-step solution

5 exam sketch notes · questions & answers

Off-centre hole in a disc

Step-by-step solution

5 exam sketch notes · questions & answers

Four masses and the least-inertia line

Step-by-step solution

5 exam sketch notes · questions & answers

A bent uniform rod

Step-by-step solution

5 exam sketch notes · questions & answers

Two joined discs

Step-by-step solution

5 exam sketch notes · questions & answers

ShunyaSaarthi · Chapter 04 Torque and angular acceleration

Rod falling from horizontal

Step-by-step solution

5 exam sketch notes · questions & answers

Speed-dependent driving torque

Step-by-step solution

5 exam sketch notes · questions & answers

Hanging mass and a fixed-axle cylinder

Step-by-step solution

5 exam sketch notes · questions & answers

Off-centre pin and near-end force

Step-by-step solution

5 exam sketch notes · questions & answers

A brake with torque proportional to angle

Step-by-step solution

5 exam sketch notes · questions & answers

ShunyaSaarthi · Chapter 05 Work and energy

Rod swinging through the bottom

Step-by-step solution

5 exam sketch notes · questions & answers

Solid sphere versus spherical shell

Step-by-step solution

5 exam sketch notes · questions & answers

Motor and opposing torque

Step-by-step solution

5 exam sketch notes · questions & answers

Cylinder rolling uphill

Step-by-step solution

5 exam sketch notes · questions & answers

Falling mass coupled to a spring and pulley