Rotational Dynamics Practice Set: NEB 12 Physics

Worked problem 1

Torque from forces at different angles

A 1.20 m uniform light bar pivots at its left end. A 30 N downward force acts at the right end. A 20 N force acts 0.40 m from the pivot at 60° above the bar. Take anticlockwise positive. Find net torque.

Model: The 30 N force is perpendicular and clockwise: τ₁=−1.20×30=−36 N m. The second force has perpendicular component 20 sin60° and produces anticlockwise torque: τ₂=0.40×20×sin60°=6.93 N m. Therefore τnet=−29.1 N m, clockwise.

Checks: The second torque cannot exceed 0.40×20=8 N m. The magnitude 29.1 N m is less than the dominant 36 N m and direction remains clockwise. Read the OpenStax torque geometry after attempting.

Worked problem 2

Moment of inertia and angular acceleration

A uniform 3.0 kg, 1.2 m rod rotates about one end. A constant net torque of 2.4 N m acts. Find I and α.

Model: For a uniform rod about an end, I=(1/3)ML²=(1/3)(3.0)(1.2²)=1.44 kg m². For a fixed axis, τnet=Iα, so α=2.4/1.44=1.67 rad s⁻².

Trap: (1/12)ML² is the centre-axis result. The object alone does not choose I; the axis does. The OpenStax inertia guide shows standard and shifted axes.

Worked problem 3

Rotational work, energy and power

A flywheel has I=5.0 kg m². It speeds from 4.0 to 10 rad s⁻¹ under a constant driving torque while 60 J is lost to friction. Find the driving work. If the angular displacement is 20 rad, find the constant driving torque.

Change in rotational energy ΔK=½I(ω₂²−ω₁²)=2.5(100−16)=210 J. Driving work must supply energy increase plus loss: Wdrive=210+60=270 J. For constant torque, W=τΔθ, so τdrive=270/20=13.5 N m.

Check: Ignoring friction would give 10.5 N m, so the required driving torque should be greater. Do not subtract the loss from the needed input. Compare the reasoning with rotational work and power.

Worked problem 4

Angular momentum with changing inertia

A student on a low-friction rotating stool has I=3.6 kg m² and ω=2.0 rad s⁻¹. Pulling two masses inward changes total I to 2.4 kg m². External torque is negligible. Find new ω and compare rotational kinetic energy.

Conservation: I₁ω₁=I₂ω₂, so ω₂=(3.6×2.0)/2.4=3.0 rad s⁻¹. Initial K=½(3.6)(2.0²)=7.2 J. Final K=½(2.4)(3.0²)=10.8 J. Energy increases by 3.6 J because internal work is done while pulling inward; angular-momentum conservation does not require constant kinetic energy.

State the system and external-torque condition. See the OpenStax angular-momentum discussion.

Worked problem 5

Rolling without slipping down a height

A solid sphere rolls without slipping from rest through vertical height 1.5 m. Ignore resistance. Find centre-of-mass speed using I=(2/5)MR² and g=9.8 m s⁻².

Energy: Mgh=½Mv²+½Iω². With ω=v/R, rotational term is ½(2/5 MR²)(v²/R²)=1/5 Mv². Thus Mgh=(7/10)Mv², and v=√(10gh/7)=√21=4.58 m s⁻¹.

Check: A sliding particle would have √(2gh)=5.42 m s⁻¹, so the rolling sphere is slower because energy is shared with rotation. The OpenStax rolling chapter gives force and energy approaches.

Independent set

Solve before checking the prompts

1. A 15 N force acts 0.50 m from a pivot at 40°. Find torque magnitude and direction from a supplied sketch.
2. Three point masses 1, 2 and 3 kg lie 0.20, 0.30 and 0.40 m from an axis. Find I.
3. A disc I=0.80 kg m² changes ω from 2 to 8 rad s⁻¹. Find ΔK.
4. Net torque 6 N m acts on I=1.5 kg m² from rest for 4 s. Find α, ω and θ.
5. A system changes I from 5 to 2 kg m² at initial ω=4 rad s⁻¹ with zero external torque. Find final ω.
6. A solid cylinder rolls from height h. Derive v in terms of g and h.
7. Explain why a force through the pivot can change linear motion of a body but has zero torque about that pivot.
8. Draw θ(t), ω(t) and α(t) for constant positive α with negative initial ω.

Answer checks, not full copying: (1) use rFsin40°; (2) sum mr²; (3) ½I(ω₂²−ω₁²); (4) use τ=Iα then constant-α kinematics; (5) Iω conservation; (6) include translation and rotation; (7) distinguish force from moment about origin; (8) slopes and sign crossings must agree.

Use the study guide to diagnose which decision failed rather than memorising the final number.

Review protocol

Turn each wrong answer into a future correct decision

Mark the first wrong step: axis, diagram, lever arm, moment of inertia, principle, constraint, sign, units, algebra or interpretation. Rewrite the specific rule, solve again without looking, then schedule a fresh problem after two days and another after one week. Record whether the original error returned. Keep the chapter inside the Class 12 Physics revision roadmap so delayed mechanics review continues while later units begin.

For derivations, reconstruct from a labelled physical model. For explanations, state mechanism and consequence. For numericals, keep symbols before substitution and add dimension, sign and limiting checks. For online or physical NEB tuition, call 9846662070. The MKS Education panel supports next-step test preparation and pre-counselling after Grade 12.

Checked sources

References and related learning

• CDC Nepal: Physics Grade 12
• CDC Nepal: Secondary Level Curriculum
• OpenStax: Rotation
• OpenStax: Moment of Inertia and Rotational Energy
• OpenStax: Torque
• OpenStax: Newton’s Second Law for Rotation
• OpenStax: Angular Momentum

Rebuild weak concepts with the topic guide, then return to fresh problems rather than repeating identical numbers. Curriculum and institutional pages were checked on 2 August 2026; follow current CDC, NEB and school notices if requirements change.

Mixed challenge

Plan a two-stage rotation problem before solving

A wheel of known I is accelerated from rest by a constant motor torque while a constant resistive torque opposes it. After a specified time the motor is switched off and resistance alone brings it to rest. Before numbers, draw two separate torque diagrams and write two angular accelerations. Stage one uses α₁=(τmotor−τresist)/I; stage two uses α₂=−τresist/I. Angular velocity at the switch connects the stages.

Now decide what extra measurements would let you determine both unknown torques experimentally: the acceleration while powered, the deceleration after switch-off and I are sufficient under the constant-torque model. Explain how bearing heating or speed-dependent drag would violate that model and appear as curvature in an ω–t graph. This challenge links dynamics, graphs, modelling and experimental evaluation rather than rewarding a single memorised equation.

For a final check, predict the areas and slopes on the piecewise ω–t graph before inserting values. The slope gives α; the area gives angular displacement. Label both with units and explain why the stopping interval need not equal the powered interval.