Vidaara.orgClass 11 · Physics
CodeVID-P11-06-ROL-01
Rotational Dynamics & Rolling — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The rotational analogue of force is:
- A.torque
- B.momentum
- C.inertia
- D.work
2.
Rotational kinetic energy equals:
- A.$I\omega$
- B.$\frac{1}{2}I\omega^2$
- C.$I\alpha$
- D.$\tau\omega$
3.
The rolling-without-slipping condition is:
- A.$v=\omega R$
- B.$v=\omega/R$
- C.$a=\omega R$
- D.$v=R/\omega$
4.
Power delivered by a torque is:
- A.$\tau\theta$
- B.$\tau\omega$
- C.$\tau\alpha$
- D.$I\omega$
5.
Down an incline the body with the highest acceleration is the:
- A.ring
- B.disc
- C.solid sphere
- D.hollow sphere
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
A torque of $8\ \text{N m}$ acts on a body with $I=4\ \text{kg m}^2$. Find $\alpha$.
7.
A flywheel with $I=0.2\ \text{kg m}^2$ spins at $10\ \text{rad/s}$. Find its rotational KE.
8.
State the rolling-without-slipping condition.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
A solid sphere of mass 1 kg rolls without slipping at $4\ \text{m/s}$. Find its total KE. ($\frac{k^2}{R^2}=\frac{2}{5}$)
10.
A ring rolls from rest down an incline of height 1.0 m. Find the speed at the bottom. ($\frac{k^2}{R^2}=1$, $g=10\ \text{m/s}^2$)
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Using energy conservation, derive the speed of a body rolling without slipping down an incline of height $h$, and use it to explain why a solid sphere beats a ring.
Answer Key
Section A — Multiple Choice Questions
- (A) torque
- (B) $\frac{1}{2}I\omega^2$
- (A) $v=\omega R$
- (B) $\tau\omega$
- (C) solid sphere
Section B — Short Answer (2 marks)
- $\alpha=2\ \text{rad/s}^2$.
- $KE=10\ \text{J}$.
- The centre speed and angular speed satisfy $v=\omega R$ (contact point at rest).
Section C — Short Answer (3 marks)
- $KE=\frac{1}{2}(1)(16)(1.4)=11.2\ \text{J}$.
- $v=\sqrt{\frac{2(10)(1)}{2}}=\sqrt{10}\approx3.16\ \text{m/s}$.
Section D — Long Answer (5 marks)
- Energy conservation gives $mgh=\frac{1}{2}mv^2(1+k^2/R^2)$, so $v=\sqrt{\frac{2gh}{1+k^2/R^2}}$; the smaller $k^2/R^2$ of the sphere ($\tfrac{2}{5}$ vs $1$) gives a larger $v$, so the sphere wins.
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