A brief note on Angular Momentum, Equilibrium and Rolling Motion . Read it before a test. Try each question at the end before you open its answer.
The big ideas Angular momentum : L = I ω L = I\omega L = I ω (body on an axis), L ⃗ = r ⃗ × p ⃗ \vec{L} = \vec{r} \times \vec{p} L = r × p (particle). Torque is its rate of change. Conservation : no external torque, so I 1 ω 1 = I 2 ω 2 I_1\omega_1 = I_2\omega_2 I 1 ω 1 = I 2 ω 2 ; kinetic energy L 2 2 I \dfrac{L^2}{2I} 2 I L 2 can change. Equilibrium : net force zero and net torque zero about every point. Rolling without slipping : v = ω R v = \omega R v = ω R ; the contact point is at rest, the top moves at 2 v 2v 2 v . Rolling energy : 1 2 M v 2 ( 1 + I M R 2 ) \tfrac12 Mv^2\left(1 + \dfrac{I}{MR^2}\right) 2 1 M v 2 ( 1 + M R 2 I ) ; down a slope a = g sin θ 1 + I / M R 2 a = \dfrac{g\sin\theta}{1 + I/MR^2} a = 1 + I / M R 2 g sin θ . The facts at a glance Body ( I / M R 2 I/MR^2 I / M R 2 ) Turning share Speed after a drop h h h Solid ball ( 2 5 \tfrac25 5 2 ) 2 7 \tfrac27 7 2 10 g h / 7 \sqrt{10gh/7} 10 g h /7 Disc ( 1 2 \tfrac12 2 1 ) 1 3 \tfrac13 3 1 4 g h / 3 \sqrt{4gh/3} 4 g h /3 Ring ( 1 1 1 ) 1 2 \tfrac12 2 1 g h \sqrt{gh} g h
A worked example A solid ball, a disc and a ring roll from rest down a slope, dropping 0.80 m 0.80\ \text{m} 0.80 m .
Ball. v = 10 × 9.8 m/s 2 × 0.80 m 7 = 3.35 m/s v = \sqrt{\dfrac{10 \times 9.8\ \text{m/s}^2 \times 0.80\ \text{m}}{7}} = 3.35\ \text{m/s} v = 7 10 × 9.8 m/s 2 × 0.80 m = 3.35 m/s . Disc. v = 4 × 9.8 m/s 2 × 0.80 m 3 = 3.23 m/s v = \sqrt{\dfrac{4 \times 9.8\ \text{m/s}^2 \times 0.80\ \text{m}}{3}} = 3.23\ \text{m/s} v = 3 4 × 9.8 m/s 2 × 0.80 m = 3.23 m/s . Ring. v = 9.8 m/s 2 × 0.80 m = 2.80 m/s v = \sqrt{9.8\ \text{m/s}^2 \times 0.80\ \text{m}} = 2.80\ \text{m/s} v = 9.8 m/s 2 × 0.80 m = 2.80 m/s . The more of its energy a body spends turning, the slower it goes. Practice with new numbers
A program to try The program works out the three speeds and the turning shares.
Read it, then run it.
Add a hollow ball, k = 2 3 k = \tfrac23 k = 3 2 . Where does it fit in the order?
Mistakes to avoid Conserving kinetic energy when a skater pulls in. Conserve angular momentum. Forgetting a body's own weight in an equilibrium problem; it acts at its centre of mass. Distances along a ladder as moment arms. Use the perpendicular (horizontal) distances. Friction doing work in rolling. Static friction at the contact point does none. Mass or size deciding the rolling race. Only I M R 2 \dfrac{I}{MR^2} M R 2 I does. Check yourself A turntable with I = 0.20 kg m 2 I = 0.20\ \text{kg m}^2 I = 0.20 kg m 2 spins at 5.0 rad/s 5.0\ \text{rad/s} 5.0 rad/s . A 0.10 kg 0.10\ \text{kg} 0.10 kg lump sticks to it 0.30 m 0.30\ \text{m} 0.30 m from the axis. What is the new spin? Stuck? Show a hint Show the answerA 40 kg 40\ \text{kg} 40 kg child sits 1.5 m 1.5\ \text{m} 1.5 m from a seesaw's pivot. Where must a 50 kg 50\ \text{kg} 50 kg child sit to balance? Stuck? Show a hint Show the answerA disc rolls at 4.0 m/s 4.0\ \text{m/s} 4.0 m/s . How fast is the top of the disc moving? Stuck? Show a hint Show the answerWhat fraction of a rolling ring's kinetic energy is in its turning? Stuck? Show a hint Show the answerWhy does a spinning top stay upright while it spins fast? Stuck? Show a hint Show the answer