AP Physics C: Mechanics

Unit 6

Energy and Momentum of Rotating Systems

8 practice questions · Free response

Notes

Rotational energy and work

  • $\displaystyle K_{rot} = \frac{1}{2}I\omega^2$. Rolling: $\displaystyle K = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$ with $v_{cm} = R\omega$ (no slipping).
  • Work by torque: $\displaystyle W = \int\tau\,d\theta$. Power: $P = \tau\omega$.
  • For a rolling object released from height $h$: $\displaystyle v = \sqrt{\frac{2gh}{1 + I/(MR^2)}}$.

Angular momentum

  • Particle: $\vec L = \vec r \times \vec p$, magnitude $mvr\sin\theta$. Rigid body: $L = I\omega$.
  • $\displaystyle \vec\tau_{net} = \frac{d\vec L}{dt}$. Angular impulse: $\displaystyle \int\tau\,dt = \Delta L$.
  • Conservation: with no net external torque, $L$ is constant.
  • A particle moving in a straight line still has angular momentum about a point not on its path.

Rolling with friction

  • Rolling without slipping: static friction does no work (the contact point isn't sliding).
  • A ball that starts sliding and spinning wrong will have kinetic friction act until $v = R\omega$; angular momentum about the contact point is conserved during that process.

Orbits

  • Angular momentum is conserved in orbits (gravity is a central force). Kepler's second law (equal areas in equal times) follows.
  • Circular orbit: $\displaystyle v = \sqrt{\frac{GM}{r}}$; total energy $\displaystyle E = -\frac{GMm}{2r}$.

Practice

A wheel has $I = 2.0$ kg·m² and spins at 3.0 rad/s. What is its angular momentum?
A solid sphere rolls without slipping from height $h$. Its speed at the bottom is:
When an object rolls without slipping down a ramp, the static friction force does:
A 0.10 kg particle moves at 5.0 m/s in a straight line that passes 2.0 m from point O at its closest. What is its angular momentum about O?
A motor delivers a constant torque of 5.0 N·m to a shaft spinning at 20 rad/s. What power does it deliver?
Clay hits and sticks to the end of a rod pivoted at its center. Which is conserved during the collision?
A satellite in a circular orbit moves to a larger orbit. Its total mechanical energy:
A torque $\tau(t) = 3t^2$ N·m acts on a wheel from rest for 2.0 s. What is the change in angular momentum?

Free response

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Free-response question

A uniform rod of mass $M$ and length $L$ is free to rotate about a frictionless pivot at its center on a horizontal table. A small ball of mass $m$ moving at speed $v_0$ perpendicular to the rod hits one end and sticks.

(a) Write an expression for the angular speed right after the collision.

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$L$ about the pivot is conserved: $\displaystyle mv_0\frac{L}{2} = \left(\frac{1}{12}ML^2 + m\frac{L^2}{4}\right)\omega \Rightarrow \omega = \frac{6mv_0}{(M + 3m)L}$.

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(b) Is kinetic energy conserved in this collision? Explain.

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No. The ball sticks (perfectly inelastic), so some kinetic energy becomes internal energy.

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(c) Explain why linear momentum of the ball-rod system is not conserved.

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The pivot exerts an external force on the rod during the collision, so the net external force isn't zero.

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