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
$L = I\omega$.
$\displaystyle I/(MR^2) = \frac{2}{5}$, so $\displaystyle v = \sqrt{\frac{2gh}{7/5}}$.
The contact point has zero velocity.
$L = mvd_\perp = 0.10(5.0)(2.0)$.
$P = \tau\omega$.
The pivot force exerts no torque about the pivot.
$\displaystyle E = -\frac{GMm}{2r}$ rises toward 0 as $r$ grows.
$\displaystyle \int_0^2 3t^2\,dt = 8$.
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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