Notes
Rotational kinetic energy
An object that rolls has both translational and rotational kinetic energy: $\displaystyle K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
- Rolling without slipping: $v_{cm} = r\omega$.
- Down a ramp, an object with larger rotational inertia (relative to $mr^2$) puts more of its energy into spinning, so it moves slower. A solid sphere beats a solid cylinder, which beats a hoop.
Work done by torque
$W = \tau\Delta\theta$, and the rotational work-energy theorem is $W_{net} = \Delta K_{rot}$.
Angular momentum
- For a rigid body: $L = I\omega$. For a point particle: $L = mvr\sin\theta$ (often $mvr$).
- Angular impulse: $\tau\Delta t = \Delta L$.
- Conservation: with no net external torque, $L$ stays constant. A skater pulling in her arms decreases $I$, so $\omega$ increases.
- In that case rotational kinetic energy increases; the skater does work pulling her arms in.
Orbits
For a satellite, angular momentum about the planet is conserved because gravity points toward the center (no torque). In an elliptical orbit it moves fastest when closest.
Practice
$\displaystyle \frac{1}{2}(0.50)(4.0)^2 = 4.0$ J.
The sphere has the smallest $I$ relative to $mr^2$, so more energy goes into translation.
$L$ is conserved: $4.0(2.0) = 1.0\omega$.
$\displaystyle K = \frac{L^2}{2I}$; with $L$ constant and $I$ smaller, $K$ increases. Her muscles do work.
$\Delta L = \tau\Delta t$.
$\displaystyle \omega = \frac{v}{r}$.
Angular momentum $mvr$ is conserved, so smaller $r$ means larger $v$.
Total $I$ increases while $L$ stays constant.
Free response
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Free-response question
A solid disk (mass 2.0 kg, radius 0.20 m, $I = \frac{1}{2}MR^2$) rolls without slipping from rest down a ramp, dropping a height of 0.60 m. Use $g = 10$ m/s².
(a) Calculate the disk's rotational inertia.
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$\displaystyle I = \frac{1}{2}(2.0)(0.20)^2 = 0.040$ kg·m².
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(b) Find the disk's speed at the bottom.
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$\displaystyle mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$. So $\displaystyle v = \sqrt{\frac{4gh}{3}} = \sqrt{8} \approx 2.8$ m/s.
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(c) A block of equal mass slides down the same height with no friction. Is it faster or slower than the disk at the bottom? Explain.
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Faster ($\displaystyle \sqrt{2gh} \approx 3.5$ m/s). All its energy becomes translational kinetic energy, while the disk puts some into rotation.
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