Notes
Rotational kinematics with calculus
$\displaystyle \omega = \frac{d\theta}{dt}$, $\displaystyle \alpha = \frac{d\omega}{dt}$. Linear and angular links: $v = r\omega$, $a_t = r\alpha$, $a_c = r\omega^2$.
Torque as a cross product
Use the right-hand rule: counterclockwise torques point out of the page.
Rotational inertia
- Point masses: $\displaystyle I = \sum mr^2$. Continuous objects: $\displaystyle I = \int r^2\,dm$.
| Object (about the axis shown) | $I$ |
|---|---|
| Thin rod, center | $\displaystyle \frac{1}{12}ML^2$ |
| Thin rod, end | $\displaystyle \frac{1}{3}ML^2$ |
| Hoop, center | $MR^2$ |
| Solid disk or cylinder, center | $\displaystyle \frac{1}{2}MR^2$ |
| Solid sphere, center | $\displaystyle \frac{2}{5}MR^2$ |
- Parallel axis theorem: $I = I_{cm} + Md^2$. Check: rod about its end, $\displaystyle \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2$.
Newton's second law for rotation
$\tau_{net} = I\alpha$.
- Pulley with mass: a hanging mass $m$ on a string wrapped around a disk pulley (mass $M$, radius $R$): $mg - T = ma$ and $\displaystyle TR = \frac{1}{2}MR^2\frac{a}{R}$, giving $\displaystyle a = \frac{mg}{m + M/2}$.
- Static equilibrium: $\displaystyle \sum F = 0$ and $\displaystyle \sum\tau = 0$ about any point.
Practice
$\omega = 6t^2 = 24$ rad/s.
$\displaystyle \frac{1}{3}(3.0)(2.0)^2 = 4.0$ kg·m².
$\displaystyle \frac{1}{2}MR^2 + MR^2$.
$\hat i \times \hat j = \hat k$.
$\displaystyle a = \frac{mg}{m + M/2} = \frac{20}{4.0}$.
$\displaystyle I = \int r^2\,dm$.
$\displaystyle \frac{2}{5}MR^2$ is the smallest of these.
$\alpha = 4.0$ rad/s², so $\displaystyle t = \frac{20}{4.0}$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A thin rod of length $L$ has a nonuniform linear density $\lambda(x) = \frac{2M}{L^2}x$, where $x$ is measured from one end.
(a) Show that the rod's total mass is $M$.
Show the answer and scoring guide
$\displaystyle \int_0^L \frac{2M}{L^2}x\,dx = \frac{2M}{L^2}\cdot\frac{L^2}{2} = M$.
Check off the points you earned:
(b) Find the rod's rotational inertia about the end at $x = 0$.
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$\displaystyle I = \int_0^L x^2\frac{2M}{L^2}x\,dx = \frac{2M}{L^2}\cdot\frac{L^4}{4} = \frac{1}{2}ML^2$.
Check off the points you earned:
(c) Is this larger or smaller than a uniform rod of the same mass about its end? Explain.
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Larger ($\displaystyle \frac{1}{2}ML^2 > \frac{1}{3}ML^2$). More mass is concentrated far from the axis.
Check off the points you earned: