Notes
Rotational kinematics
Rotation has its own versions of position, velocity and acceleration:
- Angular displacement $\Delta\theta$ (radians), angular velocity $\displaystyle \omega = \frac{\Delta\theta}{\Delta t}$, angular acceleration $\displaystyle \alpha = \frac{\Delta\omega}{\Delta t}$.
- Same equations as linear motion: $\omega = \omega_0 + \alpha t$, $\displaystyle \theta = \theta_0 + \omega_0t + \frac{1}{2}\alpha t^2$, $\omega^2 = \omega_0^2 + 2\alpha\Delta\theta$.
- Linking to linear motion at radius $r$: $v = r\omega$, $a_t = r\alpha$.
Torque
Torque is the turning effect of a force: $\tau = rF\sin\theta$, where $r$ is the distance from the axis and $\theta$ is the angle between $\vec r$ and $\vec F$.
- Push far from the hinge and perpendicular to the door for the most torque.
- Forces through the axis make no torque.
Rotational inertia
Rotational inertia $I$ measures resistance to changes in rotation. It depends on mass and how far the mass is from the axis: $\displaystyle I = \sum mr^2$.
- A hoop has more rotational inertia than a solid disk of the same mass and radius.
- Parallel axis theorem: $I = I_{cm} + Md^2$.
Newton's second law for rotation
- Rotational equilibrium: $\tau_{net} = 0$ (and translational equilibrium: $F_{net} = 0$). Balanced seesaws and static beams.
Practice
$\displaystyle \alpha = \frac{12}{4.0}$.
Torque is greatest with the largest $r$ and a perpendicular force.
$\tau = rF = 0.25(20)$.
All of the hoop's mass is at the maximum distance from the axis.
$\displaystyle \alpha = \frac{\tau}{I}$.
$30(2.0) = 40d \Rightarrow d = 1.5$ m.
$v = r\omega$.
Mass moves closer to the axis.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A uniform 4.0 m beam of mass 10 kg rests on supports at each end. A 20 kg box sits 1.0 m from the left end. Use $g = 10$ m/s².
(a) Find the force from the right support.
Show the answer and scoring guide
Pivot at the left support: $F_R(4.0) = 100(2.0) + 200(1.0) = 400$ N·m, so $F_R = 100$ N.
Check off the points you earned:
(b) Find the force from the left support.
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$F_L + F_R = 100 + 200 = 300$ N, so $F_L = 200$ N.
Check off the points you earned:
(c) The box slides toward the right end. How do the two support forces change? Explain.
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$F_R$ increases and $F_L$ decreases, while their sum stays 300 N. The box's torque about the left support grows as it moves away from it.
Check off the points you earned: