10.1 Rotational variables  (Page 3/11)

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Since $\stackrel{\to }{r}$ is constant, the term $\stackrel{\to }{\theta }\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{d\stackrel{\to }{r}}{dt}=0$ . Since $\stackrel{\to }{v}=\frac{d\stackrel{\to }{s}}{dt}$ is the tangential velocity and $\stackrel{\to }{\omega }=\frac{d\stackrel{\to }{\theta }}{dt}$ is the angular velocity, we have

$\stackrel{\to }{v}=\stackrel{\to }{\omega }\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{r}.$

That is, the tangential velocity is the cross product of the angular velocity and the position vector, as shown in [link] . From part (a) of this figure, we see that with the angular velocity in the positive z -direction, the rotation in the xy -plane is counterclockwise. In part (b), the angular velocity is in the negative z- direction, giving a clockwise rotation in the xy- plane.

Rotation of a flywheel

A flywheel rotates such that it sweeps out an angle at the rate of $\theta =\omega t=\left(45.0\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s)}t$ radians. The wheel rotates counterclockwise when viewed in the plane of the page. (a) What is the angular velocity of the flywheel? (b) What direction is the angular velocity? (c) How many radians does the flywheel rotate through in 30 s? (d) What is the tangential speed of a point on the flywheel 10 cm from the axis of rotation?

Strategy

The functional form of the angular position of the flywheel is given in the problem as $\theta \left(t\right)=\omega t$ , so by taking the derivative with respect to time, we can find the angular velocity. We use the right-hand rule to find the angular velocity. To find the angular displacement of the flywheel during 30 s, we seek the angular displacement $\text{Δ}\theta$ , where the change in angular position is between 0 and 30 s. To find the tangential speed of a point at a distance from the axis of rotation, we multiply its distance times the angular velocity of the flywheel.

Solution

1. $\omega =\frac{d\theta }{dt}=45\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}$ . We see that the angular velocity is a constant.
2. By the right-hand rule, we curl the fingers in the direction of rotation, which is counterclockwise in the plane of the page, and the thumb points in the direction of the angular velocity, which is out of the page.
3. $\text{Δ}\theta =\theta \left(30\phantom{\rule{0.2em}{0ex}}\text{s}\right)-\theta \left(0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=45.0\left(30.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)-45.0\left(0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=1350.0\phantom{\rule{0.2em}{0ex}}\text{rad}$ .
4. ${v}_{\text{t}}=r\omega =\left(0.1\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(45.0\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}\right)=4.5\phantom{\rule{0.2em}{0ex}}\text{m/s}$ .

Significance

In 30 s, the flywheel has rotated through quite a number of revolutions, about 215 if we divide the angular displacement by $2\pi$ . A massive flywheel can be used to store energy in this way, if the losses due to friction are minimal. Recent research has considered superconducting bearings on which the flywheel rests, with zero energy loss due to friction.

Angular acceleration

We have just discussed angular velocity for uniform circular motion, but not all motion is uniform. Envision an ice skater spinning with his arms outstretched—when he pulls his arms inward, his angular velocity increases. Or think about a computer’s hard disk slowing to a halt as the angular velocity decreases. We will explore these situations later, but we can already see a need to define an angular acceleration    for describing situations where $\omega$ changes. The faster the change in $\omega$ , the greater the angular acceleration. We define the instantaneous angular acceleration     $\alpha$ as the derivative of angular velocity with respect to time:

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