Question

# A rigid body rotates about a fixed axis with a variable angular velocity equal to an $a-bt$ at time t time where a and b are constants. The angle through which it rotates before it comes to stop is:

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Solution

## Given dataThe angular velocity of the body is $\omega =a-bt$Angular velocityAngular velocity is the time rate of change in the angular displacement of a body.The angular velocity of a body is defined by the form, $\omega =\frac{d\theta }{dt}$, where $\theta$ is the angular displacement of the body.Finding the angular displacementAs we know, angular velocity is $\omega =\frac{d\theta }{dt}$So, $d\theta =\omega dt\phantom{\rule{0ex}{0ex}}ord\theta =\left(a-bt\right)dt\phantom{\rule{0ex}{0ex}}or\theta =\int \left(a-bt\right)dt=at-b\frac{{t}^{2}}{2}\phantom{\rule{0ex}{0ex}}or\theta =at-b\frac{{t}^{2}}{2}..........\left(1\right)$But when the body stops rotating, its angular velocity is zero.So, $\omega =a-bt=0\phantom{\rule{0ex}{0ex}}ort=\frac{a}{b}.................\left(2\right)$From equations 1 and 2 we get,$\theta =a\left(\frac{a}{b}\right)-\frac{b}{2}\left(\frac{{a}^{2}}{{b}^{2}}\right)=\frac{{a}^{2}}{2b}\phantom{\rule{0ex}{0ex}}or\theta =\frac{{a}^{2}}{2b}.$Therefore, the angle through which it rotates before it comes to stop is $\theta =\frac{{a}^{2}}{2b}$.

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