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(A) The angular velocity is given by $\omega(t) = \alpha - \beta t$.The body comes to rest when $\omega(t) = 0$,which implies $\alpha - \beta t = 0$,s

Vedclass · Accepted Answer

$\frac{\alpha^2}{2\beta}$

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(A) The angular velocity is given by $\omega(t) = \alpha - \beta t$.The body comes to rest when $\omega(t) = 0$,which implies $\alpha - \beta t = 0$,so the time taken to come to rest is $t = \frac{\alpha}{\beta}$.The angular displacement $	heta$ is the integral of angular velocity with respect to time: $	heta = \int_{0}^{t} \omega(t) dt$.Substituting the expression for $\omega(t)$: $	heta = \int_{0}^{\alpha/\beta} (\alpha - \beta t) dt$.Integrating,we get $	heta = [\alpha t - \frac{1}{2}\beta t^2]_{0}^{\alpha/\beta}$.Evaluating at the limits: $	heta = \alpha(\frac{\alpha}{\beta}) - \frac{1}{2}\beta(\frac{\alpha}{\beta})^2 = \frac{\alpha^2}{\beta} - \frac{\alpha^2}{2\beta} = \frac{\alpha^2}{2\beta}$.

$A$ rigid body rotates about a fixed axis with variable angular velocity equal to $(\alpha - \beta t)$ at time $t,$ where $\alpha$ and $\beta$ are constants. The angle through which it rotates before it comes to rest is

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