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(A) The angular velocity is given by $\omega(t) = \alpha - \beta t$. At the time the body stops,the angular velocity $\omega = 0$. Setting $\alpha - \

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

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(A) The angular velocity is given by $\omega(t) = \alpha - \beta t$. At the time the body stops,the angular velocity $\omega = 0$. Setting $\alpha - \beta t = 0$,we find the time taken to stop: $t = \frac{\alpha}{\beta}$. 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$. Evaluating the integral: $	heta = [\alpha t - \frac{1}{2} \beta t^2]_{0}^{\alpha/\beta}$. Substituting 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}$. Thus,the angle through which it rotates before it stops is $\frac{\alpha^2}{2\beta}$.

$A$ rigid body rotates about a fixed axis with variable angular velocity equal to $\alpha - \beta t$,where $t$ is time and $\alpha, \beta$ are constants. Find the angle (in radians) through which it rotates before it stops.

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