Question 14: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

The angular velocity of the rigid body is given by $ \omega = \alpha - \beta t $, where $ \alpha $ and $ \beta $ are constants. By comparing this with the equation $ \omega = \omega_0 - \alpha t $, we can identify the initial angular velocity $ \omega_0 $ as $ \alpha $ and the angular retardation (or deceleration) as $ \beta $.
To find the angle through which the body rotates before it comes to a stop, we use the kinematic equation for rotational motion:
$ \omega^2 = \omega_0^2 + 2 \beta 	heta $
We know that when the body stops, $ \omega = 0 $. Therefore, by substituting into the equation, we get:
$ 0 = \alpha^2 - 2 \beta 	heta $
Solving for $ 	heta $, the angle rotated, we find:
$ 	heta = \frac{\alpha^2}{2 \beta} $
Thus, the angle through which the body rotates before stopping is $ \frac{\alpha^2}{2 \beta} $.

Answer

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

Answer

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

Answer

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

Answer

$ \frac{(\alpha-\beta) \alpha}{2} $

Centre of Mass and Rotational Motion

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

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