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:

Answer 1

Given data

1. The angular velocity of the body is $\omega =a-bt$

Angular velocity

1. Angular velocity is the time rate of change in the angular displacement of a body.
2. 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 displacement

As we know, angular velocity is $\omega =\frac{d\theta }{dt}$

So,

$$\begin{gathered} d\theta =\omega dt \\ ord\theta =\left(a-bt\right)dt \\ or\theta =\int \left(a-bt\right)dt=at-b\frac{t^2}{2} \\ or\theta =at-b\frac{t^2}{2}..........\left(1\right) \end{gathered}$$

But when the body stops rotating, its angular velocity is zero.

So,

$$\begin{gathered} \omega =a-bt=0 \\ ort=\frac{a}{b}.................\left(2\right) \end{gathered}$$

From equations 1 and 2 we get,

$$\begin{gathered} \theta =a\left(\frac{a}{b}\right)-\frac{b}{2}\left(\frac{a^2}{b^2}\right)=\frac{a^2}{2b} \\ or\theta =\frac{a^2}{2b}. \end{gathered}$$

Therefore, the angle through which it rotates before it comes to stop is $\theta =\frac{a^2}{2b}$.