Question

A rigid body rotates about a fixed axis with a 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 stops is:
A. ${\displaystyle \frac{{\alpha }^2}{2\beta }}$
B. ${\displaystyle \frac{{\alpha }^2-{\beta }^2}{2\alpha }}$
C. ${\displaystyle \frac{{\alpha }^2-{\beta }^2}{2\beta }}$
D. ${\displaystyle \frac{{\alpha }^3}{6{\beta }^2}}$

Answer 1

Hint: Here, the angle through which the rigid body rotates before it comes to rest is calculated by using the formula of angular velocity which is the angle per unit time. At rest, the angular velocity will be zero, therefore, we will calculate time and put it in the equation of angle.

Complete step by step answer:
It is given in the question that a rigid body is rotating about a fixed axis.
Angular velocity is defined as the rate of the velocity at which a given object or particle rotates around a given point in a given interval of time. This velocity is also known as rotational velocity. Therefore, the angular velocity, as given in the question, is given by
$\omega =\alpha -\beta t$
Where, $\omega$ is the angular velocity, $\alpha$ and $\beta$ are the constants and $t$ is the time taken.

Also, we know that angular velocity is also measured as angle per unit time, therefore,
$\omega ={\displaystyle \frac{d\theta }{dt}}$
Now, putting the value of $\omega$ in the above equation, we get
$\omega ={\displaystyle \frac{d\theta }{dt}}=\alpha -\beta t$
$\Rightarrow d\theta =\left(\alpha -\beta t\right)dt$
Now, to calculate the angle through which the rigid body rotates before it comes to rest can be calculated by integrating the above equation, we get
$\int d\theta =\int \left(\alpha -\beta t\right)dt$
$\theta =\int \left(\alpha -\beta t\right)dt$
$\theta =\alpha t-{\displaystyle \frac{\beta t^2}{2}}$

When the rigid will be at rest than the angular velocity will be zero
$\omega =0$
$\Rightarrow \alpha -\beta t=0$
$\Rightarrow \alpha =\beta t$
$\Rightarrow t={\displaystyle \frac{\alpha }{\beta }}$

Therefore, putting $t={\displaystyle \frac{\alpha }{\beta }}$ in $\theta$ , we get
Also, $\theta =\alpha \left({\displaystyle \frac{\alpha }{\beta }}\right)-{\displaystyle \frac{\beta {\left({\displaystyle \frac{\alpha }{\beta }}\right)}^2}{2}}$
$\Rightarrow \theta ={\displaystyle \frac{{\alpha }^2}{\beta }}-{\displaystyle \frac{{\alpha }^2}{2\beta }}$
$\Rightarrow \theta ={\displaystyle \frac{2{\alpha }^2-{\alpha }^2}{2\beta }}$
$\therefore \theta ={\displaystyle \frac{{\alpha }^2}{2\beta }}$
Therefore, the angle through which a rigid body rotates before it stops is ${\displaystyle \frac{{\alpha }^2}{2\beta }}$ .

Hence, option A is the correct option.

Note:In the above example, we are putting the value of $t$ in the $\theta$ equation after integrating it. But, the angle can also be calculated by integrating the equation of $\theta$ between the limits $0$ to ${\displaystyle \frac{\alpha }{\beta }}$ . The result will be the same in both cases.