Question

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

• A. $\frac{\alpha}{\beta}$
• B. $\frac{\alpha^2}{\beta}$
• C. $\frac{\alpha^2}{2\beta}$
• D. $\frac{\alpha}{2\beta}$

Hint:

This is functionally identical to the 1D kinematic equation $v^2 = u^2 + 2aS$.
Here, initial velocity $\omega_0 = \alpha$, final velocity $\omega_f = 0$, and angular acceleration is the derivative of $\omega$, which is $-\beta$.
$0^2 = \alpha^2 + 2(-\beta)\theta \implies 2\beta\theta = \alpha^2 \implies \theta = \frac{\alpha^2}{2\beta}$. Much faster!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Angular velocity $\omega = \frac{d\theta}{dt}$. To find the total angle $\theta$, we integrate $\omega$ with respect to time.

Step 2: Formula Application:
First, find the time $t$ when the body comes to rest ($\omega = 0$): $\alpha - \beta t = 0 \implies t = \alpha / \beta$.

Step 3: Explanation:
$\theta = \int_{0}^{\alpha/\beta} (\alpha - \beta t) dt$ $\theta = [ \alpha t - \frac{\beta t^2}{2} ]_{0}^{\alpha/\beta}$ $\theta = \alpha(\frac{\alpha}{\beta}) - \frac{\beta}{2}(\frac{\alpha}{\beta})^2 = \frac{\alpha^2}{\beta} - \frac{\alpha^2}{2\beta} = \frac{\alpha^2}{2\beta}$.

Step 4: Final Answer:
The total angle rotated is $\frac{\alpha^2}{2\beta}$.