Question

A particle starting from rest moves along the circumference of a circle of radius \( r \) with angular acceleration \( \alpha \). The magnitude of the average velocity, in the time it completes the small angular displacement \( \theta \), is

• A. \( r \left( \frac{2}{\alpha \theta} \right)^2 \)
• B. \( r \left( \frac{\alpha \theta}{2} \right) \)
• C. \( r \left( \frac{\alpha \theta}{2} \right)^2 \)
• D. \( r \left( \frac{\alpha \theta}{2} \right)^{1/2} \)

Hint:

In problems involving angular motion, always recall the standard equations of motion for rotational motion and use them to derive relationships between angular displacement, angular velocity, and angular acceleration.

Answer 1

The Correct Option is D

Step 1: Understanding the question.
The particle starts from rest and moves under constant angular acceleration \( \alpha \) along a circular path of radius \( r \). We need to find the magnitude of the average velocity for a small angular displacement \( \theta \).

Step 2: Formula for average velocity.
The average velocity \( v_{\text{avg}} \) in terms of angular displacement and acceleration is given by the relation: \[ v_{\text{avg}} = \frac{\text{total displacement}}{\text{time taken}} = \frac{r \theta}{t} \] From the equations of motion, we know that for angular displacement \( \theta \) under constant angular acceleration \( \alpha \), the time taken \( t \) is: \[ t = \sqrt{\frac{2 \theta}{\alpha}} \]
Step 3: Substituting the value of \( t \).
Substitute the expression for \( t \) into the formula for \( v_{\text{avg}} \): \[ v_{\text{avg}} = \frac{r \theta}{\sqrt{\frac{2 \theta}{\alpha}}} = r \left( \frac{\alpha \theta}{2} \right)^{1/2} \]
Step 4: Conclusion.
Thus, the correct expression for the average velocity is \( r \left( \frac{\alpha \theta}{2} \right)^{1/2} \), corresponding to option (D).