Question

A particle at rest starts moving with a constant angular acceleration of $4\ \text{rad}/\text{s}^2$ in a circular path. At what time the magnitudes of its tangential acceleration and centripetal acceleration will be equal?

• A. $0.5\ \text{s}$
• B. $1.0\ \text{s}$
• C. $0.25\ \text{s}$
• D. $2.0\ \text{s}$

Hint:

When tangential and centripetal accelerations are equal, the total acceleration vector makes a $45^\circ$ angle with the velocity vector. Setting $\omega^2 = \alpha$ is the fastest direct condition for this scenario.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A particle is undergoing circular motion with a constant angular acceleration. We need to find the specific time when the tangential acceleration equals the centripetal (radial) acceleration in magnitude.

Step 2: Key Formula or Approach:
The formula for tangential acceleration is $a_t = r\alpha$.
The formula for centripetal acceleration is $a_c = r\omega^2$.
The kinematic equation for angular velocity is $\omega = \omega_0 + \alpha t$.

Step 3: Detailed Explanation:
The problem states that the magnitudes of the two accelerations are equal:
$$a_c = a_t$$ $$r\omega^2 = r\alpha$$ Cancel the radius $r$ from both sides:
$$\omega^2 = \alpha$$ We are given the angular acceleration $\alpha = 4\ \text{rad}/\text{s}^2$:
$$\omega^2 = 4 \implies \omega = 2\ \text{rad}/\text{s}$$ Since the particle starts from rest, its initial angular velocity $\omega_0 = 0$. Using the kinematic equation to find the time $t$:
$$\omega = \omega_0 + \alpha t$$ $$2 = 0 + 4t$$ $$t = \frac{2}{4} = 0.5\ \text{s}$$

Step 4: Final Answer:
The required time is $0.5\ \text{s}$, matching option (A).