Question

A particle is moving with constant angular acceleration \(4\,\text{rad/s}^2\) in circular path. At what time the magnitudes of its tangential acceleration and centripetal acceleration will be equal ?

• A. \(0.2\) s
• B. \(0.4\) s
• C. \(0.5\) s
• D. \(0.6\) s

Hint:

Tangential is r alpha, centripetal is omega squared r with omega = alpha t.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For circular motion with constant angular acceleration \(\alpha\) starting from rest, \(\omega = \alpha t\). Tangential acceleration is \(a_t = r\alpha\) and centripetal acceleration is \(a_c = \omega^2 r\).

Step 2: Set them equal:
\[ r\alpha = \omega^2 r = \alpha^2t^2 r \Rightarrow \alpha t^2 = 1 \]

Step 3: Solve:
\[ t = \frac{1}{\sqrt\alpha} = \frac{1}{\sqrt4} = 0.5\ \text{s} \]

Step 4: Check:
At \(t = 0.5\) s, \(\omega = 4\times0.5 = 2\) rad/s, so \(a_c = 4r\) and \(a_t = 4r\). They are equal. At 0.2 s, \(a_c = 0.64r < a_t\), so option (A) is not it. Option (C).

Final Answer:
Equating r alpha with alpha squared t squared r gives t = 0.5 s. \[ \boxed{\text{(C) }0.5\ \text{s}} \]