Question

A particle at rest starts moving with constant angular acceleration '\(α\)' in a circular path of radius 'r'. At certain instant, the magnitude of centripetal acceleration is \((\frac{1}{3})^{rd}\) the tangential acceleration. The relation between linear speed (V) and angular acceleration (\(α\)) is

• A. \(V = \frac{rα}{3}\)
• B. \(V = α\sqrt{\frac{r}{2}}\)
• C. \(V = r\sqrt{\frac{α}{3}}\)
• D. \(V = \sqrt{\frac{αr}{3}}\)

Hint:

Set centripetal acceleration equal to one third of tangential acceleration.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A particle moving in a circle with angular acceleration \(\alpha\) has tangential acceleration \(a_t = r\alpha\) and centripetal acceleration \(a_c = \omega^2 r = V^2/r\).

Step 2: Apply the condition
\[ a_c = \frac13 a_t \Rightarrow \omega^2 r = \frac{r\alpha}{3} \Rightarrow \omega^2 = \frac{\alpha}{3} \]

Step 3: Find V
\[ V = \omega r = r\sqrt{\frac{\alpha}{3}} \]
Option (A) has no square root, option (B) has a 2 that does not appear, and (D) uses \(\sqrt{\alpha r/3}\), which is dimensionally inconsistent with the above.

Final Answer:
The speed is \(V = r\sqrt{\alpha/3}\), option (C). \[ \boxed{V = r\sqrt{\frac{\alpha}{3}}} \]