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: Write both accelerations
$a_c = V^2/r$ and $a_t = r\alpha$.

Step 2: Condition
$V^2/r = r\alpha/3$, so $V^2 = r^2\alpha/3$.

Step 3: Root
$V = r\sqrt{\alpha/3}$. Option (C).

Final Answer:
Option (C). \[ \boxed{V = r\sqrt{\frac{\alpha}{3}}} \]