Question

A particle starting from rest moves along the circumference of a circle of radius \(r=\sqrt{2}\) m with an angular acceleration \(\alpha =\frac{\pi }{9}\) rad/s\({}^{2}\). The magnitude of its average velocity and instantaneous angular velocity in the time , it completes one fourth rotation, is respectively

• A. \(\frac{2}{5}\) m/s, \(\frac{\pi }{\sqrt{2}}\) rad/s
• B. \(\frac{3}{2}\) m/s, \(\frac{\pi }{2}\) rad/s
• C. \(\frac{4}{5}\) m/s, \(\frac{\pi }{3}\) rad/s
• D. \(\frac{2}{3}\) m/s, \(\frac{\pi }{3}\) rad/s

Hint:

Find the time for a quarter turn from the angular displacement, then use the chord length for average velocity.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The particle starts from rest with constant angular acceleration, so \(\theta=\dfrac{1}{2}\alpha t^2\) and \(\omega=\alpha t\). Average velocity is displacement over time, and displacement is the chord, not the arc.

Step 2: Time for one fourth rotation
\(\theta=\dfrac{\pi}{2}\), so \(\dfrac{\pi}{2}=\dfrac{1}{2}\cdot\dfrac{\pi}{9}t^2\)
\(t^2=9\), so \(t=3\text{ s}\).

Step 3: Instantaneous angular velocity
\(\omega=\alpha t=\dfrac{\pi}{9}\times 3=\dfrac{\pi}{3}\text{ rad/s}\)

Step 4: Average velocity
The chord between two perpendicular radii is \(\sqrt{2}\,r=\sqrt{2}\times\sqrt{2}=2\text{ m}\).
\(v_{avg}=\dfrac{2}{3}\text{ m/s}\)

Final Answer:
The average velocity is \(\dfrac{2}{3}\text{ m/s}\) and the angular velocity is \(\dfrac{\pi}{3}\text{ rad/s}\). \[ \boxed{\text{(D) }\dfrac{2}{3}\text{ m/s},\ \dfrac{\pi}{3}\text{ rad/s}} \]