Question

The speed of a small object undergoing uniform circular motion is 4 m/s. The magnitude of the change in the velocity during 0.5 seconds is also 4 m/s.

• A. the angular speed of object is$\frac{2\pi }{3}$rad/sec.
• B. the centripetal acceleration of the object is$\frac{8\pi }{3}m/s^2$
• C. the radius of the circle is$\frac{6}{\pi }$metre.
• D. the radius of circle is$\frac{6}{\pi }$metre.

Answer 1

Answer: (A), (B), (C)

The correct options are
A the angular speed of object is$\frac{2\pi }{3}$rad/sec.
B the centripetal acceleration of the object is$\frac{8\pi }{3}m/s^2$
C the radius of the circle is$\frac{6}{\pi }$metre.

If the angular displacement is $/theta$ for 0.5 seconds, then the
magnitude of change in velocity is
$\Delta v=\sqrt{v^2+v^2-2v^{2}cos\theta }=2vsin\frac{\theta }{2}$
$\therefore \Delta v=2\times 4sin\frac{\theta }{2}$
$or\frac{\theta }{2}=\frac{\pi }{6}$ & $\epsilon \theta =\frac{\pi }{3}$
$\therefore \omega =\frac{\theta }{0.5}=\frac{2\pi }{3}rad/s\Rightarrow R=\frac{v}{\omega }=\frac{4\times 3}{2\pi }=\frac{6}{\pi }m$
and centripetal acceleration $=v\times \omega =4\times \frac{2\pi }{3}=\frac{8\pi }{3}m/s^2$