Question

The frequency of a particle performing circular motion changes from $60$ r.p.m to $180$ r.p.m in $20$ second. Calculate the angular acceleration. (in $rad/s^2$)

• A. $0.489$
  
• B. $0.1254$
  
• C. $0.6284$
  
• D. $0.369$
  

Answer 1

The correct option is C $0.6284$

Initial angular velocity ${\omega }_0=60\times \frac{2\pi }{60sec}=2\pi$

Final angular velocity ${\omega }_t=130\times \frac{2\pi }{60sec}=6\pi$

As we know ${\omega }_t={\omega }_0+\alpha t$

$\Rightarrow \alpha =\frac{{\omega }_t-{\omega }_0}{t}=\frac{6\pi -2\pi }{20}=0.6284rad/s^2$