Question

If the frequency of the particle performing circular motion increases from 60 RPM to 180 RPM in 20 seconds, then its angular acceleration is:
(A) $ 0.1{\text{ }}rad/{s^2} $
(B) $ 3.142rad/{s^2} $
(C) $ 0.6284rad/{s^2} $
(D) $ 0.3142rad/{s^2} $

Answer 1

Hint:
We should use the Laws of motion for Rotational Dynamics to first calculate the angular velocity. Then calculate the angular acceleration of the particle.
Formula Used: We have used the following formulae of Rotational Mechanics in this question,
 $ \omega = 2\pi \nu $
 $ \nu = \dfrac{1}{T} $
 $ \alpha = \dfrac{{\Delta \omega }}{{\Delta t}} $
Here, $ \omega $ is the angular velocity of the particle, $ \nu $ is the frequency of the particle, $ T $ is the time Period of revolution around the circular trajectory of the particle, $ \alpha $ is the angular acceleration of the particle, $ \Delta \omega $ is the change in angular velocity of the particle, and $ \Delta t $ is the time taken by the particle to change the angular velocity from its initial to final value.

Complete step by step answer:
We are already given the initial and final values of frequency of the particle and we also know the relation between frequency and angular velocity of the particle, thus we can calculate the change in the angular frequency as,
 $ \Delta \omega = 2\pi (\Delta \nu ) $
But, we are given the frequencies in terms of RPM which expands to Rotations per minute. To convert the frequencies into Rotations per Second, we should divide each frequency by 60 seconds. Hence,
 $ {\nu _1} = 60 $ RPM
 $ \Rightarrow {v_1} = \dfrac{{60}}{{60}} = 1 $ RPS
Similarly, we get,
 $ {\nu _2} = 180 $ RPM
 $ \Rightarrow {v_2} = \dfrac{{180}}{{60}} = 3 $ RPS
So, we get, Change in frequency as,
 $ \Delta \nu = {\nu _2} - {\nu _1} $
This gives,
 $ \Delta \nu = 3 - 1 = 2 $
Thus, Change in angular frequency becomes,
 $ \Delta \omega = 2\pi (\Delta \nu ) $
Putting in the values, we get,
 $
  \Delta \omega = 2\pi (2) \\
   \Rightarrow \Delta \omega = 4\pi \\
 $
Now, we are already given $ \Delta t $ as 20 seconds, So, we get angular acceleration to be,
 $ \alpha = \dfrac{{\Delta \omega }}{{\Delta t}} $
Putting the values obtained above, we get,
 $
  \alpha = \dfrac{{4\pi }}{{20}} \\
   \Rightarrow \alpha = \dfrac{\pi }{5} = 0.6284rad/{s^2} \\
 $
$ \therefore $ Option (C) is correct out of the given options.

Note:
In this question, we could have converted 20 seconds to minutes also and obtained the same result. The important point is that, always convert the units to a single homogenous system of units and use that system throughout the problem.