Question

The frequency of revolution of a particle performing circular motion changes from 60 r.p.m. to 180 r.p.m. in 20 seconds. Calculate the angular acceleration of the particle.
(pi=3.142)




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Answer 1

Answer:

Given:

Frequency of a moving particle changes from 60 rpm to 180 rpm.

Time taken = 20 seconds.

To find:

Angular acceleration

Conversion:

rpm has to be converted to radian/sec.

Initial angular Velocity

$= 60 \times \frac{2\pi}{60}$

$= 2\pi \: rad \: {s}^{ - 1}$

Final angular velocity

$= 180 \times \frac{2\pi}{60}$

$= 6\pi \: rad \: {s}^{ - 1}$

Calculation:

Let Angular acceleration be α

$\therefore \: \alpha = \dfrac{ \omega2 - \omega1}{t}$

$= > \alpha = \dfrac{(6\pi - 2\pi)}{20}$

$= > \alpha = \dfrac{4\pi}{20}$

$= > \alpha = \dfrac{\pi}{5}$

$= > \alpha = \dfrac{3.14}{5}$

$= > \alpha = 0.628$

So final answer :

$\boxed{ \green{\alpha = 0.628 \: rad \: {s}^{ - 2}}}$

Answer 2

Answer:

Final rpm = 180

Initial rpm = 60

So angular acceleration

$= \frac{180 - 60}{20}$

$= \frac{120}{20}$

$= 6 \: rpm \: per \: second$

Now converting to Rad/sec²

$6\times \frac{2\pi}{60}$

$= \frac{\pi}{5}$

$= 0.628 \: rad \: {s}^{ - 2}$