Question

Q. 45. Solve the following:
(1) The frequency of a particle performing circular
motion changes from 60 rpm to 180 rpm in
20 seconds. Calculate the angular acceleration.​

Answer 1

Explanation:

I hope my answer is satisfiable

if answer is up to your mark then mark it as brainliest answer

for more follow me

Answer 2

$\dag\:\underline{\sf AnsWer :}$

A particle performing a circular motion. The frequency of that particle changes from 60 rpm to 180 rpm i.e Initial frequency = 60 rpm and final frequency = 180 rpm and the time taken (t) by the particle is 20 seconds. We need to find the angular acceleration.

First we need to make the units same. So, let's convert the unit of frequency from rpm to rps :

$\bigstar\:\:\underline{\textbf{ Initial frequency :}}$

$:\implies \sf Initial \: frequency\:(f_1) = 60 \: rpm$

$:\implies \sf Initial \: frequency\:(f_1) = \dfrac{60}{60} \: rps$

$:\implies \underline{ \boxed{\frak{ Initial \: frequency\:(f_1) = 1 \: rps}}}$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\bigstar\:\:\underline{\textbf{Final frequency :}}$

$:\implies \sf Final \: frequency\:(f_2) = 180 \: rpm$

$:\implies \sf Final \: frequency\:(f_2) = \dfrac{180}{60} \: rps$

$:\implies \sf Final \: frequency\:(f_2) = \dfrac{18}{6} \: rps$

$:\implies \underline{ \boxed{ \frak{ Final \: frequency\:(f_2) = 3 \: rps}}}$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\bigstar\:\:\underline{\textbf{Angular Acceleration of the particle:}}$

$\dashrightarrow\:\:\sf \alpha = \dfrac{\omega -\omega _{o} }{t}$

$\qquad \qquad\tiny\dag\:\underline{\sf \omega = 2 \pi f}$

$\dashrightarrow\:\:\sf \alpha = \dfrac{2\pi f_2-2\pi f_1}{t}$

$\dashrightarrow\:\:\sf \alpha = \dfrac{2\pi (3)-2\pi (1)}{20}$

$\dashrightarrow\:\:\sf \alpha = \dfrac{6\pi -2\pi }{20}$

$\dashrightarrow\:\:\sf \alpha = \dfrac{4\pi }{20}$

$\dashrightarrow\:\:\sf \alpha = \dfrac{\pi }{5}$

$\qquad \qquad\tiny\dag\:\underline{\sf \pi = 3.14}$

$\dashrightarrow\:\:\sf \alpha = \dfrac{3.14 }{5}$

$\dashrightarrow\:\: \underline{ \boxed{\frak{\alpha = 0.628 \: rad/s^2}}}$