Question

A particle revolves with constant angular acceleration π rad/s². If the particle starts from rest, how many revolution will it make in the first 10 seconds?

Answer 1

Heya

We're going ta need few Basic Formulas !!

See ->

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$\theta = \omega_{0} ( t ) + \frac{1}{2} \alpha (t)^{2}$

♦ This above gives a description of the Angular Displacement of a body with Initial Angular Velocity Given, and given Uniform Angular Accn. as a FUNCTION OF THE TIME ELAPSED !!!

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Also, we may need :

$Number \ of \ revolutions = \frac{Total Angular Distance}{2\pi}$

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Now, after certain Calculations ->

♦ In this Question, Angular Distance covered is given by :

$\theta = ( 0 ) ( 10 ) + \frac{1}{2} \pi (10)^{2}$

$=> \theta = 50\pi$

And that, the number of revolutions hence, is given as :

$Number \ of \ revolutions = \frac{50 \pi}{2\pi}$

$n=25 rev$  

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Answer 2

➡️➡️Here is your solution⬇️⬇️

Concept:-$<b>$ Since, Acceleration is constant so we can use 'equation of motion in circular motion.

By using➡️➡️
$s = wt + \frac{1}{2} \alpha {t}^{2} \\ = 0 + \frac{1}{2} \times \pi \times {10}^{2} \\ = 50\pi$
Now
we have to find total no. of revolution➡️

–»
$number \: of \: revolution = \frac{distance \: covered}{2\pi} \\ = \frac{50\pi}{2\pi} \\ 25$

Total number of revolution is 25.✔️✔️