Question

A point on rim of disc starts circular motion from rest after time t it gained angular acceleration a=3t-t^2 what is angular velocity after 2seconds

Answer 1

Answer:

Angular Velocity after 2 seconds is 10/3 rad/sec

Explanation:

Given

Angular acceleration

$\alpha = 3t-t^2$

But we know that

rate of change of angular velocity is equal to angular acceleration

Therefore

$d\omega/dt = 3t-t^2$

$\implies d\omega = (3t-t^2)dt$

$\implies \int_0^\omega d\omega = \int_0^2(3t-t^2)dt$

$\implies \omega \Bigr |_0^\omega= (3t^2/2-t^3/3)\Bigr|_0^2$

$\implies \omega= (3\times 2^2/2-2^3/3)-0$

$\implies \omega= 6-8/3 = 10/3 \text{ rad/sec}$

Answer 2

Answer:

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