Question

Net acceleration of a circular disc after 2 seconds

Answer 1

Answer:

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

Explanation:

Given

Angular acceleration

\alpha = 3t-t^2α=3t−t2

But we know that

rate of change of angular velocity is equal to angular acceleration

Therefore

d\omega/dt = 3t-t^2dω/dt=3t−t2

\implies d\omega = (3t-t^2)dt⟹dω=(3t−t2)dt

\implies \int_0^\omega d\omega = \int_0^2(3t-t^2)dt⟹∫0ωdω=∫02(3t−t2)dt

\implies \omega \Bigr |_0^\omega= (3t^2/2-t^3/3)\Bigr|_0^2⟹ω∣∣∣0ω=(3t2/2−t3/3)∣∣∣02

\implies \omega= (3\times 2^2/2-2^3/3)-0⟹ω=(3×22/2−23/3)−0

\implies \omega= 6-8/3 = 10/3 \text{ rad/sec}⟹ω=6−8/3=10/3 rad/sec