Answer to Question #121694 in Classical Mechanics for Quanda

Question

A particle moves in a horizontal circular path of radius 2 m with uniform angular acceleration. It is observed to make 2 revolutions in the first 4 seconds of motion and 4 revolutions in the next 4 seconds. Find:
a) the initial angular velocity of the particle giving your answer in rad/s
b) the angular acceleration of the particle, giving your answer in rad per second squared.

Answer 1

Let, a particle moves in circular motion with radius $r=2m$

Now, at $t=4s$ ,angle traversed is $\theta_1=2\times2\pi=4\pi\:rad$

and at $t=8s$ , angle traversed is $\theta_2=(2+4)\times 2\pi=12\pi \: rad$

Consider, the constant angular acceleration $\alpha$ .

We know that,

$\theta=\omega_0+\frac{1}{2}\alpha t^2$

Thus,

$4\pi=\omega_0+8\alpha\hspace{1cm}(1)\\ 12\pi=\omega_0+32\alpha\hspace{1cm}(2)$

(a).

multiply by 4 in equation (1) and subtract equation (2) from (1) we get

$\omega_0=\frac{4\pi}{3}\:rad/s$

(b).

Subtract equation (1) from(2) we get,

$\alpha=\frac{\pi}{3}\:rad/s^2$