Question

A brake is applied to a rotating disk, initially moving counterclockwise, causing it to slow down with a constant angular acceleration of magnitude 5.00 rad/s2. Immediately after the brake is applied, the disk rotates through 50.4 rad during a 4.00 s time interval. What is the angular speed (in rad/s) of the disk at the end of this time interval

Answer 1

Answer:

The angular speed at the end of this time interval (t=4 s) is 12.6 rad/s.

Explanation:

When applied the brake, the velocity will be decreasing at a constant rate of 5 rad/s

$\omega=\int a\cdot dt=\int -5dt=\omega_0-5t$

We know that in the first 4 seconds, the displacement of the disk is 50.4 rad.

$\alpha=\int_0^4\omega(t)dt=\int_0^4(\omega_0-5t)dt=\omega_0t-5t^2\\\\ \alpha(0)=0\\\\\alpha(4)=50.4=\omega_04-5*4^2=4\omega_0-80\\\\ \omega_0=(50.4+80)/4=130.4/4=32.6$

Now that we know the initial angular speed, we can calculate the angular speed at t=4s

$\omega(4)=\omega_0-5t=32.6-5*4=32.6-20=12.6$

The angular speed at the end of this time interval (t=4 s) is 12.6 rad/s.