Question

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 32.8 rad. During that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) What is the instantaneous angular velocity of the disk at the end of the 5.00 s? (d) With the angular acceleration unchanged, through what additional angle (rad) will the disk turn during the next 5.00 s?

Answer 1

Answer:

Part a)

$a = 2.62 rad/s$

Part b)

$\omega_{avg} = 6.56 rad/s$

Part c)

$\omega = 13.1 rad/s$

Part d)

$\theta_2 = 98.2 rad$

Explanation:

As we know that disc rotates by 32.5 rad in 5 s from rest position

Since we know that angular acceleration of the disc is constant here so we can use the condition of kinematics

Part a)

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

$32.8 = 0 + \frac{1}{2}(a)(5^2)$

$32.8 = 12.5a$

$a = 2.62 rad/s$

Part b)

average angular speed is know as the ratio of total angular displacement and total time

so we can say

$\omega_{avg} = \frac{\theta}{t}$

$\omega_{avg} = \frac{32.8}{5} = 6.56 rad/s$

Part c)

instantaneous angular speed is given as

$\omega = \omega_0 + at$

$\omega = 0 + (2.62)(5)$

$\omega = 13.1 rad/s$

Part d)

Total angle turn by the disc in total t = 10 s

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

$\theta' = 0 + \frac{1}{2}(2.62)(10^2)$

$\theta' = 131 rad$

now the angle turned in next 5 s will be

$\theta_2 = \theta' - \theta$

$\theta_2 = 131 - 32.8 = 98.2 rad$