Question

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 8.00 s, it rotates 17.5 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 8.00 s? (d) With the angular acceleration unchanged, through what additional angle (rad) will the disk turn during the next 8.00 s?

Answer 1

Given,

time = 8 s

θ = 17.5 rad

initial angular velocity = 0 rad/s

Using rotational motion equation

`\theta=\omega_(0) t+0.5 \alpha t^(2)`

`17.5=0+0.5 \alpha(8)^(2)`

`\alpha = 0.546 \ rad/s^2`

a. angular acceleration
`\alpha=0.546 \ rad/s^2`

b. Average angular velocity = total angle/total time taken

=
`(17.5)/(8)`= 2.187 rad/s

c. we have,

`\omega=\omega_(0)+a t`

`=0+0.546 * 8`

Angular velocity at end of 8 seconds
`=\omega=4.368` rad/s

d. we have, additional angle in next 8 seconds:

`\theta=\omega_(0) t+0.5 \alpha {t}^(2)`

`=4.368 * 8+0.5 * 0.546 * 8^(2)`

`=52.416` rads