Question

A wheel has a constant angular acceleration of $3.0rad/{s^2} $, During a certain $4.0s $interval, it turns through an angle of $120rad $. Assuming that at $t = 0 $, angular speed ${\omega _0} = 3rad/s $ how long is motion at the start of this $4.0 $ second interval?
(A) $7s $
(B) $9s $
(C) $4s $
(D) $10s $

Answer 1

Hint The angle of rotation of the wheel can be determined using the equations of circular motion. It is known that the difference between the initial and final angles is $120rad $, this can be put into the second equation of circular motion and solved for the time.
Formula used:
 $\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2} $
Where $\theta $ is the angle covered by the rotating object.
 ${\omega _0} $ is the initial angular velocity.
 $\alpha $ is the angular acceleration.
 $t $is the time taken.

Complete Step by step answer
It is given in the question that,
The angular acceleration of the wheel is constant and is equal to $\alpha = 3.0rad/{s^2} $
The initial angular speed, ${\omega _0} = 3rad/s $
Angle turned by the wheel in the given time interval, $\theta = 120rad $
Time interval given in the question is $4\sec $.
Let the start of the $4\sec $ motion be at time $t $.
Then the end of this motion happens at the time $t + 4 $
Let the initial angle of the wheel be ${\theta _1} $ and the final angle of the wheel be ${\theta _2} $.
The total angle turned by the wheel can be written as-
 $120 = {\theta _2} - {\theta _1} $
From the equation of circular motion, we know that-
 $\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2} $
For ${\theta _1} $,
 ${\theta _1} = 3t + \dfrac{1}{2} \times 3{t^2} $
For ${\theta _2} $,
 ${\theta _2} = 3\left( {t + 4} \right) + \dfrac{1}{2} \times 3{\left( {t + 4} \right)^2} $
Combining both of these equations by subtracting ${\theta _1} $ from ${\theta _2} $,
 ${\theta _2} - {\theta _1} = \left( {3(t + 4) - 3t} \right) + \left( {\dfrac{3}{2}\left( {{{(t + 4)}^2} - {t^2}} \right)} \right) $
 ${\theta _2} - {\theta _1} = 12 + \left( {\dfrac{3}{2}\left( {({t^2} + 8t + 16) - {t^2}} \right)} \right) $
 ${\theta _2} - {\theta _1} = 12 + \left( {\dfrac{3}{2}\left( {8t + 16} \right)} \right) $
 $120 = 12 + \dfrac{3}{2}\left( {8t + 16} \right) $
 $\left( {108 \times \dfrac{2}{3}} \right) - 16 = 8t $
 $t = \dfrac{{72 - 16}}{8} = \dfrac{{56}}{8} = 7 $

The motion of the wheel had started $7 $ seconds before the $4s $ time interval.

Note The equations of circular motion are similar to the equations of linear motion. Only the linear components are replaced with their angular counterparts. Like the distance in linear motion is replaced with the angle covered, linear velocity is replaced with angular velocity and the linear acceleration is replaced with angular acceleration.