Question

A wheel has a constant angular acceleration of $3rad/{s^2}$. During a 4s interval, it turns through an angle of 60 rad. If the wheel started from rest, how long has it been in motion before the start of this 4s interval?
$
  (a){\text{ 3s}} \\
  (b){\text{ 6s}} \\
  (c){\text{ 9s}} \\
  (d){\text{ 12s}} \\
$

Answer 1

Hint: In this question use the direct formula that is $\theta = {\omega _1}t + \dfrac{1}{2}\alpha {t^2}$ where $\theta $ is the angle of turning, $\omega $ is the angular velocity, $\alpha $ is the angular acceleration. This will help finding the value of ${\omega _1}$, then use the first equation of rotational kinematics that is ${\omega _1} = {\omega _o} + \alpha t'$. This will help finding the right answer.

Complete step-by-step solution -
Given data:
Angular acceleration of the wheel = 3 ${rad/s^{2}}$.
During a 4 seconds interval.
In this it turns through an angle of 60 radians.
Now if the wheel started from the rest then we have to find out how long it will take to be in motion before the start of this 4s interval.
Now according to rotational kinematics we have,
Angle of turning $\left( \theta \right)$ = angular velocity (${\omega _1}$) times time + half times angular acceleration $\left( \alpha \right)$ time square of time interval
$ \Rightarrow \theta = {\omega _1}t + \dfrac{1}{2}\alpha {t^2}$
Now using this formula calculate the angular velocity of the wheel, here $\theta = 60$ rad, $t = 4s$ and
\[\alpha = 3\]rad/$s^2$.
Now substitute the values we have,
$ \Rightarrow 60 = {\omega _1}\left( 4 \right) + \dfrac{1}{2}\left( 3 \right){4^2}$
Now simplify this we have,
$ \Rightarrow 60 = 4{\omega _1} + 24$
$ \Rightarrow 4{\omega _1} = 60 - 24 = 36$
$ \Rightarrow {\omega _1} = \dfrac{{36}}{4} = 9$ rad/s.
Now it is also given that the wheel starts from the rest so the initial angular velocity of the wheel is zero.
$ \Rightarrow {\omega _o} = 0$ rad/s, where, ${\omega _o}$ = initial angular velocity.
Now according to first equation of the rotational kinematics we have,
$ \Rightarrow {\omega _1} = {\omega _o} + \alpha t'$
Where, ${\omega _1}$ = angular velocity of the wheel after time interval t’
Now substitute the values in this equation we have,
$ \Rightarrow 9 = 0 + \left( 3 \right)t'$
Now simplify this we have,
$ \Rightarrow t' = \dfrac{9}{3} = 3$ Seconds.
So this is the required time it will take to be in motion before the start of the 4s interval.
Hence option (A) is the correct answer.

Note: There are three laws of rotational kinematics that is${\omega _1} = {\omega _o} + \alpha t'$, $\theta = {\omega _1}t + \dfrac{1}{2}\alpha {t^2}$ and ${\omega _1}^2 - {\omega _0}^2 = 2\alpha \theta $. These equations resembles same as that of three equations of linear kinematics that is $v = u + at$, $s = ut + \dfrac{1}{2}a{t^2}$ and ${v^2} - {u^2} = 2as$. It is advised to remember these equations as it helps saving a lot of time while dealing with problems of this kind.