Question

The shaft of an electric motor starts from rest and on the application of the torque, it gains an angular acceleration given by $\alpha = 3t - {t^2}$, during the first $2$ seconds after it starts after which $\alpha = 0$. The angular velocity after $6$ seconds will be
(a) $10/3$ rad/sec
(b) $20/3$ rad/sec
(c) $5/3$ rad/sec
(d) $1/3$ rad/sec

Answer 1

Hint: To find the required solution of the given question we will apply the concept of Newton’s first law of motion. First, we need to deduce the expression for angular velocity after $2$ seconds and then integrate it to get the final answer.

Complete step-by-step answer:
It is given that the shaft of an electric motor starts from rest and on the application of the torque, it gains an angular acceleration given by $\alpha = 3t - {t^2}$,
Final time of observation is given $6$ seconds.
As we know, angular velocity is represented by the symbol $\omega $ Since, it starts from rest, so initial angular velocity would be zero. i.e. ${\omega _0} = 0$
Now, velocity after $2$ seconds
$\alpha = 3t - {t^2}$; only for first $2$ seconds
$\int_0^{{\omega _1}} {d\omega } = \int_0^2 {\alpha dt} $; Since, acceleration is given by, $\alpha = \dfrac{{d\omega }}{{dt}}$
$ \Rightarrow {\omega _1} = \int_0^2 {\left( {3t - {t^2}} \right)} dt$
$ \Rightarrow {\omega _1} = \int_0^2 {3tdt} - \int_0^2 {{t^2}dt} $
$ \Rightarrow {\omega _1} = {\left[ {\dfrac{{3{t^2}}}{2} - \dfrac{{{t^3}}}{3}} \right]^2}_0$
$ \Rightarrow {\omega _1} = \left[ {\dfrac{{3{{(2)}^2}}}{2} - \dfrac{{{{(2)}^3}}}{3}} \right]$ $ = \dfrac{{12}}{2} - \dfrac{8}{3} = \dfrac{{10}}{3}$ rad/sec
The angular velocity after $2$ seconds is $\dfrac{{10}}{3}$ rad/sec According to the Newton’s law of motion, which states that every object remains at rest or in a uniform motion in a straight line unless and until compelled to change its state by the action of an external force. This will continue even after the end of $6$seconds of the start of the motion.
So, $w$ after total $6$ seconds will be, $w = \dfrac{{10}}{3}$ rad/sec

So, the correct answer is “Option A”.

Note: We know that angular velocity is defined as the rate of velocity at which an object or a particle is rotating a specific point in a given time period. It is rotational equivalent to the linear velocity. And angular acceleration is defined as the non-constant velocity. Constant angular velocity in a circle is also known as uniform circular motion.