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Question: The shaft of an electric motor starts from rest and on the application of the torque, it gains an an...

The shaft of an electric motor starts from rest and on the application of the torque, it gains an angular acceleration given by α=3t−t2\alpha = 3t - {t^2}, during the first 22 seconds after it starts after which α=0\alpha = 0. The angular velocity after 66 seconds will be
(a) 10/310/3 rad/sec
(b) 20/320/3 rad/sec
(c) 5/35/3 rad/sec
(d) 1/31/3 rad/sec

Explanation

Solution

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 22 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 α=3t−t2\alpha = 3t - {t^2},
Final time of observation is given 66 seconds.
As we know, angular velocity is represented by the symbol ω\omega Since, it starts from rest, so initial angular velocity would be zero. i.e. ω0=0{\omega _0} = 0
Now, velocity after 22 seconds
α=3t−t2\alpha = 3t - {t^2}; only for first 22 seconds
∫0ω1dω=∫02αdt\int_0^{{\omega _1}} {d\omega } = \int_0^2 {\alpha dt} ; Since, acceleration is given by, α=dωdt\alpha = \dfrac{{d\omega }}{{dt}}
⇒ω1=∫02(3t−t2)dt\Rightarrow {\omega _1} = \int_0^2 {\left( {3t - {t^2}} \right)} dt
⇒ω1=∫023tdt−∫02t2dt\Rightarrow {\omega _1} = \int_0^2 {3tdt} - \int_0^2 {{t^2}dt}
⇒ω1=[3t22−t33]20\Rightarrow {\omega _1} = {\left[ {\dfrac{{3{t^2}}}{2} - \dfrac{{{t^3}}}{3}} \right]^2}_0
⇒ω1=[3(2)22−(2)33]\Rightarrow {\omega _1} = \left[ {\dfrac{{3{{(2)}^2}}}{2} - \dfrac{{{{(2)}^3}}}{3}} \right] =122−83=103 = \dfrac{{12}}{2} - \dfrac{8}{3} = \dfrac{{10}}{3} rad/sec
The angular velocity after 22 seconds is 103\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 66seconds of the start of the motion.
So, ww after total 66 seconds will be, w=103w = \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.