Question

4. The shaft of an electric motor starts from rest

and on the application of a torque, it gains an angular

acceleration given by α = 3t - t2

during the first 2

seconds after it starts after which α = 0 . The angular

velocity after 6 sec will be​

Answer 1

Answer:

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Answer 2

$\omega_f=4\ rad.s^{-1}$ was the angular velocity at the end of 2 second when the acceleration stopped and now this will continue to be even after the end of 6 seconds of the start of motion according to the Newton's first law of motion.

Explanation:

Given:

• initial angular speed of the shaft, $\omega_i=0\ rad.s^{-1}$

• angular acceleration follows the function, $\alpha=3t-t^2$

• time for which the acceleration is followed, $t_a=2\ s$

• final time of observation, $t=6\ s$

Now, the angular acceleration at the end of 2 seconds:

$\alpha=3\times 2-2^2$

$\alpha=2\ rad.s^{-2}$

According to the question, the acceleration after 2 seconds is zero. So, it will continue the speed which was at the end of 2 seconds.

So the angular velocity after 6 seconds:

Using eq. of motion.

$\omega_f=\omega_i+\alpha.t$

Now we substitute the conditions of the end of 2 seconds

$\omega_f=0+2\times 2$

$\omega_f=4\ rad.s^{-1}$ was the angular velocity at the end of 2 second when the acceleration stopped and now this will continue to be even after the end of 6 seconds of the start of motion according to the Newton's first law of motion.