A cylinder having moment of inertia $I,$ which is free to rotate about its axis, receives an angular impulse of \[Jkg{{m}^{2}}/s\] initially, followed by similar impulse after every 4s. What is the angular speed of the cylinder 30s after the initial impulse?

$A.\dfrac{7J}{I}$
$B.\dfrac{8J}{I}$
$C.\dfrac{J}{I}$
$D.0$

Hint:
This is a basic question on rotational kinematics. Knowledge of torque, angular momentum, moment of inertia and angular velocity is required. The relations between them is enough to solve this question.

Step by step solution:
We know that a cylinder has moment of inertia $I,$ and it is being given angular impulses every 4 seconds. We are to find the case at the end of 30 seconds.

So let’s consider the first impulse at \[~t=0s\]

The next will be at \[~t=4s\] and then at \[~t=8s\] and so on till \[~t=28s.\]

Hence, the total number of times angular impulse is given to the cylinder is 8.

Now, we will use the relationship between torque $(\tau )$ and angular momentum $(L)$,$\tau =\dfrac{dL}{dt}.$

$\tau dt=dL$

$\tau dt$is also known as angular impulse. Therefore, $\int\limits_{0}^{30}{\tau dt}=\sum\limits_{i=0}^{8}{{{J}_{i}}}=\int{dL}$

Since the given value of angular impulse is \[Jkg{{m}^{2}}/s,\] we will add it each time the impulses were given in the time limits. We know, $L=I\omega $

Therefore, \[\sum\limits_{0}^{8}{{{J}_{i}}}=I\int\limits_{0}^{\omega }{d\omega }\]

$8J=I\omega $

Therefore the angular velocity at the end of 30s is $\omega =\dfrac{8J}{I}.$
Note:
Even though the question states about Inertia, don’t start to derive the value of Inertia for a cylinder as it isn’t required in this question.
The change from integral to summation is necessary, since we aren’t given the value of torque in the question and additionally the value of impulse is readily given making the summation easy to solve.