Question

Pulley of radius 2m is rotated about its axis by a force $F = \left( {20t - 5{t^2}} \right)$ Newton (where $t$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $10kg{m^2}$, the number of rotations made by the pulley before its direction of motion is reversed, is:
(A) less than 3
(B) more than 3 but less than 6
(C) more than 6 but less than 9
(D) more than 9

Answer 1

We can find the angular acceleration of the pulley from the equations of its torque. Then using that value of the angular acceleration we can calculate the angular speed. Now by equating the angular speed to zero we calculate the time at which the angular speed becomes zero. From there by substituting the limits, we can find the angular displacement in radians. The angular displacement divided by $2\pi$ will give us the number of rotations.
Formula Used: In this solution, we will be using the following formula,
$\Rightarrow \tau = FR$ where $\tau$ is the torque,
$F$ is the force and $R$ is the radius of the pulley.
$\Rightarrow \tau = I\alpha$
where $I$ is the moment of inertia and $\alpha$ is the angular acceleration.

Complete step by step solution:
In the question, we are given that the pulley is rotated about an axis by a force. Now from the equations of the torque that is acting on the pulley, we can write,
$\Rightarrow \tau = FR$ and $\tau = I\alpha$
So on equating these two equations we get,
$\Rightarrow FR = I\alpha$
Therefore, from here we can write the angular acceleration as,
$\Rightarrow \alpha = \dfrac{{FR}}{I}$
In the question we are given, $F = \left( {20t - 5{t^2}} \right)$, $R = 2m$ and $I = 10kg{m^2}$
So substituting we get,
$\Rightarrow \alpha = \dfrac{{\left( {20t - 5{t^2}} \right) \times 2}}{{10}}$
Hence we get on calculating, $\alpha = 4t - {t^2}$
Now the angular acceleration can be written as the differentiation of the angular speed with time. So we get,
$\Rightarrow \alpha = \dfrac{{d\omega }}{{dt}} = 4t - {t^2}$
On taking the $dt$ to the RHS,
$\Rightarrow d\omega = \left( {4t - {t^2}} \right)dt$
Now on integrating on both the sides we get,
$\Rightarrow \int\limits_0^\omega {d\omega } = \int\limits_0^t {\left( {4t - {t^2}} \right)dt}$
On integrating we get,
$\Rightarrow \omega = \dfrac{{4{t^2}}}{2} - \dfrac{{{t^3}}}{3} = 2{t^2} - \dfrac{{{t^3}}}{3}$….(1)
Now if we set the value of the angular speed to zero,
$\Rightarrow 2{t^2} - \dfrac{{{t^3}}}{3} = 0$
On taking common we have,
$\Rightarrow {t^2}\left( {2 - \dfrac{t}{3}} \right) = 0$
Therefore we get 2 values of time when the angular sped is zero. They are,
$\Rightarrow {t^2} = 0$, that is $t = 0$ and
$\Rightarrow 2 - \dfrac{t}{3} = 0$, that is, $t = 2 \times 3 = 6s$
Now the angular speed can be written as the differentiation of the angular displacement with respect to time.
$\Rightarrow \omega = \dfrac{{d\theta }}{{dt}}$
So substituting from equation 1 and then taking $dt$ to the RHS we get,
$\Rightarrow d\theta = \left( {2{t^2} - \dfrac{{{t^3}}}{3}} \right)dt$
On integrating both the sides, where we can set the limit in the RHS from 0 to 6 we get,
$\Rightarrow \int\limits_0^\theta {d\theta } = \int\limits_0^6 {\left( {2{t^2} - \dfrac{{{t^3}}}{3}} \right)dt}$
On integrating we get,
$\Rightarrow \theta = \dfrac{2}{3}\left. {{t^3}} \right|_0^6 - \dfrac{1}{{12}}\left. {{t^4}} \right|_0^6$
On calculating we get,
$\Rightarrow \theta = \dfrac{2}{3} \times 216 - \dfrac{1}{{12}} \times 1296 = 144 - 108$
So we get the angular displacement as,
$\Rightarrow \theta = 36rad$
So the number of rotations will be angular displacement divided by $2\pi$.
So we have,
$\Rightarrow n = \dfrac{{36}}{{2\pi }}$
The value of this is less than 6. Hence the number of rotations will be more than 3 but less than 6.
So the correct answer is option (B).

Note:
The torque on a body is the amount of force that can cause a body to rotate about an axis. It can be described as the rotational equivalent of the linear force. The point about which the body rotates is called the axis of rotation. The SI unit of torque is Newton-meter.