Question

A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm. The torque required to stop after $2\pi$ revolution is
A. $2\times {{10}^{-6}}Nm$
B. $2\times {{10}^{-3}}Nm$
C. $12\times {{10}^{-4}}Nm$
D. $2\times {{10}^{6}}Nm$

Answer 1

Hint: Even if it is a rotational motion, they are obeying the laws of translational motion. The difference is, in rotational motion the axis is fixed and in translational motion, the direction is fixed. We can find the torque from the conservation of energy and work-energy theorem.

Formula used:
$W=\dfrac{1}{2}I(\omega _{f}^{2}-\omega _{i}^{2})$, where $I$ is the moment of inertia, ${{\omega }_{f}}\And {{\omega }_{i}}$ are the corresponding final and initial angular velocities respectively.
$\text{angular displacement = 2 }\\!\\!\pi\\!\\!\text{ }\\!\\!\times\\!\\!\text{ revolution}$
$I=\dfrac{m{{r}^{2}}}{2}$, where m is the mass and r is the radius of the solid cylinder.
$\dfrac{1}{2}I{{\omega }^{2}}=\tau \times \theta$, where $\theta$ is the angular displacement, $\tau$ is the torque, $\omega$ is the angular velocity and $I$ is the moment of inertia.

Complete step-by-step answer:

According to the work-energy theorem, the total work done acting on an object will be equal to the change in kinetic energy. Here we are dealing with angular motion. So we have to consider inertia instead of mass and linear velocity will be the angular velocity. Thus we can write the work-energy theorem as,
$W=\dfrac{1}{2}I(\omega _{f}^{2}-\omega _{i}^{2})$, where $I$ is the moment of inertia, ${{\omega }_{f}}\And {{\omega }_{i}}$ are the corresponding final and initial angular velocities respectively.
$\text{angular displacement = 2 }\\!\\!\pi\\!\\!\text{ }\\!\\!\times\\!\\!\text{ revolution}$
Here we are considering the only $2\pi$ revolution. So the angular displacement will be

& \text{angular displacement = 2 }\\!\\!\pi\\!\\!\text{ }\\!\\!\times\\!\\!\text{ 2 }\\!\\!\pi\\!\\!\text{ } \\\ & =4{{\text{ }\\!\\!\pi\\!\\!\text{ }}^{2}} \\\ \end{aligned}$$ Next, we can find out the moment of inertia. $$I=\dfrac{m{{r}^{2}}}{2}$$, where m is the mass and r is the radius of the solid cylinder. Here the mass of the cylinder is 4 kg and the radius of the solid cylinder is 4 cm. So the moment of inertia will be, $$I=\dfrac{1}{2}\times 2\times {{(0.04)}^{2}}$$ $$I=1.6\times {{10}^{-3}}kg{{m}^{2}}$$ Next, we have to find out the angular velocity. Here we are trying to stop the rotation. Thus the final angular velocity will be zero. The initial angular velocity is, $${{\omega }_{i}}=3\text{rpm}$$ $${{\omega }_{i}}=3\times \dfrac{2\pi }{60}rad/s$$ $${{\omega }_{i}}=\dfrac{\pi }{10}rad/s$$ From the conservation of energy theorem, We can say that change in rotational kinetic energy will be equal to the product of torque and angular displacement. It can be mathematically represented as, $$\dfrac{1}{2}I{{\omega }^{2}}=\tau \times \theta $$, where $$\theta $$ is the angular displacement, $$\tau $$ is the torque, $$\omega $$ is the angular velocity and $$I$$ is the moment of inertia. We can put the values into the equation. $$\dfrac{1}{2}\times 1.6\times {{10}^{-3}}\times {{\left( \dfrac{\pi }{10} \right)}^{2}}=\tau \times 4{{\pi }^{2}}$$ $$\tau =\dfrac{\dfrac{1}{2}\times 1.6\times {{10}^{-3}}\times {{\left( \dfrac{\pi }{10} \right)}^{2}}}{4{{\pi }^{2}}}$$ $$\tau =2\times {{10}^{-6}}Nm$$ Hence the correct option is A. Note: Angular frequency and angular velocity are the same. They are represented by the $$\omega $$. So you can use either angular velocity or angular frequency in the solution. angular frequency is the angular displacement per unit time. The given options look similar. So you have to check the answer once before choosing the answer. Rotation motion can be easily transferred from the pure translational motion by using rotational motion quantities.