Question

A solid cylinder of mass $50\,{\text{kg}}$ and radius $0.5\,{\text{m}}$ is free to rotate about horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of $2\,{\text{revolutions}} \cdot {{\text{s}}^{ - 2}}$ is:
A. $25\,{\text{N}}$
B. $50\,{\text{N}}$
C. $78.5\,{\text{N}}$
D. $157\,{\text{N}}$

Answer 1

Use the formulae for moment of inertia of the solid cylinder, torque due to a force and torque in terms of angular acceleration and moment of inertia. First determine the toque with the moment of inertia and angular acceleration and then determine the tension in the string using obtained torque.

Formulae used:
The expression for the moment of inertia $I$ of a solid cylinder is
$I = \dfrac{{M{R^2}}}{2}$ …… (1)
Here, $M$ is the mass of the cylinder and $R$ is the radius of the cylinder.
The expression for the torque $\tau$ due to a force $F$ is
$\tau = Fr$ …… (2)
Here, $r$ is the perpendicular distance between the point of action of the force and the centre of the torque.
The expression for the torque $\tau$ is
$\tau = I\alpha$ …… (3)
Here, $I$ is the moment of inertia and $\alpha$ is the angular acceleration.

Complete step by step answer:
It is given that the solid cylinder has mass $50\,{\text{kg}}$ and radius $0.5\,{\text{m}}$ which is rotating about its horizontal axis.
$M = 50\,{\text{kg}}$
$R = 0.5\,{\text{m}}$
The angular acceleration is $2\,{\text{revolutions}} \cdot {{\text{s}}^{ - 2}}$.
$\alpha = 2\,{\text{revolutions}} \cdot {{\text{s}}^{ - 2}}$
Convert the unit of the angular acceleration in the SI system of units.
$\alpha = \left( {2\,{\text{revolutions}} \cdot {{\text{s}}^{ - 2}}} \right)\left( {\dfrac{{2\pi \,{\text{rad}}}}{{1\,{\text{revolution}}}}} \right)$
$\Rightarrow \alpha = 4\pi \,{\text{rad}} \cdot {{\text{s}}^{ - 2}}$
Determine the torque $\tau$ of the disc due to its rotational motion.
Substitute $\dfrac{{M{R^2}}}{2}$ for $I$ and $4\pi \,{\text{rad}} \cdot {{\text{s}}^{ - 2}}$ for $\alpha$ in equation (3).
$\tau = \left( {\dfrac{{M{R^2}}}{2}} \right)\left( {4\pi \,{\text{rad}} \cdot {{\text{s}}^{ - 2}}} \right)$
Substitute $50\,{\text{kg}}$ for $M$, $0.5\,{\text{m}}$ for $R$ and $3.14$ for $\pi$ in the above equation.
$\tau = \left( {\dfrac{{\left( {50\,{\text{kg}}} \right){{\left( {0.5\,{\text{m}}} \right)}^2}}}{2}} \right)\left( {4\left( {3.14} \right)\,{\text{rad}} \cdot {{\text{s}}^{ - 2}}} \right)$
$\Rightarrow \tau = 78.5\,{\text{N}} \cdot {\text{m}}$
Hence, the torque is $78.5\,{\text{N}} \cdot {\text{m}}$.
Now determine the torque on the cylinder due to tension in the string.
Substitute $T$ for $F$ and $R$ for $r$ in equation (2).
$\tau = TR$
$\Rightarrow T = \dfrac{\tau }{R}$
Substitute $78.5\,{\text{N}} \cdot {\text{m}}$ for $\tau$ and $0.5\,{\text{m}}$ for $R$ in the above equation.
$T = \dfrac{{78.5\,{\text{N}} \cdot {\text{m}}}}{{0.5\,{\text{m}}}}$
$\therefore T = 157\,{\text{N}}$
Therefore, the tension in the string is $157\,{\text{N}}$.

Hence, the correct option is D.

Note: The angular acceleration of the solid cylinder is given in revolutions per second square but it is not the SI unit of the angular acceleration. Don’t forget to convert the unit of angular acceleration of the solid cylinder in the SI system of units as all the remaining physical quantities are used with their SI unit.