Question

A rope is wound around a hollow cylinder of mass $3kg$ and radius $40cm$. What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 N$?

Answer 1

Hint: When the rope is being pulled, on account of its whirling a force acts on the cylinder. This force has the tendency to make the cylinder move around in a circular motion, so this is the torque. Torque is defined as any force which acts on a body and due to that that body starts rotating around some axis.

Complete step by step solution:
Mass of the cylinder, M= 3kg, Radius of the circular part, R= 40 cm= 0.4m
In this case the force and the radius are perpendicular to each other, thus torque is given by as:
$\overrightarrow{\tau }=\overrightarrow{R}\times \overrightarrow{F} \\
\Rightarrow \tau =RF\sin 90 \\
\Rightarrow \tau =RF \\
\Rightarrow \tau =0.4\times 30 \\
\therefore \tau =12Nm \\$
But torque is given by \[\tau =I\alpha \], where I is the moment of inertia and alpha is angular acceleration.
For cylinder moment of inertia, \[I=M{{R}^{2}}\]
$\Rightarrow I\alpha =12 \\
\Rightarrow \alpha =\dfrac{12}{M{{R}^{2}}} \\
\Rightarrow \alpha =\dfrac{12}{3\times {{0.4}^{2}}} \\
\therefore \alpha =25rad/{{s}^{2}} $

Hence, the angular acceleration of the cylinder is $25rad/{{s}^{2}}$.

Note: While substituting the values all the units must be in the same notation that is standard SI units. torque is a vector quantity and is given by the cross product of force and perpendicular distance from the axis of rotation. Both the vectors were perpendicular to each other in this problem. Also, moment of inertia here is calculated about an axis passing through the centre of mass.The moment of force popularly known as torque, is a measure of the ability of the force acting on a particular body to rotate it around a given axis.