QJEEM23Phaeda: JEE Main 2023 Physics - Class 11 JEEM23Phaeda Solution: A light rope is wound around a hollow cylinder of mass 5 kg and …

In this problem, the net force on the cylinder is the tension $T$ in the rope, which is equal to the force applied to the rope:

$F=T=52.5~\mathrm{N}$

The force causes the cylinder to accelerate with an angular acceleration $\alpha$ , which is related to its linear acceleration $a$ and the radius of the cylinder $R$ by the equation:

$\alpha=\frac{a}{R}$

The linear acceleration $a$ of the cylinder can be found using the formula $F=ma$ :

$ma=F=52.5~\mathrm{N}$

where $m=5~\mathrm{kg}$ is the mass of the cylinder. Solving for $a$ , we get:

$a=\frac{F}{m}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}}=10.5~\mathrm{m/s^2}$

Substituting this value of $a$ into the equation for $\alpha$ , we get:

$\alpha=\frac{a}{R}=\frac{10.5~\mathrm{m/s^2}}{0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}$

Therefore, the angular acceleration of the cylinder is $15~\mathrm{rad/s^2}$ .

Alternate Method:

Let's first draw a free body diagram of the cylinder. The force $F$ applied to the rope creates a tension in the rope, which in turn exerts a force on the cylinder in the opposite direction. This force is given by:

$T=F$

where $T$ is the tension in the rope. The cylinder also experiences a torque due to the tension in the rope, which causes it to rotate. The torque is given by:

$\tau=TR$

where $R$ is the radius of the cylinder.

The net torque on the cylinder is equal to the product of the moment of inertia $I$ of the cylinder and its angular acceleration $\alpha$ :

$\tau=I\alpha$

The moment of inertia of a hollow cylinder about its geometrical axis which is parallel to its length is given by:

$I=MR^2$

where $M$ is the mass of the cylinder.

Substituting the given values, we get:

$\tau=TR=I\alpha=MR^2\alpha$

Solving for $\alpha$ , we get:

$\alpha=\frac{T}{MR}=\frac{F}{MR}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}\cdot0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}$

Therefore, the angular acceleration of the cylinder is $15~\mathrm{rad/s^2}$ .