JEE Main 2023 — Rotational Motion Question with Solution

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:

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

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

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

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

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:

$$I=M{R}^2$$

where $$M$$ is the mass of the cylinder.

Substituting the given values, we get:

$$\tau =TR=I\alpha =M{R}^2\alpha$$

Solving for $$\alpha$$, we get:

$$\alpha =\frac{T}{MR}=\frac{F}{MR}=\frac{52.5\mathrm{N}}{5\mathrm{kg}\cdot 0.7\mathrm{m}}=\boxed{15rad/{\mathrm{s}}^2}$$

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