Question

A light rope is wound around a hollow cylinder of mass 5 kg and radius 70 cm. The rope is pulled with a force of 52.5 N. The angular acceleration of the cylinder will be rad s^-2

Answer 1

Correct Answer: 15

The torque acting on the hollow cylinder is given by: \[ \tau = F R, \] where \( F = 52.5 \, \text{N} \) is the force, and \( R = 70 \, \text{cm} = 0.7 \, \text{m} \) is the radius. The moment of inertia (\( I \)) of a hollow cylinder about its axis is: \[ I = m R^2, \] where \( m = 5 \, \text{kg} \) is the mass of the cylinder. The angular acceleration \( \alpha \) is given by: \[ \alpha = \frac{\tau}{I}. \] Substitute \( \tau = F R \) and \( I = m R^2 \): \[ \alpha = \frac{F R}{m R^2}. \] Simplify: \[ \alpha = \frac{F}{m R}. \] Substitute the values: \[ \alpha = \frac{52.5}{5 \cdot 0.7} = \frac{52.5}{3.5} = 15 \, \text{rad/s}^2. \] Thus, the angular acceleration of the cylinder is \( \boxed{15 \, \text{rad/s}^2} \).