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

Question

A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

Answer 1

To find the angular acceleration of the cylinder, let's first understand the physics behind it. Angular acceleration (\(\alpha\)) can be determined by applying the relation between torque (\(\tau\)) and angular acceleration:

\(\tau = I \cdot \alpha\)

Where \(I\) is the moment of inertia of the hollow cylinder.

For a hollow cylinder, the moment of inertia \(I\) is given by:

I = m \cdot r^2

Where \(m\) is the mass of the cylinder and \(r\) is its radius.

Given:

Mass (\(m\)) = 3 kg
Radius (\(r\)) = 40 cm = 0.4 m
Force applied (\(F\)) = 30 N

The torque (\(\tau\)) can be calculated as the product of the force and the radius, because the force is applied tangentially:

\(\tau = F \cdot r\)

Now, calculate the torque:

\(\tau = 30 \, \text{N} \cdot 0.4 \, \text{m} = 12 \, \text{Nm}\)

Next, calculate the moment of inertia of the hollow cylinder:

I = 3 \, \text{kg} \cdot (0.4 \, \text{m})^2 = 3 \cdot 0.16 = 0.48 \, \text{kg}\cdot\text{m}^2

Now, we have to find the angular acceleration (\(\alpha\)) using the equation:

\(\alpha = \frac{\tau}{I} = \frac{12 \, \text{Nm}}{0.48 \, \text{kg}\cdot\text{m}^2}\)

Performing the calculation:

\(\alpha = 25 \, \text{rad/s}^2\)

Thus, the angular acceleration of the cylinder is \(25 \, \text{rad/s}^2\).

Therefore, the correct answer is 25 rad/s².

Answer 2