A mass $$m$$ hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass $$m$$ and radius $$R$$. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass $$m$$, if the string does not slip on the pulley, is:

To find the acceleration of the hanging mass, we can analyze the forces and torques acting on the system.

1. Analyze the hanging mass m:

   Let the tension in the string be T and the downward linear acceleration of the mass be a.

   The forces acting on the hanging mass are gravity acting downwards and tension acting upwards. Applying Newton's second law:

   $$mg - T = ma$$
2. Analyze the pulley:

   The pulley is a uniform circular disc of mass m and radius R. Its moment of inertia I about its central axis is:

   $$I = \frac{1}{2}mR^2$$

The torque tau on the pulley is provided by the tension T of the string at radius R:

$$\tau = TR$$

Applying the rotational form of Newton's second law ($$\tau = I\alpha$$, where $$\alpha$$ is the angular acceleration):

$$TR = I\alpha$$

$$TR = \left(\frac{1}{2}mR^2\right)\alpha$$

1. Relate linear and angular acceleration:

   Since the string does not slip on the pulley, the linear acceleration a of the mass is related to the angular acceleration $$\alpha$$ of the pulley by:

   $$a = \alpha R$$
2. $$\alpha = \frac{a}{R}$$
3. Substitute alpha into the torque equation:$$TR = \left(\frac{1}{2}mR^2\right)\left(\frac{a}{R}\right)$$
4. $$TR = \frac{1}{2}mRa$$

Dividing both sides by R gives the expression for tension:

$$T = \frac{1}{2}ma$$

1. Substitute T back into the mass equation:$$mg - \frac{1}{2}ma = ma$$

Move all terms containing acceleration a to one side:

$$mg = ma + \frac{1}{2}ma$$

$$mg = \frac{3}{2}ma$$

Cancel the mass m from both sides:

$$g = \frac{3}{2}a$$

Solve for acceleration a:

$$a = \frac{2}{3}g$$