a mass m is supported by a massless string wound around a uniform hollow cylinder of mass m & radius R . If the string does not slip on the cylinder , with what acceleration will the mass fall on release?

To determine the acceleration of the mass as it falls when supported by a massless string wound around a uniform hollow cylinder, we need to analyze the forces and torques acting on the system. This involves applying Newton's second law for both the falling mass and the rotational motion of the cylinder.

Understanding the Forces at Play

Let’s denote the mass of the object as m, the radius of the cylinder as R, and the acceleration due to gravity as g. When the mass is released, it will experience a downward force due to gravity, which can be expressed as:

• Weight of the mass: F_gravity = m * g

As the mass falls, it will also exert a tension T in the string, which opposes its motion. The net force acting on the mass can be described by Newton's second law:

• Net force: F_net = m * a = m * g - T

Torque and Rotational Motion

Now, let’s consider the hollow cylinder. The tension in the string will create a torque about the center of the cylinder. The torque τ can be expressed as:

• τ = T * R

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

• I = m * R²

According to Newton's second law for rotation, the torque is also related to the angular acceleration α:

• τ = I * α

Since the string does not slip, the linear acceleration a of the mass is related to the angular acceleration of the cylinder by the equation:

• a = α * R

Setting Up the Equations

Now we can combine these equations. From the torque equation, we have:

• T * R = (m * R²) * (a / R)
• This simplifies to: T = (m * a)

Now we can substitute this expression for tension back into the net force equation for the mass:

• m * a = m * g - (m * a)

Solving for Acceleration

Rearranging the equation gives us:

• m * a + m * a = m * g
• 2 * m * a = m * g
• a = g / 2

Final Result

Thus, the acceleration of the mass as it falls is:

• a = g / 2

This means that the mass will fall with an acceleration equal to half the acceleration due to gravity. This result illustrates the interplay between linear and rotational dynamics in a system involving a mass and a rotating object.