A string is wound around a uniform disk of radius R and mass M. The disk is released from

rest when the string is vertical and its top end is tied to a fixed bar.
(a) Draw a free body diagram indicating all the forces acting on the disk.
(b) Calculate the acceleration of the center of mass of the disk.
(c) Calculate the tension in the string.
(d) Calculate the speed of the center of mass as the disk descends a distance h. (Do not use energy
conservation)
(e) Use energy conservation to verify your result in part (d)
4

To answer these questions, we will need to use some principles of physics, such as Newton's laws of motion and the concepts of torque and rotational motion.

(a) To draw a free body diagram, we need to identify all the forces acting on the disk. In this case, we have:

- Gravitational force (mg) acting downward at the center of mass.
- Tension force (T) exerted by the string, directed upward.
- Normal force (N) exerted by the fixed bar, perpendicular to the surface of the disk.
- Friction force (f) between the disk and the surface it is rolling on, assuming there is friction present.
- Air resistance (if applicable) acting opposite to the direction of motion.

(b) To calculate the acceleration of the center of mass of the disk, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F = ma).

In this case, the only net force acting on the disk is the tension force, T. Therefore, we have:

T - mg = Ma

Solving for acceleration (a), we get:

a = (T - mg) / M

(c) To calculate the tension in the string, we need to consider the rotational motion of the disk. Tension creates a torque that causes the disk to accelerate.

The net torque acting on the disk is equal to the product of the tension force (T) and the radius of the disk (R). Therefore, we have:

net torque = T * R

The net torque is also equal to the moment of inertia (I) of the disk multiplied by its angular acceleration (α). For a uniform disk, the moment of inertia is given by 1/2 * MR^2, and the angular acceleration is equal to the linear acceleration divided by the radius (α = a / R).

So we can write:

T * R = (1/2 * MR^2) * (a / R)

Simplifying, we find:

T = (1/2 * MR) * a

(d) To calculate the speed of the center of mass as the disk descends a distance h, we can use kinematic equations. Specifically, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

In our case, the initial velocity is zero, as the disk starts from rest. The acceleration is given by the expression we derived in part (b). The distance traveled is h.

Plugging the values into the equation, we have:

v^2 = 0 + 2a * h

Simplifying, we find:

v = √(2ah)

(e) We can use energy conservation to verify our result in part (d). In this system, the potential energy is converted into kinetic energy as the disk descends.

The potential energy of the disk when it is at a height h above its starting point is given by mgh, where m is the mass of the disk and g is the acceleration due to gravity.

At the bottom of the descent, the potential energy is zero, and all the initial potential energy is converted into kinetic energy. The kinetic energy is given by (1/2)mv^2, where v is the speed of the center of mass that we calculated in part (d).

Setting these equations equal to each other, we have:

mgh = (1/2)mv^2

Canceling out the mass (m), we get:

gh = (1/2)v^2

Simplifying, we find:

v = √(2gh)

Comparing this result with our previous calculation in part (d), we can verify that the speed of the center of mass calculated using energy conservation is consistent with our earlier result.