Question

A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can rotate with negligible friction about a stationary horizontal axis. The drum is not a uniform cylinder and has unknown moment of inertia. When you release the bucket from rest, you find that it has a downward acceleration of magnitude a. What is the tension in the cable between the drum and the bucket

Answer 1

The tension in the cable connecting the drum to the bucket is equal to the mass of the bucket times the difference between the acceleration due to gravity and the bucket's downward acceleration.

Step-by-step explanation:

To determine the tension in the cable between the drum and the bucket, we can apply Newton's second law to the hanging mass. The net force acting on the bucket is the difference between the weight of the bucket and the tension in the rope:

`F_(net)` = mg - T

Since we know the bucket has a downward acceleration a, we can write Newton's second law as:

ma = mg - T

Where m is the mass of the bucket, g is the acceleration due to gravity, and T is the tension in the rope. We can rearrange the equation to solve for T:

T = mg - ma

T = m(g - a)

So the tension in the cable is equal to the mass of the bucket times the difference between the acceleration due to gravity and the downward acceleration of the bucket.

Answer 2

T = mg - (m²g/(I/R² + m))

Step-by-step explanation:

Let T be the tension in the cable between the drum and the bucket

Now, by applying newton's second law of gravity on the downward movement of the bucket, we will obtain;

mg - T = ma - - - - (eq1)

Now, on the drum , a torque of TR will be acting which will create an angular acceleration of "α" in it.

Where R is the radius.

Let "I" denote the moment of inertia of the drum. Thus, we have;

TR = Iα

Now, the angular acceleration is expressed in the form;

α = a/R

Where a is the linear downward acceleration.

Thus;

TR = Ia/ R

T = Ia/ R²

Let's put Ia/ R² for T into equation 1 to give;

mg - Ia/R² = ma

Ia/R² + ma = mg

a( I/R² + m) = mg

a = mg/(I/R² +m)

Now putting mg/(I/R² +m) for a in eq 1 gives;

mg - T = m(mg/(I/R² +m))

T = mg - m(mg/(I/R² +m))

T = mg - m²g/(I/R² + m)