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

Answer:

T = mg - (m²g/(I/R² + m))

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)