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
Question
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)