find angular acceleration?

Question

Imagine a frictionless pulley (a solid cylinder) of unknown mass M and radius r = 0.22 m is used to draw water from a well. A bucket of mass m = 1.46 kg is attached to a cord wrapped around the cylinder. A bucket starts from rest at the top of the well and falls for t = 2.01 s before hitting the water h = 7.48 m below the top of the well. Neglect the mass of the cord.

Find angular acceleration, and tension of the cord

Answer 1

It doesn't say initial or final tension or acceleration, but these turn out to be constant. Assuming them so, the bucket travels distance h under constant accleration a, so:
h = (1/2)a*t^2
a = 2h / t^2

The net force on the bucket is (mg-T), where T is the tension.
mg - T = ma
T = mg - ma = mg - mh / t^2

The angular acceleration dw/dt = d(v/r)/dt = (dv/dt)/r = a/r.

If you don't assume from the problem statement that tension is constant, you can derive it by solving for the unknown mass M instead.

The work done by gravity is mgh, producing a velocity v for the bucket, and an angular velocity w = v/r of the cylinder. So the work creates a combined linear and rotational kinetic energy:
mgh = (1/2)m*v^2 + (1/2)I*w^2 = (1/2)m*v^2 + (1/2)I*(v/r)^2
2mgh = (m + I/r^2)v^2

(w=v/r is angular velocity of the cylinder...imagine that w is the "omega" that I don't know how to type in this editor. :^)

But v=dh/dt, so:
dh/dt = sqrt(h) * sqrt(2mg / (m + I/r^2))
dh / sqrt(h) = dt * sqrt(2mg / (m + I/r^2))

Integrate from (t=0,h=0 to t=t,t=h) and you have an equation where I is the only unknown. Solve for I and then use I to find M. Without grinding this out, however, note that sqrt(h) will be proportional to t, hence h is proportional to t^2 and we had constant acceleration (hence constant tension) all along. The original solution works.