Problem solving...?

Question

In the cylinder with a radius R=0.2m with an inertia moment I=0.1kgm^2 it is rolled a cord in the bottom of which it is hanged up a solid with a weight 0.5 kg. Before that the cylinder started rolling, the solid was standing in the height h=1m from the floor. Find:
a) The time in which the solid will reach the floor.
b) The kinetic energy in the moment when the solid touches the floor.
c) The tension (T) of the cord. ( Do not include in the result the frictional forces)
Answers: t=1s, Kinetic energy=0.81J, T=4.1N

Answer 1

Sorry, I'm not sure of the setup. I'll make some guesses. The ends of the cylinder sit on 2 tables. The cord with the weight hangs down between the tables. So that makes the cylinder roll along the table while the cord unwinds.

At the start, the weight (in English Physics classes, it would be called a mass) posses potential energy of
m*g*h = 0.5 kg*(9.8 m/s^2)*1 m

At the end (just before touch-down) the original energy of the system will be shared between the KE of the mass, the translational KE of the cylinder, and the rotational KE of the cylinder. Hmmm, I just noticed that the mass of the cylinder is not given. That would be required to know the translational KE of the cylinder. If I assume it's a solid cylinder
(I=(M*R^2)/2) then M=10 kg.

You can write equations to express the relationship between the vertical velocity of the mass, the horizontal velocity of the cylinder and the rotational velocity of the cylinder. (Remember when calculating the KEs, that the mass will have a horizontal velocity and vertical velocity. Determine the resultant velocity.) Then set the sum of the 3 KEs equal to the original potential energy.

The discussion above should help you get the answer to b). a) since you now know the mass's vertical velocity, you can use
y = 1 m = ((Vo + V) / 2)*t
c) You can use
V = Vo + a*t
to get the acceleration.
Then
Fnet = m*a
to get the net force that caused that acceleration. And then
Fnet = m*g - T
hope that helps