An inertia physics problems?

Question

A cable is wrapped around a secured pulley in the shape of a solid cylinder (1/2mr^2) with rotational inertia I= 0.250 kg-m^2 and radius R = 0.500 m. The cylinder rotates without friction. The free of the end of the cable is connected to a vertically hanging block of mass m = 1.0 kg

a.) What is the magnitude a of the block's downward acceleration?
b.) what is the tension T in the cable?

c.) Suppose the blocks drops from rest through a vertical distance h = 2.0 m. What is the speed v at the speed of this motion?

Answer 1

One has to draw the free body diagram for the hanging mass and the rotating cylinder. Note that the tension, T in the string acts downward on the cylinder of Moment of Inertia I an radius r at its rim that is 'r' distance away from the axis. Also that on the mass the tension T acts upwards. It has been given that frictional force is not there so it need not be shown. So we get

Tr = I x (alpha); alpha = angular acceleration
mg - T = ma; g = acceleration due to gravity and
a = acceleration of the hung mass

we know that alpha = a/r

eliminating T we get a = [g/(1 + I/mr^2)]
a = 9.8/(1+ 0.250/0.25) = 4.9 m/s^2

T = Ia/r^2 = 0.250x4.9/(.5^2) = 4.9 N

v after dropping 2.0 m = SQRT(2x4.9x2) = 4.43 m/s, using third kinematic formula for motion with constant acceleration.

Answer 2

that is not that straightforward, Jayman. it quite is merely real because of the fact the centre of gravity of the hollow sphere is comparable to that of the forged sphere, and IF the balls have a similar diameter. began with a similar kinetic enegy, they are going to finally end up with a similar skill power , i.e centres of gravity on a similar height. So (d) is the right answer.