A mass of 12 kg is ties to a string that is wound around a large cylinder of rad

Question

A mass of 12 kg is ties to a string that is wound around a large cylinder of radius .5 m such that the cylinder turns as the mass falls. The mass is released 4 m above the ground. Just before it hits the ground the mass has a speed of 5 m/s. What is the angular velocity of the cylinder just before the mass hits the ground? What is the moment of inertia of the cylinder? (use conservation of energy) What is the angular momentum of the cylinder just before the mass hits the ground? The apparatus is disassembled and the mass of the cylinder measured to be 51.3 kg. Is the cylinder below, or more likely to be solid?

Explanation / Answer

a) speed of rope just before hitting is 5 m/s .

for cylinder,

v = w*r

5 = w * 0.5

w = 10 rad/s

b) using work-energy theorem,

work done by gravity = change in KE

mgh = mv^2 /2 + Iw^2 /2

12*9.81*4 = (12 * 5^2 /2 ) + ( I * 10^2 / 2)

I = 6.42 kg m^2

c) angular momentum = Iw = 6.42 * 10 = 64.2 kg m^2 /s

d) if cylinder is hollow then I = mr^2 = 51.3 * 0.5^2 = 12.82

so its not hollow.

if it is solid then

I = m r^2 /2 = 12.82/2 = 6.41 kg m^2

so this cylinder is solid.

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