Two masses, ma=32.0 kg and mb=38.0kg, are connected by a rope that hangs over a
ID: 1287571 • Letter: T
Question
Two masses, ma=32.0 kg and mb=38.0kg, are connected by a rope that hangs over a pulley.The pulley is a uniform cylinder of radius 0.311m and mass 3.1kg . Initially ma is on the ground and mb rests 2.5m above the ground. If the system is released, use conservation of energy to determine the speed of mb just before it strikes the ground. Assume a frictional torque in the pulley bearings of 2.57Nm.
Two masses, ma=32.0 kg and mb=38.0kg, are connected by a rope that hangs over a pulley.The pulley is a uniform cylinder of radius 0.311m and mass 3.1kg . Initially ma is on the ground and mb rests 2.5m above the ground. If the system is released, use conservation of energy to determine the speed of mb just before it strikes the ground. Assume a frictional torque in the pulley bearings of 2.57Nm.Explanation / Answer
The initial Potential energy of m2 equals the final potential energy of m1 plus the final kinetic energies of m1, m2 and the rotating pulley;
m2gh = m1gh + (1/2)m1v^2 + (1/2)m2v^2 + (1/2)Iw^2
The moment of inertia of a cylinder of radius R is;
I = (1/2)MR^2
The angular velocity "w" of the pulley is related to the linear velocity "v" of the rim, which is the same as the masses, as;
w = v/R
m2gh = m1gh + (1/2)m1v^2 + (1/2)m2v^2 + (1/4)Mv^2
Now just solve this for "v".
You might want to check the mass' because you said m1 was the same as M, which might be ok but seems a little unusual.
Using the masses given ,( and h=2.1),I get;