Two masses hang on either side of a cylindrical pulley, which is rotating about

Question

Two masses hang on either side of a cylindrical pulley, which is rotating about its center (with

mass M1, radius R and moment of interia of I =1/2M1R^2 about its center). The mass hanging off

the right side of the pulley has mass 2m2 and the mass handing o the left side of the pulley has

mass m2. Initially, the masses and pulley start from rest and the masses are both height h above

the ground. Then the masses begin moving until one of them hits the ground (assume the string

suspending them is massless, unstretchable, and long enough so that one mass can reach the ground

without pulling the other over the pulley). Find an expression for the speed of the mass that hits the ground, just before it hits the

ground. Your expression may include M1, R, m2, g, h and numbers, but nothing else. ) If M1 = 10 kg, R = 0.2 m, m2 = 2 kg, and h = 1 m , what is the nal speed of the falling mass? Include units all the way through your calculations!

Explanation / Answer

FBD of each body is

Assuming equal acceleration for blocks and pulley

Block on right

2m2g - T1 = 2m2a

Block on left

T2 - m2g = -m2a

Torque equation on pulley

(T1 - T2)R = Ia/R = M1Ra/2

T1 - T2 = M1a/2

Adding equations for blocks we get

m2g + T2 - T1 = m2a

Using T1 - T2 = M1a/2

m2g - M1a/2 = m2a

a = m2g/[m2 + M1/2]

Now according to equations of motion

v^2 = u^2 + 2ah

Since, u=0

v^2 = 2m2gh/[m2 + M1/2]

v = (2m2gh/[m2 + M1/2])^.5

Substituting values gives

v = 2.37 m/s

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