In the following figure ( Figure 1 ) , the cylinder and pulley turn without fric

Question

In the following figure (Figure 1) , the cylinder and pulley turn without friction about stationary horizontal axles that pass through their centers. A light rope is wrapped around the cylinder, passes over the pulley, and has a 3.00-kg box suspended from its free end. There is no slipping between the rope and the pulley surface. The uniform cylinder has mass 5.00 kg and radius 40.0 cm. The pulley is a uniform disk with mass 2.00 kg and radius 20.0 cm. The box is released from rest and descends as the rope unwraps from the cylinder. Find the speed of the box when it has fallen 2.50

m.

Explanation / Answer

Law of conservation of energy: Uin + Ekin = Ufi + Ekfi.
Uin = Mb g h = 3*9.8*2.5 = 73.5 J
Ekin = 0
Ufi = 0
Ekfi = (1/2)(Mb v^2 + Ic wc^2 + Ip wp^2)
wc = v/Rc ; wp = v/Rp
The moment of inertia of the cylinder Ic = (1/2)Mc Rc^2
the moment of inertia of the pulley Ip = (1/2)Mp Rp^2
then Ekfi =(1/2)(Mb +Mc/2 +Mp/2) v^2
So v^2 = 2*73.5/(3+2.5+1) =22.615 (m/s)^2
v = 4.755m/s

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