In the following figure, the cylinder and the pully turn without friction about

Question

In the following figure, the cylinder and the pully turn without friction about stationary horizontal axles that pass through their center. A light rope is wrapped about the cylinder, pass over the pulley, and has a 3.00kp box suspended from its free end. There is no sloipping between the rope and the rope and the pulley surface. The box accelerates down a ramp is mew(k) = 0.100. The uniform cylinder has a mass 5.00kg and radius 40.0cm. The pulley is a uniform disk with mass 2.00kg and radius 20.0cm. The box is released from rest and slides down the ramp as the rope unwraps from the cylinder. Use energy methods to calculate the angular velocity of the pulley when the box has slid 2.50m along the ramp. You may use the fact that the normal force is mgcos(theta) when on a ramp.

Pleas setup with diagram, and show how you solved it please. I will give you full points!!

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) =22.61 (m/s)^2
v = 4.75 m/s

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