In Fig. the cylinder and pulley turn without friction about stationary horizonta
ID: 1489199 • Letter: I
Question
In Fig. 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
Use law o conservation of energy
PEi + KEi = PEf + KEf
Since vi=0m/s and hf=0m KEi = PEf = 0 J
PEi = KEf
mbgh = ½*mbvb2 + ½*Ip*wp2 + ½*Ic*wc2
Ip =1/2*mp*rp2 , Ic =1/2*mc*rc2 , wp= vb/rp and wc= vb/rc
Plugging these in above equation and sinpmlifying
mbgh = ½*[mb+1/2mp+1/2mc]*vb2
vb2 = [2(3*9.8*2.50)]/[3*0.5*2 + 0.5*5] = 26.72 m^2/s^2
vb = 5.17 m/s