A light, flexible rope is wrapped several times around a hollow cylinder with a weight of 43.0 N and a radius of 0.25 m, that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligible moment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force P for a distance of 7.00 m, at which point the end of the rope is moving at 6.00 m/s. If the rope does not slip on the cylinder, what is the value of P?
2 Answers

KE of rolling drum = 1/2(I)(Ω)²
{where I = mR², Ω = (V/R)(2π)}
V = 6.00 m/s
R = 0.25 m
m = 43.0/g = 43.0/9.81 = 4.383 kg
I = (4.383)(0.25)² = 0.274 kgm/s
Ω = (6.00/0.25)(2π) = 150.8 rad/s
KE of rolling drum = (0.5)(0.274)(150.8)² = 3115.5 J
Work done in pulling rope off of drum = 3115.5 J = P x 7.00
p = 3115.5/7.00 = 445 N ANS

K=1/2Iw^2=Pd
w=V/R I=MR^2
P=(1/2Mv^2)/d
P=[1/2*(43/g)*6^2]/7