In the figure below, the hanging object has a mass of m 1 = 0.445 kg; the slidin
ID: 2032024 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.445 kg; the sliding block has a mass of m2 = 0.805 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R1 = 0.020 0 m, and an outer radius of R2 = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is ?k = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of vi = 0.820 m/s toward the pulley when it passes a reference point on the table.
Explanation / Answer
h = x = 0.7 m
work done by gravitational force Wg = m1*g*h
work done by frictioanl force Wf = -uk*m2*g*x
moment of inertia of pulley I = (1/2)*M*(R1^2 + R2^2) = (1/2)*0.445*(0.03^2+0.02^2) = 2.895*10^-4 kg m^2
initial kinetic energy of system = Ki = (1/2)*m1*vi^2 + (1/2)*m2*vi^2 + (1/2)*I*wi^2
wi = vi/R2
initial kinetic energy of system = Ki = (1/2)*(m1+m2)*vi^2 + (1/2)*I*(vi/R2)^2
final kinetic energy of system = Kf = (1/2)*(m1+m2)*vf^2 + (1/2)*I*(vf/R2)^2
work energy relation
work done = change in KE
Wg + Wf = Kf - Ki
m1*g*h - uk*m2*g*x = (1/2)*(m1+m2)*(vf^2-vi^2) + (1/2)*I*(vf^2-vi^2)/R2^2
0.445*9.8*0.7 - 0.25*0.805*9.8*0.7 = (1/2)*(0.445+0.805)*(vf^2-0.82^2) + (1/2)*2.8925*10^-4*(vf^2-0.82^2)/0.03^2
vf = 1.67 m/s
(b)
angular speed wf = vf/R2 = 1.67/0.03 = 55.67 rad/s