In the figure below, the hanging object has a mass of m1 = 0.355 kg; the sliding
ID: 1442510 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.355 kg; the sliding block has a mass of m2 = 0.795 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. (a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away. m/s (b) Find the angular speed of the pulley at the same moment. rad/sExplanation / Answer
force of kinetic friction against movement of block = (0.250)(0.860)(9.81) = 2.109 N
I of pulley = 1/2M(R1² + R2²) = (0.5)(0.350)(0.020²+0.030²) = 2.275E-4 kgm²
at ref point pulley's w = V/R = 0.820/R2 = 0.820/0.030 = 27.333 rad/s
at ref point pulley's angular KE = 1/2Iw² = (0.5)(2.275E-4)(27.333)² = 850E-4 J
at ref point m2's KE = 1/2(m2)(0.820)² = (0.5)(0.795)(0.820)² = 0.2672 J
at ref point m1's KE = 1/2(m1)(0.820)² = (0.5)(0.355)(0.820)² = 0.1193 J
at ref point the system's total KE = 0.0850 + 0.2672 + 0.1193 = 0.47155 J
system's KE is split:
0.0850/0.47155 = 18.02% {pulley's motion}
0.1193/0.47155 = 25.299% {m1's motion}
0.2672/0.47155 = 56.664% {m2's motion}
at ref point m1's GPE = (m1)gh = (0.3550)(9.81)(0.700) = 2.437 J
solution reasoning:
after m2 moves 0.700 m from ref point, the system will gain in KE the GPE of m1, but will lose the energy of the frictional work done = 0.25*0.795*9.81(0.700) = 1.364 J.
ie the system NET KE gain = (2.437 - 1.364) = 1.072 J
after m1 & m2 move a total distance = 0.700 m the system's total KE = 0.47155 + 1.072 = 1.543 J
system's KE is split:
pulley's motion = (0.18)(1.543) = 0.2778 J
m2's motion = (0.566)(1.543) = 0.8733 J
m1's motion = (0.25)(1.543) = 0.385 J
after m1 & m2 move from ref point a distance = 0.700 m:
speed of m2 = [2KE/m2] = [2(0.8733)/0.795] = 1.48 m/s ANS (a)
w of pulley = [2KE/I] = [2(0.2778)/2.275E-4] = = 0.4943E2 = 49.4 rad/s ANS (b)