In the figure below, the hanging object has a mass of m_1 = 0.435 Kg; the slidin

Question

In the figure below, the hanging object has a mass of m_1 = 0.435 Kg; the sliding block has a mass of m_2 = 0.875 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 Kg, an inner radius of R_1 = 0.020 0 m, and an outer radius of R_2 = 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 jA = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and doest not slip on the pulley. The block has a velocity of V_1 = 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 second point, 0.700 m away. (b) Find the angular speed of the pulley at the same moment.

Explanation / Answer

force of kinetic friction against movement of block = (0.250)(0.875)(9.81) = 2.14 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)(0.860)(0.820)² = 0.2891 J
at ref point m1's KE = 1/2(m1)(0.820)² = (0.5)(0.360)(0.820)² = 0.1210 J
at ref point the system's total KE = 0.0850 + 0.2891 + 0.1210 = 0.4951 J
system's KE is split:
0.0850/0.4951 = 17.2% {pulley's motion}
0.2891/0.4951 = 58.4% {m2's motion}
0.1210/0.4951 = 24.4% {m1's motion}

at ref point m1's GPE = (m1)gh = (0.435)(9.81)(0.700) = 2.987J

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 = 2.14(0.700) = 1.498 J.
ie the system NET KE gain = ( 2.987 - 1.498) = 1.489 J
after m1 & m2 move a total distance = 0.700 m the system's total KE = 0.4951 + 1.489 = 1.9841 J

system's KE is split:
pulley's motion = (0.172)(1.9841) = 0.0866 J
m2's motion = (0.584)(1.9841) = 0.2943 J
m1's motion = (0.244)(1.9841) = 0.1229 J

after m1 & m2 move from ref point a distance = 0.700 m:
speed of m2 = [2KE/m2] = [2( 0.2943)/0.875] = 0.820 m/s ANS (a)
w of pulley = [2KE/I] = [2(0.0866)/2.275E-4] = 0.2233846E4 = 27.59 rad/s ANS (b)

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