In the figure below, the hanging object has a mass of m_1 = 0.480 kg; the slidin
ID: 1503459 • Letter: I
Question
In the figure below, the hanging object has a mass of m_1 = 0.480 kg; the sliding block has a mass of m_2 = 0.860 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 mu 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 v_i = 0.820 m/s toward the pulley when It passes a reference point on the table. Use energy methods to predict its speed after it has moved to a second point, 0.700 m away. Find the angular speed of the pulley at the same moment.Explanation / 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)(0.860)(0.820)² = 0.2891 J
at ref point m1's KE = 1/2(m1)(0.820)² = (0.5)(0.480)(0.820)² = 0.1613 J
at ref point the system's total KE = 0.0850 + 0.2891 + 0.1613 = 0.5354 J
system's KE is split:
0.0850/0.5354 = 15.8% {pulley's motion}
0.2891/0.5354 = 54% {m2's motion}
0.1613/0.5354 = 30.12% {m1's motion}
at ref point m1's GPE = (m1)gh = (0.480)(9.81)(0.700) = 3.2961 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 = 2.190(0.700) = 1.490 J.
ie the system NET KE gain = (3.2961 - 1.490) = 1.80616 J
after m1 & m2 move a total distance = 0.700 m the system's total KE = 0.5354 + 1.80616 = 2.34156 J
system's KE is split:
pulley's motion = (0.158)( 2.34156) = 0.36996 J
m2's motion = (0.54)( 2.34156) = 1.2644 J
m1's motion = (0.301)( 2.34156) = 0.7048 J
after m1 & m2 move from ref point a distance = 0.700 m:
speed of m2 = [2KE/m2] = [2(1.2644)/0.860] = 0.7687 m/s ANS for (a)
w of pulley = [2KE/I] = [2(0.36996)/2.275E-4] = 57.029 rad/s ANS for (b)