In the figure below, the hanging object has a mass of m 1 = 0.370 kg; the slidin
ID: 1446619 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.370 kg; the sliding block has a mass of m2 = 0.810 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.
(b) Find the angular speed of the pulley at the same moment
Explanation / Answer
here,
force of kinetic friction against movement of block = (0.250)(0.810)(9.81) = 1.987 N
I of pulley = 1/2M(R1^2 + R2^2)
I = (0.5)(0.350)(0.020^2+0.030^2)
I = 2.275 * 10^-4 kgm^2
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 = 1/2Iw^2 = (0.5)(2.275 * 10^-4)(27.333)^2 = 850 * 10^-4 J
at ref point m2's KE = 1/2(m2)(0.820)^2 = (0.5)(0.810)(0.820)^2 = 0.2723 J
at ref point m1's KE = 1/2(m1)(0.820)^2 = (0.5)(0.370)(0.820)^2 = 0.1244 J
at ref point the system's total KE = 0.0850 + 0.2723 + 0.1244 = 0.4817 J
system's KE is split:
0.0850/0.4817 = 17.6 %
0.2723/0.4817 = 55 %
0.1244/0.4817 = 25.8 % {m1's motion}
at ref point m1's GPE = (m1)gh = (0.370)(9.81)(0.700) = 2.538 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 = 2.538 (0.700) = 1.777 J.
ie the system NET KE gain = (2.538 - 1.777) = 0.76 J
after m1 & m2 move a total distance = 0.700 m the system's total KE = 0.4817 + 0.76 = 1.243 J
system's KE is split:
pulley's motion = (0.176)(1.243) = 0.2187 J
m2's motion = (0.55)(1.243) = 0.684 J
m1's motion = (0.258)(1.243) = 0.321 J
after m1 & m2 move from ref point a distance = 0.700 m:
(a)
speed of m2 = sqrt[2KE/m2] = sqrt[2(0.684)/0.810] = 1.3 m/s
(b)
angular speed of pulley , w = sqrt[2KE/I] = sqrt[2(0.2187)/2.275*10^-4]
w = 43.85 rad/s