In the figure below, the hanging object has a mass of m1 = 0.495 kg; the sliding
ID: 1449854 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.495 kg; the sliding block has a mass of m2 = 0.780 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
I = moment of inertia of pulley = (0.5) M (R12 + R22) = (0.5) (0.35) ((0.02)2 + (0.03)2) = 0.0002275
f = frictional force acting = k m2 g = 0.25 x 0.78 x 9.8 = 1.91 N
Vi = initial speed = 0.820 m/s
Wi = initial angular speed for pulley = Vi/R2 = 0.82 / 0.03 = 27.33 rad/s
d = distance travelled
using conservation of energy
m1gd + (0.5) m1 Vi2 + (0.5) m2 Vi2 + (0.5) I Wi2 = (0.5) m1 Vf2 + (0.5) m2 Vf2 + (0.5) I Wf2 + fd
(0.495) (9.8) (0.7) + (0.5) (0.495) (0.82)2 + (0.5) (0.78) (0.82)2 + (0.5) (0.0002275) (27.33)2 = (0.5) (0.495) Vf2 + (0.5) (0.78) Vf2 + (0.5) (0.0002275) (Vf/0.03)2 + (1.91) (0.7)
Vf = 1.84 m/s
b)
Wf = Vf/R2 = 1.84 / 0.03 = 61.33 rad/s