In the figure below, the hanging object has a mass of m_1 = 0.365 kg; the slidin
ID: 1449549 • Letter: I
Question
In the figure below, the hanging object has a mass of m_1 = 0.365 kg; the sliding block has a mass of m_2 = 0.840 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 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. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s Find the angular speed of the pulley at the same moment.Explanation / Answer
a) friction force = uk N = uk m2 g
moment of inertia of pulley = M (R1^2 + R2^2) /2
= 0.350 (0.02^2 + 0.03^2) /2 = 2.275 x 10^-4 kg m^2
Using work eenergy theorem,
work done by gravity + work done by friction = change in KE
(m1 g d) + (- uk m2 g d ) = [ (m1 vf^2 /2 ) + (m2 vf^2 /2 ) + (I wf^2 )/2 ] - [ (m1 v^2 /2 ) + (m2 v^2 /2 ) + (I w^2 )/2 ]
(0.365 x 9.8 x 0.7 ) + (-0.25 x 0.84 x 9.8 x 0.7) = [ (0.365 + 0.840)vf^2 /2) + (2.275 x 10^-4 (vf / 0.03)^2 /2 )]
- [ (0.365 + 0.840)(0.820^2) /2) + (2.275 x 10^-4 (0.820 / 0.03)^2 /2 )
1.0633 = [ 0.6025 vf^2 + 0.1264vf^2] - [ 0.405 + 0.085]
vf = 1.46 m/s
b) angular speed, w = vf /R2 = 1.46 / 0.03 = 48.66 rad/s