Question

In the figure below, the hanging object has a mass of m₁ = 0.400 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 μ_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.

(a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away.
m/s

(b) Find the angular speed of the pulley at the same moment.
rad/s

Answer 1

Solution:

a) The total kinetic energy of the block M2 + mass M1 + Pulley = total work done by the net force .

Initial Kinetic energy of the block M2 = 1/2 M2vi^2 = 1/2 (0.810)(0.82)^2 = 0.272 j

Initial kinetic energy of the mass M1 = 1/2 M1vi^2 = 1/2 ( 0.4) (0.82)^2 = 0.134 j

Total initial kinetic energy = 0.272 + 0.134 = 0.406 j

Final kinetic energy of M2 and M1 = 1/2 (M1+M2)vf^2 = 1/2(0.81 + 0.0.40) vf^2 = 0.605 vf²

Initial kinetic energy of the pulley = 1/2 I ^2 = 1/2 [ M(R1^2 +R2^2)] (vi/ R2)^2 =

1/2 (0.35)(0.02^2 + 0.03^2)](0.82)^2 /(0.03)^2= 0.17 j

Total initial kinetic energy = 0.406 + 0.17 = 0.576 j

Total Final kinetic energy = 0.605 vf^2 + 1/2(0.35)[0.02^2+0.03^2) vf^2 / R2^2 = 1.22 vf^2

potential energy of M1 = M1gh = (0.40)(9.81)(0.70) = 2.744 j

work done by friction = mgd =- (0.250)(0.4)(9.81)(0.70) = -0.686 j (negative since it is frictional force work)

Net work = 2.744 -0.686 = 2.058 j

Change in kinetic energy = net work done

=> 1.22 vf2 - 0.676 = 2.058

=> 1.22 vf^2 = 2.058 + 0.676 = 2.734

=> vf = Final velocity = 1.49 m/s

b) Angular velocity = = vf / R₂ = ( 1.49) / 0.03 = 49.8 rad/s