In the figure below, the hanging object has a mass of 1 0.425 kg; the sliding bl

Question

In the figure below, the hanging object has a mass of 1 0.425 kg; the sliding block has a mass of m2 0.860 kg; and the pulley s hollow cylinder with ? mass of M-0.350 kg, an inner radius of R1-0.020 0 m, and an outer edi s of R2 = 0.030 0 m. Assume the mass of the spokes is negligible. netic friction between the back and the horizontal surface ise 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 0.820 m/s toward the pulley when it passes a reference point on the table The coefficient of ki (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. red/s

Explanation / Answer

h = x = 0.7 m

work done by gravitational force Wg = m1*g*h

work done by frictioanl force Wf = -uk*m2*g*x

moment of inertia of pulley I = (1/2)*M*(R1^2 + R2^2) = (1/2)*0.35*(0.02^2+0.03^2) = 22.75*10^-5 kg m^2

initial kinetic energy of system = Ki = (1/2)*m1*vi^2 + (1/2)*m2*vi^2 + (1/2)*I*wi^2

wi = vi/R2

initial kinetic energy of system = Ki = (1/2)*(m1+m2)*vi^2 + (1/2)*I*(vi/R2)^2

final kinetic energy of system = Kf = (1/2)*(m1+m2)*vf^2 + (1/2)*I*(vf/R2)^2

work energy relation

work done = change in KE

Wg + Wf = Kf - Ki

m1*g*h - uk*m2*g*x = (1/2)*(m1+m2)*(vf^2-vi^2) + (1/2)*I*(vf^2-vi^2)/R2^2

0.425*9.8*0.7 - 0.25*0.860*9.8*0.7 = (1/2)*(0.425+0.860)*(vf^2-0.82^2) + (1/2)*22.75*10^-5*(vf^2-0.82^2)/0.03^2

vf = 1.6 m/s

(b)

angular speed wf = vf/R2 = 1.6/0.03 = 53.3 rad/s

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