In the figure below, the hanging object has a mass of m1 = 0.360 kg; the sliding
ID: 1684521 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.360 kg; the sliding block has a mass of m2 = 0.795 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 µk = 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. (m/s) (b) Find the angular speed of the pulley at the same moment. (rad/s) On this one please show all your steps, I'm trying to figure out what I'm doing wrong. I'm trying to calculate net torque using m2 and m1. The numbers I get seem good, but it's a no go.Explanation / Answer
m1 = 0.360 kg, m2 = 0.795 kg, M = 0.350 kg, R1 = 0.0200 m, R2 = 0.0300 m. µk = 0.250, vi = 0.820 m/s, moment of inertia I = M(R1^2 + R2^2)/2 = 2.275*10^-4 kgm^2 angular speed = w (a) Use energy methods to predict its speed vf after it has moved to a second point, d = 0.700 m away. (m/s) initial energy = m1^vi^2/2 + m2^vi^2/2 + I*wi^/2 = (m1 + m2)*vi^2/2 + I*(vi/R2)^2/2 = (m1 + m2 + I/R2^2)*vi^2/2 final energy = (m1 + m2 + I/R2^2)*vf^2/2 - m1*g*d work done by friction W = -uk*m2*gd so -uk*m2*gd = (m1 + m2 + I/R2^2)*vf^2/2 - m1*g*d - (m1 + m2 + I/R2^2)*vi^2/2 (m1 + m2 + I/R2^2)*vf^2 = (m1 + m2 + I/R2^2)*vi^2 + 2(m1 - uk*m2)*gd vf = sqrt[vi^2 + 2(m1 - uk*m2)*gd/(m1 + m2 + I/R2^2)] = 1.50 m/s (b) Find the angular speed wf of the pulley at the same moment. (rad/s) wf = vf/R2 = 49.9 rad/s