A m1 = 15.0 kg object and a m2 = 10.5 kg object are suspended, joined by a cord

Question

A m1 = 15.0 kg object and a m2 = 10.5 kg object are suspended, joined by a cord that passes over a pulley with a radius of 10.0 cm and a mass of 3.00 kg (Fig. P10.46). The cord has a negligible mass and does not slip on the pulley. The pulley rotates on its axis without friction. The objects start from rest 3.00 m apart. Treating the pulley as a uniform disk,determine the speeds of the two objects as they pass each other. (Hint: There are two ways to do this problem: (1) The easy method is with conservation of energy; (2) The hard method is with torque and angular acceleration. If you choose this more challenging method, note that the tensions on either side of the pulley are NOT the same.)

Explanation / Answer

Let the speed of the masses be v.

Rotational Kinetoc Energy of Pulley = 1/2*I*^2
Rotational Kinetoc Energy of Pulley = 1/2 * 1/2 * M*R^2 * v^2/R^2 = 1/4 * M*v^2

Using Energy Conservation,

Initial Potential Energy + Initial Kinetic Energy  = Final Kinetic Energy + Final Potential Energy + Rotational Kinetoc Energy of Pulley

m1*g*h + 0 = 1/2*m1*v^2 + 1/2*m2*v^2 + m1*g*h/2 + m2*g*h/2 + 1/4 * M*v^2
Substituing Values,
15.0 * 9.8 * 3.0 = 1/2 * 15.0 * v^2 + 1/2 * 10.5 * v^2 + 15.0*9.8*1.5 + 10.5*9.8 * 1.5 + 1/4 * 3.0 * v^2
Solvong for v
v = 2.21 m/s

Speed as they pass each other, v = 2.21 m/s

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