4. Question Details A m1 = 13.0 kg object and a m2 = 11.0 kg object are suspende

Question

4. Question Details A m1 = 13.0 kg object and a m2 = 11.0 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

given,

m1 = 13 kg

m2 = 11 kg

radius of pully = 10 cm

mass of pully = 3 kg

masses displaced by = 3 m

moment of inertia = 0.5 * mass * radius^2

moment of inertia = 0.5 * 3 * 0.1^2

moment of inertia = 0.015 kg m^2

initial energy of the system = m1 * gh

initial energy of the system = 13 * 9.8 * 3

final energy of the system = m1 * gh/2 + m2 * gh/2 + 0.5 * (m1 + m2) * v^2 + 0.5 * moment of inertia * (v/r)^2

final energy of the system = 13 * 9.8 * 3/2 + 11 * 9.8 * 3/2 + 0.5 * (13 + 11) * v^2 + 0.5 * 0.015 * (v/0.1)^2

by conservation of energy

initial energy = final energy

13 * 9.8 * 3 = 13 * 9.8 * 3/2 + 11 * 9.8 * 3/2 + 0.5 * (13 + 11) * v^2 + 0.5 * 0.015 * (v/0.1)^2

v = 1.518 m/s

speed of the two objects = 1.518 m/s

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