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a pulley on a friction- less axle has the shape of a uniform solid disk of mass 2.50 kg and ra- dius 20.0 cm. a 1.50 kg stone is attached to a very light wire that is wrapped around the rim of the pul- ley (fig. e9.47), and the system is released from rest. (a) how far must the stone fall so that the pulley has 4.50 j of kinetic energy? (b) what percent of the total kinetic energy does the pulley have?



Answer :

a)The stone fall so that the pulley has 4.50 j of kinetic energy is  0.673 m

b) Percent of the total kinetic energy does the pulley have 45.5%

Given, the mass of the pulley disk mₚ = 2.50 kg

Its radius r = 20.0 cm = 0.2 m

Its moment of inertia I = (1/2)mₚr²

                               I = (1/2)(2.50 kg)(0.2 m)²

                               I = 0.05 kgm²

Given, the kinetic energy of the pulley disk = 4.50 J

If w the angular velocity of the pulley,

                              (1/2)Iw² = 4.50 J

                              (1/2)(0.05 kgm²)w²= 4.50 J

                              w² = 9/0.05 rad²/s²

                              w² =180 rad²/s²

                               v² = 180(0.2)^2 m²/s²

                              v² = 7.2 m²/s²

Given, the mass of the stone mₛ = 1.50 kg

Its kinetic energy = (1/2)(1.50 kg)(7.2 m²/s²) = 5.4 J

Potential energy loss of the stone = (1.50 kg)(9.80 m/s²)h

(a) Therefore, from equation (1), w get

     (1.50 kg)(9.80 m/s²)h = 4.5 J + 5.4 J

                                    h = (9.9 J)/(1.50 kg)(9.80 m/s²)

                                    h = 0.673 m

So, the. stone. mus.t fall by a distan.ce of  0.673 m.

(b) Total kinetic. energy of the. system. = 9.9 J

The kinetic energy. of the pulley = 4.50 J

The pulley’s fraction. of kinetic energy. = 4.50 J/9.9 J

                                                          = 0.455 = 45.5%

Therefore, the pulley has 45.5% of the total kinetic energy.

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