Question

 A frictionless pulley has the shape of a uniform solid disk of mass 2.40 kg and radius 10 cm. A 1.20 kg stone is attached to a very light wire that is wrapped around the rim of the pulley (the figure ), and the system is released from rest.

 Part A

 How far must the stone fall so that the pulley has 4.80 J of kinetic energy?

 Part B

 What percent of the total kinetic energy does the pulley have?

Answer 1

part A) by applying conservation of energy we have

P.E=K.E

mgh=1/2mv²+4.8............1)

we have rotational kinetic energy is

k.e=1/2I²................2)

=v²/R² and I=mR²/2

now eqn 2) becomes

I=1/4*m*v²

4.8=1/4*2.4*v²

v²=8

so now using eqn 1)

2.4*9.8*h=1/2*1.2*8+4.8

h=0.41m

so the answer is 0.41 m or 0.408 m

part B) so

Kp=4.80 J

Ktotal=1/2*mv²+4.8=0.5*1.2*8+4.8=9.6 J

kp/kt=4.80/9.60 *100=50%

so the answer is 50 or 50.0