A frictionless pulley has the shape of a uniform solid disk of mass 2.20kg and r

Question

A frictionless pulley has the shape of a uniform solid disk of mass 2.20kg and radius 10 cm. A 1.90kg stone is attached to a very light wire that is wrapped around the rim of the pulley (the figure (Figure 1) ), and the system is released from rest.

Part A) How far must the stone fall so that the pulley has 5.50J of kinetic energy?

Part B) What percent of the total kinetic energy does the pulley have?

Exercise 9.47 Part A A frictionless pulley has the shape of a uniform solid disk of mass 2.20kg and radius 10 cm. A 1.90kg stone is attached to a very light wire that is wrapped around the rim of the pulley (the figure (Figure 1)), and the system is released from rest. How far must the stone fall so that the pulley has 5.50J of kinetic energy? Submit My Answers Give Up Incorrect, Try Again, 9 attempts remaining Figure 1 of 1 Part B What percent of the total kinetic energy does the pulley have? pulley tot Submit My Answers Give Up stone

Explanation / Answer

Given that

A frictionless pulley has the shape of a uniform solid disk of mass (M) = 2.20kg

The radius of the disk is (R) = 10 cm.

A stone of mass is attached to a very light wire (m) =1.90kg

We know that the moment of inertia of the disk is (I disk)= (1/2)MR2

Now the kinetic energy aquired by the stone is given by

KE = (1/2) I w2
= (1/2) (1/2)MR2 * w2
= (1/4) Mv2 ( Since v =Rw)
5.50 = (1/4) * 2.20 * v2
v2 =10 ====> v =3.162m/s

The distance moved by the stone fall so that the pulley has 5.50J of kinetic energy is given by

Here the loss in potential energy in stone = gain in kinetic energy of the stone +pulley kinetic energy

mgh =(1/2)mv2 +5.50J

1.90*9.81*(h) =0.5*1.90*(3.162m/s)2+5.50

1.90*9.81*(h) =9.498+5.50

1.90*9.81*h =14.998

Then h =0.8046m

B)

The percent of the total kinetic energy does the pulley have is

=[KE of pulley/(KE of pulley+KE of stone)]*100

= [5.50 / (5.50+ 9.498)]*100

= 36.67%.

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