A frictionless pulley has the shape of a uniform solid disk of mass 2.60 kg and

Question

A frictionless pulley has the shape of a uniform solid disk of mass 2.60 kg and radius 10 cm . A 1.40 kg 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 3.60 J of kinetic energy?

Part B

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

Please help since I am having difficulty answering and comprehending this question. Many Thanks in advance

Explanation / Answer

from the conservation of the energy of the system

E = E0

( mstone v2 ) / 2 + Kpulley = mstone g h

where the speed of the stone is

v2 = 2 a h

then

( mstone 2 a h ) / 2 + Kpulley = mstone g h

mstone a h + Kpulley = mstone g h

so

h = Kpulley / ( mstone g - mstone a )

* The force on the stone are

T - mstone g = - mstone a --->   T = mstone g - mstone a (1)

* The net torque on the pulley is

= R T = I ---> T = [ I ] / R

where the moment of inertiaof a solid disk is

I = ( mpulley R2 ) / 2

then

T = [ ( mpulley R2 ) / 2 ) * ] / R

T = ( mpulley R ) / 2

T = ( mpulley a ) / 2 (2)

combining equation (1) and (2) the accerelation of the system is

( mpulley a ) / 2 = mstone g - mstone a

mpulley a = 2 mstone g - 2 mstone a

a = ( 2 mstone g ) / ( mpulley + 2 mstone )

a = ( 2 * 1.40 kg * 9.81 m/s2 ) / ( 2.60 kg + 2 * 1.40 kg )

a = 5.09 m/s2

Hence

h = Kpulley / ( mstone g - mstone a )

h = ( 3.60 J ) / ( 1.40 kg * 9.81 m/s2 - 1.40 kg * 5.09 m/s2 )

h = 0.545 m

the kinetic energy of the system is

K = Kstone + Kpulley

K = ( mstone v2 ) / 2 + Kpulley

K = mstone a h + Kpulley

K = 1.40 kg * 5.09 m/s2 * 0.545 m + 3.60 J

K = 7.42 J

the percentage is ( Kpulley / K ) * 100 = ( 3.60 J / 7.42 J ) * 100 = 48.5%

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