A frictionless pulley has the shape of a uniform solid disk of mass 2.60 kg and
ID: 1491356 • Letter: A
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%