Energy in rotational motion
Hi everyone,
I have come across a problem which I initially thought was straightforward but now I find that no matter how I try to solve it I cannot get the correct answer that is shown at the back of my Physics textbook. I just can't see where I am going wrong. PROBLEM
A friction less pulley has a shape of a uniform solid disk of mass 2.50 kg and a radius of 20.0 cm. A 1.50 kg stone is attached to a very light wire that is wrapped around the rim of the pulley, 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 ?
there is a part b) to this problem but I will omit it as it is part a) that I have the problem with.
Initially I found the Moment of Inertia of the pulley using .. I = 1/2 MR(sqrd),
where I is it's Moment of Inertia and M is the mass of the pulley and R is the radius of the disk (I took the disk to be a solid cylinder).
from this I = 0.05 kgm(sqrd)
I then used this value for Moment of inertia to find rotational velocity (w), using ..
K = I w(sqrd)
where K is the Kinetic energy of the pulley (=4.50 J) and w = rotaional velocity.
from this w = 13.4 rad/s
I then found the a value for the rotational acceleration .. a(tan)  r. alpha
where a(tan) is the tangential acceleration of the pulley and alpha the rotational acceleration of the pulley
and noting that a(tan) is equal to the acceleration of the stone under gravity so a(tan) = 9.8 m's(sqrd)
this gives alpha = 49.0 rad/s(sqrd)
Next I used w(sqrd) = w(0)(sqrd + 2 alpha (thetatheta(0))
to find theta (the angle in rads that is completed by the rotating pulley)
and where w(0) is initial angular velocity (= 0, from rest) and theta(0) initial angle value (= 0 )#
from this theta = 18.36 rad
finally I used the equation d = r . theta
where d = distance traveled, r = radius of pulley and theta is the angle of rotation as before.
so I get d= 3.66 m ( 366 cm)
However the answer in the book says d = 67.3 cm ?
Can anyone point out where I have gone wrong KI can't see it ?
Thanks.
Jackthehat
