To DAMON.. (RE: Phys question u solved yesterday)
posted by Stacu on .
YOU DID
The tension in the string is the force exerted down by the falling disk and up by whatever is holding the disk. They are equal and opposite. If the disk is a yoyo, that force is exerted by your finger on the string to give the tension.
Force up on disk = T
Force down on disk = m g
a = acceleration down = (mg T)/m
Torque on disk = T r
Torque = I alpha
so
alpha = T r/I
but alpha = a/r by geometry
T r/I = (m g  T)/(m r)
m T r^2 = (m gT) I
T (m r^2 + I) = m g I
but I = .5 m r^2
T (1.5 m r^2) = .5 m^2 g r^2
T = (.5/1.5) m g = (1/3) m g
now you should be able to do the rest
for the energy part
KE = .5 m v^2 + .5 I omega^2
but omega = v/r
_____________
for the part where u have
T r/I = (m g  T)/(m r)..
where did that come from?? (the r in "mr"?
I get that a = mg  T / m
but the r?

becuz when i equate them
i ended up getting
(mg  T) / m = (Torque r) I