March 27, 2015

Homework Help: To DAMON.. (RE: Phys question u solved yesterday)

Posted by Stacu on Tuesday, December 16, 2008 at 6:19pm.


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 yo-yo, 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
alpha = T r/I
but alpha = a/r by geometry
T r/I = (m g - T)/(m r)
m T r^2 = (m g-T) 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?

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