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March 28, 2017

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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 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
so
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?

  • To DAMON.. (RE: Phys question u solved yesterday) - ,

    becuz when i equate them

    i ended up getting

    (mg - T) / m = (Torque r) I

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