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
what are the forces acting on a string that is wounded around a uniform disk of Radius R and mass R? i know tension is in the string...what else?
and I of the disk is 1/2 MR2... so how would you show that the tension in the string is one third the weight of the disk?
and...show that the magnitude of the acceleration of the center of mass is 2g/3
and show that the linear speed v, of the centre of mass is (4gh/3) ^1/2.. how would u derive ur answer for the linear speed by using instead energy conservation principles
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