- #1
thenewbosco
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I have a diagram for this at http://snipurl.com/diagr
There is a uniform disk with radius R, mass M, it has a string wrapped around it and is attached to a fixed rod.
Part 1: Show the tension in the string is 1/3 the weight of the disk
what i have done for this is
Torque=Fd
and setting the force equal to Tension
Torque=TR
then \sum Torque=I\frac{a}{R} and i used I=\frac{1}{2}MR^2for the moment of inertia
Then TR=\frac{1}{2}MRa
T=\frac{1}{2}Ma
and then acceleration is just g, so i get the tension equals 1/2 the weight. where have i gone wrong?
There is a uniform disk with radius R, mass M, it has a string wrapped around it and is attached to a fixed rod.
Part 1: Show the tension in the string is 1/3 the weight of the disk
what i have done for this is
Torque=Fd
and setting the force equal to Tension
Torque=TR
then \sum Torque=I\frac{a}{R} and i used I=\frac{1}{2}MR^2for the moment of inertia
Then TR=\frac{1}{2}MRa
T=\frac{1}{2}Ma
and then acceleration is just g, so i get the tension equals 1/2 the weight. where have i gone wrong?
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