Posted by **Goldberg** on Saturday, May 5, 2012 at 9:07pm.

One end of the string in the diagram at right is wrapped around the cylinder. The other end is tied to the ceiling. The cylinder is released from rest. How fast is it moving after it falls 3.0m?

- Physics 121 -
**Goldberg**, Saturday, May 5, 2012 at 9:08pm
Please show the formula and how you do it. Thx

- Physics 121 -
**bobpursley**, Saturday, May 5, 2012 at 9:13pm
Look at energy:

potential energy lost= kinetic energy gained

mgh=1/2 m vf^2 +1/2 I w^2

mgh=1/2 m vf^2 + 1/2 I (vf/r)^2

Now, you have to decide the moment of inertia for the cylinder. Solid, or not. Lets just assume it is hollow for this demonstration.

I= m r^2

then

mgh= 1/2 m vf^2+1/2 m vf^2

Now, average velocity going down was Vf/2, so in three seconds it went 1.5Vf

h= 1.5Vf

1.5 mg Vf= 1/2 m Vf^2+ 1/2 m Vf^2

you can solve for Vf.

I would divide by mVf, and solve

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