Use conservation of energy to determine the angular speed of the spool shown in the figure below after the 3.00 kg bucket has fallen 4.60 m, starting from rest. The light string attached to the bucket is wrapped around the spool and does not slip as it unwinds.
(rad/s)
http://i199.photobucket.com/albums/a...0116/p8-36.gif

Are you familiar with the formula for kinetic energy of (a) a falling mass and (b) a spinning wheel? Add them together and set equal to the change in potential energy as the bucket falls. Post back with your attempt and we'll check it for you.

When I tried the question I got the following formula by adding the two equations together:
mgh=(1/2)(m)(v^2)+(1/2)(I)(w^2)

since v=(w)(r) I substituted this in for v^2

mgh=(1/2)(m)(wr)^2+(1/2)(I)(w^2)

Right here I would substitute I in for the moment of inertia for the spool but I am not a hundred percent sure what it would be, would it be (m)(r^2)?

Then as well I get confused as to which mass is used for the equation, the mass of the spool or the mass of the bucket, or do you use both?

Quote:

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Originally Posted by Echo901

When I tried the question I got the following formula by adding the two equations together:
mgh=(1/2)(m)(v^2)+(1/2)(I)(w^2)

since v=(w)(r) I substituted this in for v^2

mgh=(1/2)(m)(wr)^2+(1/2)(I)(w^2)

Right here I would substitute I in for the moment of inertia for the spool but I am not a hundred percent sure what it would be, would it be (m)(r^2)?

Then as well I get confused as to which mass is used for the equation, the mass of the spool or the mass of the bucket, or do you use both?

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Thanks for your help, got the answer I was looking for!