Calculating Gravitational Energy of 100 lb Mass in Rotation

[Pinon1977]

Homework Statement

I have a 100 lb mass that is attached to a disk which is rotating freely about an axle. How do I calculate the gravitational energy contributed buy this 100 pound mass as it rotates from the 12 position to the six position

Homework Equations

No clue maybe. M x g x h?

The Attempt at a Solution

No clue

 

[BvU]

Depends on the orientation of the axle ... but otherwise you are correct. Pay attention to the units !

 

[DrClaude]

The Attempt at a Solution

No clue

This is not acceptable, per PF rules. Using @BvU's help, your next post should be your attempt at a solution.

 

[Pinon1977]

Ok. My attempt would look like this:

KE = 100lbs x 9.8ms2 x 10ft = 1354J

But that is if the weight it's Free Falling, correct? This particular scenario involves the weight being attached to a disc that is 10 feet in diameter. The disc weighs approximately 100 pounds as well. So wouldn't those two variables be taken into consideration?

 

[BvU]

the gravitational energy contributed by this 100 pound mass

I should think not.

You still haven't told us about the orientation nor about the location of the axis of rotation

 

[Pinon1977]

My apologies. The 100lbs is permanently mounted on the disc 5 feet from the axel ( or at the outer edge of the disk). As the disc rotates, the orientation of the weight changes respectively (as the weight is hard mounted to the disc which is free spinning).

 

[CWatters]

How do I calculate the gravitational energy contributed buy this 100 pound mass as it rotates from the 12 position to the six position

Homework Equations

No clue maybe. M x g x h?

Yes that's a relevant but incomplete equation. M*g*h = ??

 

[Pinon1977]

Ke = mgh. So, 100 pounds attached at the Outer Perimeter of a 10-foot diameter free Spinning Disk (which has a mass of 100 lb itself) would still be 100 times 9.8 times 10?

 

[BvU]

Watch your units. Is $g$ 9.81 thingies in your non-SI units ?

 

[CWatters]

Ke = mgh.

Not KE. The change in Potential Energy (PE) = mgh

So, 100 pounds attached at the Outer Perimeter of a 10-foot diameter free Spinning Disk (which has a mass of 100 lb itself) would still be 100 times 9.8 times 10?

What BvU said about units. g is only 9.81 m/s/s if you are working in SI units..

Working in SI units the description becomes... "45kg attached to the outer perimeter of a 3 meter diameter spinning disc"

To rotate the disk so that the 45kg mass moves from the bottom/6 o'clock position to the top/12 o'clock position would require..
ΔPE = 45 * 9.81 * 3 = 1324 Joules

To rotate the disk so that the 45kg mass moves from the top/12 o'clock position to the bottom/6 o'clock position would require..
ΔPE = 45 * 9.81 * -3 = -1324 Joules

eg you get the same number of joules back again.

Note: We're assuming the 100lbs/45kg is all lumped together in one place on the perimeter not distributed around the perimeter of the disk.