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Weird disk rotating on pivot

by dkgojackets
Tags: disk, pivot, rotating, weird
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dkgojackets
#1
Feb19-07, 03:48 PM
P: 38
1. The problem statement, all variables and given/known data

A uniform solid disk of radius (r) 6.71 m and mass (m) 38 kg is free to rotate on a frictionless pivot through a point on its rim. If facing the disk, the pivot is on the left side. The disk is then released. What is the speed of the center of mass when the disk reaches a position such that the pivot is now at the top (lowest point in swing)?

What is the speed of the lowest point of the disk in said position?

Repeat part one for a uniform hoop.

2. Relevant equations

Looks like conservation of energy here. I know that the disk drops down one radius, so my initial energy was just mgr. At the bottom it has translational and rotational KE. .5mv^2 + .5Iw^2.

3. The attempt at a solution

Not sure, I'm looking for a step in the right direction. Probably need to use the parallel axis theorem somehow since it isn't rotating through the center?
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AlephZero
#2
Feb19-07, 04:31 PM
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If your formula KE = .5mv^2 + .5Iw^2 , v is the linear velocity of the CG of the disk, and I is the inertia about the CG.

The disk is pivoting about a point on the circumference, and that gives you a relation between the velocity at the CG and the rotational speed w.
dkgojackets
#3
Feb19-07, 05:36 PM
P: 38
I know I can substitute v/r for w. I have mgr = .5mv^2 + .5I(v/r)^2, with v being velocity at the center of mass.

I have r and m, I just don't know how to adjust the moment of inertia for it rotating around the edge.

Doc Al
#4
Feb19-07, 05:54 PM
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Weird disk rotating on pivot

If you, as AlephZero suggests, treat the total KE as the sum of the translational KE of the cm (.5mv^2) plus the rotational KE about the cm (.5Iw^2), then I is about the cm.

You can also treat the disk as in pure rotation about the pivot, in which case the KE = .5Iw^2, where I is now about the pivot point not the cm. (Use the parallel axis theorem to find I.)
AlephZero
#5
Feb19-07, 06:07 PM
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If you use the motion of the CG (translation and rotation) as you have done, then you don't need to adjust the moment of inertia.

The KE of a body is always the sum of two parts:

Translational energy of the CG (.5mv^2)
+ Rotational energy of the body rotating about the CG (.5Iw^2).

---------------------------

You could do the problem a different way, and consider the body rotating about the pivot. In that case you would use the parallel axis theorem to get the moment of inertia about the pivot = I (about the CG) + mr^2.

If you do it that way, there is no translation energy (because the pivot is not moving), but the additional rotational energy of .5(mr^2)w^2 is the same as the translational energy of the CG in the first method, because v = rw.

Do it whichever way you find easiest to think about.
dkgojackets
#6
Feb19-07, 06:13 PM
P: 38
Got it. I was just combining the two approaches, using the parallel axis theorem to find MoI in the first equation.


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