# Rigid body rotation

1. Dec 12, 2011

### dttah

1. The problem statement, all variables and given/known data

A system is made of the following things:

A homogeneous disk of mass m = 4kg.
A point P (m = 2kg) free to rotate around a horizontal axis which is perpendicular to the disk in point A.
AO = OP/2
If the system starts from rest and AP forms a 90° angle with the vertical axis, find the angular velocity and the action on the axis (not sure what this means) in point A, when the system is passing through its equilibrium position.

2. Relevant equations

3. The attempt at a solution

******Completely not sure about the following:

I am just starting out with rigid bodies, so I will be a lil' slow. As long as I understood I should find the center of mass, so uhm.
The disk is homogenous so the center of mass would be the center.

Let O be the origin of our system.

The center of mass would be:

$\frac{0*4+0.8*2}{6}$

So the center of mass would be (0,26,0);
Uhm, now I thought I would have found the radius of the circumference having center A and radius Ax (center of mass).
So I did:

r = OP/2 + 0.26 = 0.66.

And now I am completely stuck.
Could you please give away some hints, and check if I did correctly 'till now or I completely missed it? Thanks.

The picture is this one, forget about the italian, it'd be the circumference on the bottom:

2. Dec 12, 2011

### Simon Bridge

Oh all right - it's poorly described isn't it: the disk and mass form a pendulum.
Point P is on the perimeter of the disk - has an extra point mass m.
Point A is on the line through OP but on the opposite side of O to P
|OA|=|OP|/2 = R/2

Point A is the pivot ... so this is all about rotation about a point other than the center.

When AP is 90deg to vertical, is the position shown ... AP is horizontal.

If we say vertical-down is zero-degrees, then this is the equilibrium position for the pendulum... zero degrees.

Initially $Mg(R/2)+mg(3R/2)=I\alpha$ ... or something... M is the mass of the disk.

I don't know what "action on the axis" means either.

3. Dec 12, 2011

### dttah

Holy, that's kind of unexpected, I mean I don't see any wire nor talking about gravity, so yeah, the idea didn't even pass by my mind.

I don't get why you say that "This is all about rotation about a point other than the center"
Why's that?

4. Dec 12, 2011

### Simon Bridge

Point A is the pivot - it is not at the center of mass is it? Nor is it at the center of the disk.
This affects the moment of inertia.

I for the point mass is easy, it will be m(3R/2)^2
I for the disk is trickier - if the pivot was at O then it would be (1/2)MR^2 - but it ain't.
How is the moment of inertia affected when the pivot is off-center?

5. Dec 12, 2011

### D H

Staff Emeritus
Not really all that poorly described. You just have to infer that gravity is present, and "la posizione di equilibrio" gives a big clue that this is the case.

6. Dec 13, 2011

### dttah

I am sorry if I am a little slow but as I said I am just starting out with these.
The disk isn't rotation around its center of mass (the center), so we have to apply the integral of r^2dm
in order to find out the moment of inertia?
Or maybe I could apply the parallel axis theorem?

7. Dec 13, 2011

### D H

Staff Emeritus
Try that.

8. Dec 13, 2011

### dttah

Alright, so the center of mass would be the center of the disk itself.

So it'd be..

1/2 MR^2 + MD^2.

Where the distance here would be OP/2.

So 1/2 (4kg) * 0.8m + 4kg*0.4m = 3,2.

Now I found the moment of inertia (if I did it right, lol) of the disk rotating alone point A.

The moment that the whole system is set free it has no initial kinetic energy, but it has potential energy.

Then everything is converted in kinetik energy of rotation.
So it'd be something like...

mgh (disk) + mgh(mass) = 1/2Iw^2 (disk) + 1/2Iw^2 (mass).

I am not completely sure though. Is this right? >.<

9. Dec 13, 2011

### D H

Staff Emeritus
You didn't. Moment of inertia should have units of mass*length2.

I suggest that you get in the habit of defining short symbolic names for key items. For example, you could denote the radius of the disk as r, the mass of the disk as M, and the mass of the point mass as m. Oftentimes (and this is one of those times), it is better to keep things symbolic as long as possible.

Correct. How much does the center of mass of the disk move vertically from the initial position to the equilibrium position? How about of the point mass?

10. Dec 13, 2011

### dttah

Alright, thanks for the suggestion, I'll try to stick to that from now on.Here is the numerical correction, I = 1.28 + 0.64 = 1.92.

Now uhm.
The center of mass of the disk is the center of the disk, which is rotating alone point A.
Radius would be OP/2. It'd move pi/2 radians, shoudln't it? And same does the other point there.

So in mgh of the disk I'd stick in ...
And in the point I'd stick:

11. Dec 13, 2011

### D H

Staff Emeritus
You need to distinguish between the mass of the disk and the mass of the point mass. I suggested capital M and lower case m. Another choice is a simple subscript, md and mp.

More importantly, the equilibrium position of the point mass would be one disk radius below the initial position if the disk was rotating about its center, which it isn't. The disk is rotating about the point A, which is off-center.

12. Dec 13, 2011

### dttah

Got it, so it'd be like:

Let
M be the mass of the disk
m be the mass of the point

For the disk: