# Rotational Imbalance

1. Jun 22, 2004

### e(ho0n3

Hallo Alle,

I wish I had a scanner to scan the figure that comes with this problem but anyways...there is a thin massless rigid rod with masses m and M at the ends. This rod is attached to an axle at its center of mass and makes an angle $\phi$ with the axle. The distance of m and M from the point of attachment (call this point the origin O) is r and R respectively. The axle is turning with angular velocity $\omega$. Oh, and there are bearings at the end of the axle. I hope I've verbalized this properly.

So here is the question: Suppose that M = 0. If the bearings are each a distance d from the point of attachment of the rod and axle, what are the forces at the upper and lower bearings respectively (ignore gravity)?

Now, as I understand it, the angular momentum of this system is always changing direction causing the axle to move around but this movement is 'countered' by the bearings and so I have to calculate the force that will counteract the movement on each bearing.

The angular momentum of the system is
$$\vec{L} = \vec{r} \times \vec{p}$$​
where
$$\vec{p} = m\vec{v} = m\vec{r} \times \vec{\omega} = mr\omega\sin{\phi}$$​
I can find the torque using $\tau = dL/dt$ and then using $\tau = F_{\bot}r$ to find the forces. My concern is that since the bearings are the same distance d from O, then the forces will have the same magnitude (because of $\tau = F_{\bot}r$). But I know this can't be since the axle will wobble more on one end that the other if the bearings were not present meaning that one of the bearings much exert more force than other. I just don't get it. Any tips?

2. Jun 24, 2004

### e(ho0n3

A Picture

Looks like my verbalization of the problem has discouraged any response. Perhaps a picture might help. Hopefully I'm better at drawing than writing.

#### Attached Files:

• ###### axle.jpg
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3. Jun 24, 2004

### Staff: Mentor

some questions

The picture does help, but I'm still not completely clear on the setup. I would have assumed that the rod is rigidly fixed to the axle (think welded) at that angle $\phi$, and it's the axle that rotates at $\omega$. If my understanding is correct, the angular momentum vector would be parallel to the axle, not pointing off into space. Also, do you really mean to ignore gravity?

4. Jun 24, 2004

### kuengb

Hallo

As I see it you're simply forgetting the movement of the center of mass! (You wrote that the c.o.m lies on the axle, but for M=0 this makes no sense) For a rigid body, which this rod basically is, we have dL/dt=M and dP/dt=F where P is the total linear momentum.

In the case M=0 choose as coordinate system the one with origin O at the level of m (i. e. at the level of the c.o.m.), then you have L=const. For the forces on the bearings: their resulting torque (with respect to O) has to be zero, and (F1 + F2)=dP/dt.

5. Jun 24, 2004

### e(ho0n3

The problem ignores gravity. And the L in the picture represents the sum of the angular momentums of m and M. The velocity vector of m say in the picture is pointing into the picture. Using the right hand rule, the angular momentum vector will be perpendicular to the position vector of m (and this is not parallel to the angular velocity vector).

6. Jun 24, 2004

### e(ho0n3

How do you get dL/dt = M? That makes no utter sense.
I never though of moving the origin. Good idea. I'll try that.

7. Jun 24, 2004

### kuengb

Sorry, I'm used to taking the letter M for the torque.

Last edited: Jun 24, 2004
8. Jun 24, 2004

### Gokul43201

Staff Emeritus
The torque is also refered to as the moment of a force, or simply the Moment. The reqirements that the resultant torque be zero, for no angular acceleration is also know as the Law of Equal Moments.

PS : Why is 'Sie' always capitalized ?

9. Jun 24, 2004

### e(ho0n3

I got confused there too since I'm using M for one of the masses. The only language I know where M is used for torque is German (sprechen Sie Deutsch?).

I just noted that I forgot to mention that the bearings are a distance d apart from the point of attachment of the massless rod and the axle.

Since the torque is zero then $F_1(d-r\cos\phi) = F_2(d + r\cos\phi)$ right? The change in linear momentum is due to m, so I derived $dP/dt = mr\omega^2\sin\phi$. Solving gives me the answer.

Vielen Dank.

10. Jun 24, 2004

### e(ho0n3

Right.
In german sie can me she, you (plural) and you (formal). The formal is always capitalized to avoid confusion.

11. Jun 25, 2004

### kuengb

Ja, tatsaechlich. Herzliche Gruesse aus der Schweiz

12. Jun 25, 2004

### Gokul43201

Staff Emeritus
Aah, okay.

So, 'sie' can only mean 3 different things - not 4 ! What a relief !

13. Jun 25, 2004

### e(ho0n3

Actually 'sie' has one more meaning. Depending on the mode (dative I think), it can mean her.