# Find the period of a pendulum

1. Apr 19, 2007

### anand

1. The problem statement, all variables and given/known data
Find the period of a pendulum consisting of a disk of mass M and radius R fixed to the end of a rod of length l and mass m.How does the period change if the disk is mounted to the rod by a frictionless bearing so that it is perfectly free to spin?

2. Relevant equations
torque=moment of inertia*alpha=r X F

3. The attempt at a solution

I know how to do the first part,but I can't do the second part.What happens when the disk is free to spin?

2. Apr 19, 2007

### denverdoc

maybe the moment of inertia changes wrt motion of pendulum? Interesting problem. Can you post the soln to first part?

3. Apr 19, 2007

### Dick

If the disk is free to rotate then it will, in fact, not rotate while the pendulum is swinging. Hence for moment purposes I think it can be treated as a point mass.

4. Apr 19, 2007

### denverdoc

Thats very interesting Dick, and I should read more carefully, I assumed it was spinning and hence had to take some average moment of inertia. So the pendulum is just falling into the energetically most favorable situation? If the disk were perpendicular to the motion of travel initially would it remain there indefinitely (ignoring air resistance)?

5. Apr 19, 2007

### Dick

Assuming the bearing is at the center of mass of the disk - then there aren't any forces around that could cause a torque. If it's already spinning, I don't think that causes any complication as long as the spin axis of the disk is perpendicular to the plane of the pendulum. If it's not then it's a gyroscope and all bets are off.

6. Apr 20, 2007

### anand

Here is how I did the first part:

First find the total moment of inertia around an axis passing through the pivot.That would be 1/3 ml^2+1/2MR^2+Ml^2

The centre of mass of the system can be found,which is
C=(ml/2+Ml)/(m+M)

The component of the weight tangential to motion is (M+m)g sin(theta)
Torque is vector from origin X perpendicular component of force:
-(m+M)g *sin(theta) *C

But it is also equal to moment of inertia*angular acceleration:

(1/3 ml^2+1/2MR^2+Ml^2) * alpha

Equating the two,we get an equation for SHM,and
omega is (1/3ml^2+1/2MR^2+Ml^2)/(ml/2+Ml)g

From what you have said,the disk wouldn't spin,once the bearings are in place.Would it kind of keep pointing in the same direction,instead of being carried along with the rod?Do you mean to say that the term 1/2 MR^2 should be dropped?Please explain.

7. Apr 20, 2007

### denverdoc

the part i struggle with is assume its a long rod and not a disk just for clarity, it seems to me there would be a big difference if the rod is within the plane of the pendulums motion vs one that is perpendicular. Just hits the ole counterintuitive nerve.

8. Apr 20, 2007

### anand

9. Apr 20, 2007

### shamalex

hi

thanks
thanks
thanks
thanks

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10. Apr 20, 2007

### Dick

I think we have to assume that the plane of rotation of the disk is is the same as the plane of the pendulum. And then, yes, the disk will kind of keep pointing in the same direction (zero torque). Hence you can drop it's center of mass inertia (1/2mR^2) and treat it as a point mass (Ml^2).

11. Apr 20, 2007

### anand

Thanks a lot!