# Pendulum Problem

1. Dec 9, 2013

### Clearik

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

A pendulum is composed of a thin rod of length L = 50 cm, a moment of inertia of I = 0.2kgm2 with respect to its center and mass M = 400g. Attached to the bottom of the rod is a point mass (m = 200g). The pendulum swings in a vertical plane, attached by the top end to the ceiling.

2. Relevant equations

T = 2∏√(I/Mg x d)
*I think*

3. The attempt at a solution

T = 2∏√(.2/(.4)(10) x .25)

No radius was given for the point mass, so I assumed the moment of inertia of center of mass is .2kgm2 and therefore the distance of center of mass (d) would be L/2 or .25m

Did I go about solving this correctly?

Also, the second problem asks: How long will it take for the pendulum to perform 100 oscillations?

Would I simply multiple the period I found from the first part by 100?

Thank you all for your time and consideration.

Last edited: Dec 9, 2013
2. Dec 10, 2013

### Simon Bridge

Don't guess.
It helps to sketch the situation out:

- Which part of the rod is attached to the ceiling?
- Which part of the rod has the point mass attached to it?

You are given the moment of inertia for a rod pivoted at the center of mass - but that is not where the rod is pivoted - so you need to adjust this (hint: parallel axis theorem).

... yes.
The question is basically checking if you understand what "one period" means.

It's good to see you thinking beyond what's in the question - just remember that everything you want to assert has to be justified by something said inside the problem statement.

3. Dec 10, 2013

### Clearik

Hmm, I see.

The equation for moment of inertia at end of rod is Iend = mL2/3

Therefore, Iend = 1.33 kgm2

From my understanding of using the parallel axis theorem to calculate moment of inertia of a system, one would simply add up the inertias.

Isystem = Icom + Iend of rod + Ipoint mass

Isystem = .2kgm2 + 1.33kgm2 + 2.00kgm2

Isystem = 3.53kgm2

Does this sound more reasonable?

4. Dec 10, 2013

### Simon Bridge

That's not how the parallel axis theorem works.

You should only have two terms to calculate inertia: $I_{tot}=I_{rod}+I_{point}$

You can just look up the inertia for the rod-pivoted-about-one-end.
However, since the question tells you the com inertia, they are expecting you to use that and the parallel axis theorem. Either way is correct but both together is not.

Aside: interesting... the PF spell-checker does not accept either "inertias" or "inertiæ" (accepted plurals for "inertia").

Last edited: Dec 10, 2013