Pendulum Problem: Solve 100 Oscillations in 50cm Rod

[Clearik]

Homework Statement

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.

Homework Equations

T = 2∏√(I/Mg x d)
*I think*

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.

 

[Simon Bridge]

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

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).

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

... 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.

 

[Clearik]

Hmm, I see.

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

Therefore, I_end = 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.

I_system = I_com + I_{end of rod} + I_{point mass}

I_system = .2kgm² + 1.33kgm² + 2.00kgm²

I_system = 3.53kgm²

Does this sound more reasonable?

 

[Simon Bridge]

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

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").

 

[HalfLostAndSearching]

Yes, your approach to solving the first part of the problem is correct. The period of a pendulum is given by the equation T = 2∏√(I/Mgd), where I is the moment of inertia, M is the mass, g is the acceleration due to gravity, and d is the distance from the center of mass to the pivot point. In this case, the distance d is equal to half of the length of the rod, since the point mass is attached at the bottom. Therefore, your calculation for the period is correct.

For the second part of the problem, you are correct in multiplying the period by 100 to find the time it takes for the pendulum to perform 100 oscillations. This is because the period is the time it takes for one complete oscillation, so multiplying by 100 gives the total time for 100 oscillations. Keep in mind that the period may change slightly as the pendulum loses energy due to friction, so the time for 100 oscillations may not be exactly 100 times the period you calculated. However, it will be very close.