Pendulum Problem: Solve 100 Oscillations in 50cm Rod

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In summary, the homework statement states that a pendulum is composed of a thin rod of length 50 cm, a moment of inertia of 0.2kgm2 with respect to its center and mass 400g. The pendulum swings in a vertical plane, attached by the top end to the ceiling. The Attempt at a Solution states that the equation for moment of inertia at end of rod is Iend = mL2/3 and therefore the distance of center of mass (d) would be L/2 or .25m.
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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.
 
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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.
 
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  • #3
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
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").
 
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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.
 

FAQ: Pendulum Problem: Solve 100 Oscillations in 50cm Rod

1. How do you calculate the period of a pendulum?

The period of a pendulum can be calculated using the formula T = 2π√(l/g), where T is the period in seconds, l is the length of the pendulum in meters, and g is the acceleration due to gravity (9.8 m/s^2).

2. What is the relationship between the length of a pendulum and its period?

The length of a pendulum is directly proportional to its period. This means that as the length of the pendulum increases, the period also increases. This relationship is known as the "Pendulum Law".

3. How many oscillations can a pendulum complete in a given time?

The number of oscillations a pendulum can complete in a given time depends on the length of the pendulum and the initial angle at which it is released. For a 50cm rod, the pendulum can complete 100 oscillations in 50 seconds if it is released at a perfect angle.

4. How does the mass of a pendulum affect its period?

The mass of a pendulum does not affect its period. The period of a pendulum is only dependent on its length and the acceleration due to gravity. This means that two pendulums with different masses but the same length will have the same period.

5. How can I ensure the accuracy of my results when solving the pendulum problem?

To ensure the accuracy of your results when solving the pendulum problem, it is important to use precise measurements for the length of the pendulum and the time it takes to complete the oscillations. It is also important to minimize any outside factors that could affect the pendulum's motion, such as air resistance or friction. Repeating the experiment multiple times and taking an average of the results can also help improve accuracy.

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