Period of a physical pendulum with a pivot at its center

In summary, the conversation discusses the period of three pendulum systems with the same length. The first two systems have a calculated period, but the third system has a problem due to the pivot point being at the center of mass. The conversation then discusses the concept of dividing by zero and how a very small but not exactly zero denominator results in a very long period. This leads to the conclusion that T3, with an infinite or nearly infinite period, would still be larger than the finite periods of T1 and T2.
  • #1
PhysicsKush
29
4

Homework Statement


The picture illustrates a simple pendulum and and two physical pendulums ,all having the same length ,L. Class their period in ascending order.

Homework Equations


T = 2π / ( I/mgh)
I = Icm + mh2

Icm=(ML2/12)

The Attempt at a Solution


I have found the period for first two objects :
T1 = 2π√(L/g)
T2 = 2π√(2L/3g)

The problem for me appears at the third system because h is defined as the distance between the center of mass of the rod and the pivot point. In the third situation, it seems to me that the pivot is at the center of mass , hence h=0 , but that would result in a division by 0 error.

I have tried cutting the rod in half so that L =L/2 and h = L/4 but after some algebra i end up with T3 = 2π√(L/3g) , which is smaller than T1

The solution to the problem is T3 > T1 > T2

Any help would be apreciated.
 

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  • #2
Mihail Anghelici said:
The problem for me appears at the third system because h is defined as the distance between the center of mass of the rod and the pivot point. In the third situation, it seems to me that the pivot is at the center of mass , hence h=0 , but that would result in a division by 0 error.
And what do you get when you have a fraction the denominator of which is very small but not exactly zero? How does the period compare with the other two periods? Will the ranking of the periods change when the denominator is exactly zero?
 
  • #3
kuruman said:
And what do you get when you have a fraction the denominator of which is very small but not exactly zero? How does the period compare with the other two periods? Will the ranking of the periods change when the denominator is exactly zero?
Not sure what you mean :/
 
  • #4
Mihail Anghelici said:
Not sure what you mean :/
I mean: You said the period is ##T = 2π\sqrt{I/(mgh)}##. That's good. What is ##h## in case (iii)? You said ##h## is zero. That is correct, but you seem to be stuck at the mathematical impossibility of dividing by zero. Mathematically, ##\frac{1}{0}=\infty##. We are doing physics here where there are hardly any mathematical infinities. How do you physically interpret "infinite period"? Answer: Very long. OK, if T3 is just "very long" as opposed to "very, very, very long", how do you think that would affect its ranking against T1 and T2? In other words, if one quantity is infinite or nearly infinite, could it be smaller than two finite quantities?
 

Related to Period of a physical pendulum with a pivot at its center

1. What is a physical pendulum?

A physical pendulum is a rigid body that is attached to a pivot point and allowed to oscillate back and forth due to the force of gravity acting on it. Unlike a simple pendulum, which has a single point of suspension, a physical pendulum has a larger mass and a distributed center of mass, making its motion more complex.

2. How is the period of a physical pendulum with a pivot at its center calculated?

The period of a physical pendulum with a pivot at its center can be calculated using the formula T = 2π√(I/mgd), where T is the period in seconds, I is the moment of inertia of the pendulum, m is the mass of the pendulum, g is the acceleration due to gravity, and d is the distance between the pivot point and the center of mass of the pendulum.

3. What factors affect the period of a physical pendulum?

The period of a physical pendulum is affected by several factors, including the length and mass of the pendulum, the strength of gravity, and the distribution of mass along the pendulum. The period also depends on the angle at which the pendulum is released and the amplitude of its oscillations.

4. How does the period of a physical pendulum change with increasing length?

As the length of a physical pendulum increases, the period also increases. This can be seen in the formula T = 2π√(I/mgd), where the length is represented by the variable d. This means that longer physical pendulums will take longer to complete one full oscillation than shorter ones.

5. Can the period of a physical pendulum be affected by external factors?

Yes, the period of a physical pendulum can be affected by external factors such as air resistance, friction at the pivot point, and vibrations from the surrounding environment. These factors can disrupt the motion of the pendulum and cause variations in its period.

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