# Extreme Period for a Physical Pendulum

• whitetiger
In summary: Take a derivative wrt d and set it to zero to see if there is an extremum. If there is one, determine if it is a max or min. If you have not taken calculus, graph it.
whitetiger

## Homework Statement

http://img120.imageshack.us/img120/808/periodicemotionhd3.jpg solid, uniform disk of mass M and radius a may be rotated about any axis parallel to the disk axis, at variable distances from the center of the disk

If you use this disk as a pendulum bob, what is T(d), the period of the pendulum, if the axis is a distance d from the center of mass of the disk?

and

The period of the pendulum has an extremum (a local maximum or a local minimum) for some value of d between zero and infinity. Is it a local maximum or a local minimum?

## Homework Equations

From the picture, I come up with the moment of inertia of the solid disk around its center of mass
I = 1/2Ma^2
From the question, we are asked to find the period of the pendulum if the axis distance d from the center of mass.

The period T for this is P= 2pi (sqrt L/g) where g is the gravitation force
and L is the lenght.
From my understanding is that because of the new lenght, we need to use the Parallel Theorem to find the new lenght

Iend = Icm + Md^2
Iend = 1/2Ma^2 + Md^2

So the period is P = 2pi (sqrt(( a^2 +d^2)/g))
But this is not correct.

Thank

Last edited by a moderator:
Good thoughts, but your expression for the period doesn't seem correct. You may want to look at this: http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html" . Your period expression has to include the moment of inertia.

Last edited by a moderator:
Good thoughts, but your expression for the period doesn't seem correct. You may want to look at this: http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html" . Your period expression has to include the moment of inertia.

So is this right :

I = 1/2Ma^2

P = P= 2pi (sqrt 1/2Ma^2/(1/2Mgd^2)) == then we can cancel out the M to get

P = 2pi (sqrt 1/2a^2/(1/2gD^2)) === cancel 1/2 ===

P = sqrt(a^2/gd^2)

Last edited by a moderator:
whitetiger said:
So is this right :

I = 1/2Ma^2

P = P= 2pi (sqrt 1/2Ma^2/(1/2Mgd^2)) == then we can cancel out the M to get

P = 2pi (sqrt 1/2a^2/(1/2gD^2)) === cancel 1/2 ===

P = sqrt(a^2/gd^2)

You need to use the parallel axis theorem to find the moment of inertia of the disk about the pivot point. I don't see why you have a 1/2 in the denominator of the fraction. What became of the 2pi?

http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#pax

Last edited:
OlderDan said:
You need to use the parallel axis theorem to find the moment of inertia of the disk about the pivot point. I don't see why you have a 1/2 in the denominator of the fraction. What became of the 2pi?

http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#pax

Thank for the useful info.

Is this the correct moment of inertia of the disk about the pivot point.

I= 1/2Ma^2 + Md^2 = M ( 1/2a^2 + d^2)

So the period of the disk is

P = 2pi (sqrt (M(1/2a^2 + d^2))

whitetiger said:
Thank for the useful info.

Is this the correct moment of inertia of the disk about the pivot point.

I= 1/2Ma^2 + Md^2 = M ( 1/2a^2 + d^2)

So the period of the disk is

P = 2pi (sqrt (M(1/2a^2 + d^2))

No. Check the formula for the period of a physical pendulum

http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

OlderDan said:
No. Check the formula for the period of a physical pendulum

http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

I think I've got it.

P = 2pi(sqrt(a^2/2gd + d/g))

If you look at this, how can you determine whether it has a local maximum or local minimum for some value of d?

whitetiger said:
I think I've got it.

P = 2pi(sqrt(a^2/2gd + d/g))

If you look at this, how can you determine whether it has a local maximum or local minimum for some value of d?

Take a derivative wrt d and set it to zero to see if there is an extremum. If there is one, determine if it is a max or min. If you have not taken calculus, graph it.

## 1. What is an extreme period for a physical pendulum?

The extreme period for a physical pendulum is the amount of time it takes for the pendulum to complete one full oscillation at its maximum angle of displacement. This is also known as the maximum period or the longest period of oscillation for the pendulum.

## 2. How is the extreme period calculated for a physical pendulum?

The extreme period for a physical pendulum can be calculated using the equation T = 2π√(I/mgd), where T is the period, I is the moment of inertia, 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 extreme period of a physical pendulum?

The extreme period of a physical pendulum is affected by the length of the pendulum, the mass of the pendulum, the distance between the pivot point and the center of mass, and the acceleration due to gravity. These factors influence the moment of inertia and the oscillation period of the pendulum.

## 4. Why is the extreme period important in studying physical pendulums?

The extreme period is important in studying physical pendulums because it provides a benchmark for the maximum possible period of oscillation for the pendulum. This can be used to compare with the actual period of oscillation and determine the accuracy of the pendulum's motion.

## 5. How does the extreme period differ from the simple harmonic motion period of a physical pendulum?

The extreme period is the longest possible period of oscillation for a physical pendulum at its maximum angle of displacement, while the simple harmonic motion period is the period of oscillation for small angles of displacement. The extreme period is usually longer than the simple harmonic motion period due to the increased moment of inertia at larger angles of displacement.

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