# Extreme Period for a Physical Pendulum

by whitetiger
Tags: extreme, pendulum, period, physical
 P: 22 1. The problem statement, all variables and given/known data 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? 2. Relevant 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 I am not sure about this, so hope someone can help 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
 HW Helper P: 3,224 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.
P: 22
 Quote by radou 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)

HW Helper
Sci Advisor
P: 3,033

## Extreme Period for a Physical Pendulum

 Quote by whitetiger 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...parax.html#pax
P: 22
 Quote by OlderDan 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...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))
HW Helper
Sci Advisor
P: 3,033
 Quote by whitetiger 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
P: 22
 Quote by OlderDan 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?
HW Helper
Sci Advisor
P: 3,033
 Quote by whitetiger 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.

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