Extreme Period for a Physical Pendulum

[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

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

 

[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.

 

[whitetiger]

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)

 

[OlderDan]

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

 

[whitetiger]

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))

 

[OlderDan]

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

 

[whitetiger]

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?

 

[OlderDan]

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.