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Moment of inertia physical pendulum

by arierreF
Tags: inertia, moment, pendulum, physical
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arierreF
#1
Jan5-13, 08:56 PM
P: 62
1. The problem statement, all variables and given/known data

I have a physical pendulum that is rotating about an fixed axis.

The period is:

[itex]T=2\pi \sqrt{\frac{I}{mgd}}[/itex]

I = moment of inertia

d = distance between the center of mass and the axis.


The problem is:

If you add a mass in the end of the pendulum. The period is going to increase or decrease.


My attempt:

the inertia of a rod (physical pendulum)
[itex]I = md^{2} + \frac{1}{12}mL^{2}[/itex]


Without the mass, the moment inertia is :

[itex]d = \frac {L}{2}[/itex] the center of mass in in the middle of the physical pendulum.

[itex]I = m\frac {L^{2}}{4} + \frac{1}{12}mL^{2}[/itex]
[itex]I = \frac{1}{3} mL^{2}[/itex]


if i add a mass in the end of the pendulum then

[itex]I = mL^{2} + \frac{1}{12}mL^{2}[/itex]

[itex]I = \frac {13}{12} mL^{2}[/itex]


without mass:

[itex]T=2\pi \sqrt{\frac{\frac{1}{3} mL^{2}}{mg\frac{L}{2}}}[/itex]

[itex] T=2\pi \sqrt{\frac{2L}{3g}} [/itex]

with mass:

[itex]T=2\pi \sqrt{\frac{\frac {13}{12} mL^{2}}{mgL}}[/itex]

[itex]T=2\pi \sqrt{\frac {13L}{12g} }[/itex]




If we add a mass in the end of the physical pendulum, then the period is going to increase.


Can somebody check if my answer is correct?
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SammyS
#2
Jan5-13, 09:14 PM
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Quote Quote by arierreF View Post
1. The problem statement, all variables and given/known data

I have a physical pendulum that is rotating about an fixed axis.

The period is:

[itex]T=2\pi \sqrt{\frac{I}{mgd}}[/itex]

I = moment of inertia

d = distance between the center of mass and the axis.

The problem is:

If you add a mass in the end of the pendulum. The period is going to increase or decrease.

My attempt:

the inertia of a rod (physical pendulum)
[itex]I = md^{2} + \frac{1}{12}mL^{2}[/itex]

Without the mass, the moment inertia is :

[itex]d = \frac {L}{2}[/itex] the center of mass in in the middle of the physical pendulum.

[itex]I = m\frac {L^{2}}{4} + \frac{1}{12}mL^{2}[/itex]
[itex]I = \frac{1}{3} mL^{2}[/itex]

if i add a mass in the end of the pendulum then

[itex]I = mL^{2} + \frac{1}{12}mL^{2}[/itex]

[itex]I = \frac {13}{12} mL^{2}[/itex]

without mass:

[itex]T=2\pi \sqrt{\frac{\frac{1}{3} mL^{2}}{mg\frac{L}{2}}}[/itex]

[itex] T=2\pi \sqrt{\frac{2L}{3g}} [/itex]

with mass:

[itex]T=2\pi \sqrt{\frac{\frac {13}{12} mL^{2}}{mgL}}[/itex]

[itex]T=2\pi \sqrt{\frac {13L}{12g} }[/itex]

If we add a mass in the end of the physical pendulum, then the period is going to increase.

Can somebody check if my answer is correct?
I is the moment of inertia about the fixed axis.

Do you know where the fixed axis is ?
arierreF
#3
Jan6-13, 07:08 AM
P: 62
Quote Quote by SammyS View Post
I is the moment of inertia about the fixed axis.

Do you know where the fixed axis is ?
The fixed axis is where the pendulum swings (pivot). (Beginning of the pendulum)

Doc Al
#4
Jan6-13, 07:55 AM
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Moment of inertia physical pendulum

Quote Quote by arierreF View Post
My attempt:

the inertia of a rod (physical pendulum)
[itex]I = md^{2} + \frac{1}{12}mL^{2}[/itex]


Without the mass, the moment inertia is :

[itex]d = \frac {L}{2}[/itex] the center of mass in in the middle of the physical pendulum.

[itex]I = m\frac {L^{2}}{4} + \frac{1}{12}mL^{2}[/itex]
[itex]I = \frac{1}{3} mL^{2}[/itex]
OK. Your physical pendulum is just a thin rod of mass m. (Is that what you had in mind? Or were you supposed to consider any physical pendulum?)


if i add a mass in the end of the pendulum then

[itex]I = mL^{2} + \frac{1}{12}mL^{2}[/itex]

[itex]I = \frac {13}{12} mL^{2}[/itex]
Is the added mass equal to the mass of the rod? Since you want the rotational inertia about the axis, why are you adding the rotational inertia of the rod about its center of mass?

Don't forget that when you add mass the location of the center of mass changes.
arierreF
#5
Jan6-13, 09:12 AM
P: 62
Quote Quote by Doc Al View Post
OK. Your physical pendulum is just a thin rod of mass m. (Is that what you had in mind? Or were you supposed to consider any physical pendulum?)



Is the added mass equal to the mass of the rod? Since you want the rotational inertia about the axis, why are you adding the rotational inertia of the rod about its center of mass?

Don't forget that when you add mass the location of the center of mass changes.
it is the a thin rod.


consider the mass of the (added mass) = to the mass of the rod.



because d it's the distance from the center of mass and the pivot, then when we add a mass in the end of pendulum then d = L (the center of mass is in the end of the pendulum)

if the mass is in the center then

d = L^2/2
Doc Al
#6
Jan6-13, 09:43 AM
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Quote Quote by arierreF View Post
it is the a thin rod.


consider the mass of the (added mass) = to the mass of the rod.
OK. But you still need to correct your new rotational inertia.
because d it's the distance from the center of mass and the pivot, then when we add a mass in the end of pendulum then d = L (the center of mass is in the end of the pendulum)

if the mass is in the center then

d = L^2/2
Before you add the additional mass, the center of mass is at d = L/2. When you add a mass 'm' to the end of the rod, where is the new center of mass of the system?
arierreF
#7
Jan6-13, 09:59 AM
P: 62
Quote Quote by Doc Al View Post
OK. But you still need to correct your new rotational inertia.

Before you add the additional mass, the center of mass is at d = L/2. When you add a mass 'm' to the end of the rod, where is the new center of mass of the system?
When i add it to the end the d= L

so it is [itex]I = mL^{2} + \frac{1}{12}mL^{2} = \frac{13}{13}mL^{2}[/itex] considering the mass of the added mass = mass of the thin rod.
Doc Al
#8
Jan6-13, 10:16 AM
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Quote Quote by arierreF View Post
When i add it to the end the d= L
No. The center of mass would be at d = L only if all the mass were at the end.

so it is [itex]I = mL^{2} + \frac{1}{12}mL^{2} = \frac{13}{13}mL^{2}[/itex] considering the mass of the added mass = mass of the thin rod.
No. What's the rotational inertia of the rod about its end? (You calculated that in post #1, but then made no use of it.) That's what you should be adding.
arierreF
#9
Jan6-13, 10:21 AM
P: 62
Quote Quote by Doc Al View Post
No. The center of mass would be at d = L only if all the mass were at the end.


No. What's the rotational inertia of the rod about its end? (You calculated that in post #1, but then made no use of it.) That's what you should be adding.
hum it is. 1/3ML^2

so my calculus are wrong, because the mass is not all in the end. But if the mass of (added mass) is much greater than the mass of the rod, then the center of the mass is in the end right?
Doc Al
#10
Jan6-13, 10:51 AM
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Quote Quote by arierreF View Post
hum it is. 1/3ML^2
Good.
so my calculus are wrong, because the mass is not all in the end.
Yes.
But if the mass of (added mass) is much greater than the mass of the rod, then the center of the mass is in the end right?
Sure. But then you will be treating it just like a simple pendulum, ignoring the mass of the rod.
arierreF
#11
Jan6-13, 07:16 PM
P: 62
Quote Quote by Doc Al View Post
Good.

Yes.

Sure. But then you will be treating it just like a simple pendulum, ignoring the mass of the rod.
So i can conclude that if we had a significant mass in the end of the rod (physical pendulum), the moment of inertia is going to increase, as the period.
Doc Al
#12
Jan7-13, 12:29 PM
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Quote Quote by arierreF View Post
So i can conclude that if we had a significant mass in the end of the rod (physical pendulum), the moment of inertia is going to increase, as the period.
The moment of inertia and the distance from pivot to center of mass will both increase. To see the effect on the period, just compare the results for the thin rod pendulum to that for the simple pendulum. (But yes, the period will increase.)


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