# Moment of inertia physical pendulum

• arierreF
In summary, the problem involves a physical pendulum rotating about a fixed axis and the question is whether adding a mass at the end of the pendulum will increase or decrease the period. To find the moment of inertia, the formula I = md^2 + (1/12)mL^2 is used, where I is the moment of inertia, m is the mass of the rod, d is the distance between the center of mass and the axis, and L is the length of the rod. Without the added mass, the moment of inertia is (1/3)mL^2, and with the added mass, it becomes (13/12)mL^2. However, the center of mass changes when the mass is added
arierreF

## Homework Statement

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

The period is:

$T=2\pi \sqrt{\frac{I}{mgd}}$

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)
$I = md^{2} + \frac{1}{12}mL^{2}$Without the mass, the moment inertia is :

$d = \frac {L}{2}$ the center of mass in in the middle of the physical pendulum.

$I = m\frac {L^{2}}{4} + \frac{1}{12}mL^{2}$
$I = \frac{1}{3} mL^{2}$if i add a mass in the end of the pendulum then

$I = mL^{2} + \frac{1}{12}mL^{2}$

$I = \frac {13}{12} mL^{2}$without mass:

$T=2\pi \sqrt{\frac{\frac{1}{3} mL^{2}}{mg\frac{L}{2}}}$

$T=2\pi \sqrt{\frac{2L}{3g}}$

with mass:

$T=2\pi \sqrt{\frac{\frac {13}{12} mL^{2}}{mgL}}$

$T=2\pi \sqrt{\frac {13L}{12g} }$

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?

arierreF said:

## Homework Statement

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

The period is:

$T=2\pi \sqrt{\frac{I}{mgd}}$

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)
$I = md^{2} + \frac{1}{12}mL^{2}$

Without the mass, the moment inertia is :

$d = \frac {L}{2}$ the center of mass in in the middle of the physical pendulum.

$I = m\frac {L^{2}}{4} + \frac{1}{12}mL^{2}$
$I = \frac{1}{3} mL^{2}$

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

$I = mL^{2} + \frac{1}{12}mL^{2}$

$I = \frac {13}{12} mL^{2}$

without mass:

$T=2\pi \sqrt{\frac{\frac{1}{3} mL^{2}}{mg\frac{L}{2}}}$

$T=2\pi \sqrt{\frac{2L}{3g}}$

with mass:

$T=2\pi \sqrt{\frac{\frac {13}{12} mL^{2}}{mgL}}$

$T=2\pi \sqrt{\frac {13L}{12g} }$

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 ?

SammyS said:
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)

arierreF said:
My attempt:

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

Without the mass, the moment inertia is :

$d = \frac {L}{2}$ the center of mass in in the middle of the physical pendulum.

$I = m\frac {L^{2}}{4} + \frac{1}{12}mL^{2}$
$I = \frac{1}{3} mL^{2}$
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

$I = mL^{2} + \frac{1}{12}mL^{2}$

$I = \frac {13}{12} mL^{2}$
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.

Doc Al said:
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

arierreF said:
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?

Doc Al said:
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 $I = mL^{2} + \frac{1}{12}mL^{2} = \frac{13}{13}mL^{2}$ considering the mass of the added mass = mass of the thin rod.

arierreF said:
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 $I = mL^{2} + \frac{1}{12}mL^{2} = \frac{13}{13}mL^{2}$ 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.

Doc Al said:
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?

arierreF said:
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.

Doc Al said:
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.

Last edited:
arierreF said:
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.)

## 1. What is moment of inertia physical pendulum?

Moment of inertia physical pendulum is a measure of an object's resistance to changes in its rotational motion. In the case of a physical pendulum, it is the measure of its resistance to changes in its swinging motion.

## 2. How is moment of inertia physical pendulum calculated?

Moment of inertia physical pendulum is calculated by multiplying the mass of the object by the square of its distance from the pivot point, and then adding the product of the object's mass and the square of its radius of gyration.

## 3. What factors affect the moment of inertia physical pendulum?

The moment of inertia physical pendulum is affected by the mass of the object, the distance of the object from the pivot point, and the shape and distribution of mass of the object.

## 4. Why is moment of inertia physical pendulum important?

Moment of inertia physical pendulum is important because it helps us understand how objects behave when they are rotating around a fixed point, such as a pendulum. It also helps us calculate the amount of energy needed to overcome an object's resistance to changes in its rotational motion.

## 5. How does moment of inertia physical pendulum differ from moment of inertia for a point mass?

Moment of inertia physical pendulum takes into account the distribution of mass in an object, while moment of inertia for a point mass only considers the object's mass and distance from the pivot point. In other words, moment of inertia physical pendulum is a more accurate representation of an object's resistance to changes in its rotational motion.

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