How do you integrate r^2 dm for to find the moment of inertia?

Question

Is there a way to calculate the inertia of an object when an object is moving linearly or is that just the mass? If it's just the mass, why can't you just use the mass at the COM to find the moment of inertia?

Integrate r^2 dm. I would pull dm outside the integral and integrate r^2 as r^3/3, leaving me with dmr^3/3.

If someone could explain how to do this and exactly what the moment of inertia is I would be very grateful

Answer 1

"Is there a way to calculate the inertia of an object when an object is moving linearly or is that just the mass?"

Not sure what you're asking here. Most of your question is about MOMENT of inertia, which is a different thing from just plain "inertia". Sometimes inertia means momentum. Sometimes it means the mass.



"If it's just the mass, why can't you just use the mass at the COM to find the moment of inertia?"

Because moment of inertia is defined as the sum of mr^2 for every little piece of mass. It depends on not only the mass but how the mass is distributed. Mass far from the axis takes more energy and more torque to get spinning. That's why r matters.



I guess this is related to your first question, where you are confusing "moment of inertia" with "inertia". Two different things.



"Integrate r^2 dm. I would pull dm outside the integral and integrate r^2 as r^3/3, leaving me with dmr^3/3."



I hope not. "Integrate r^2 dm" means "Integrate r^2 with respect to m". You can't pull dm out of the integral, it's the variable of integration. r^3/3 is the integral of r^2 with respect to r. But you aren't integrating with respect to r.



The way you solve those integrals depends on the distribution of mass. You usually end up rewriting dm as some function times dr and THEN you have an expression in r being integrated with respect to dr.

Answer 2

Moment of Inertia is used for rotational mechanics only. Mass is the linear equivalent of it. The moment of inertia is defined with respect to an axis of rotation and that is why we need the "r" term.



Here's where the I = mr^2 comes from...

Torque = I*alpha = F*r



With respect to linear motion for an unchanging mass, F = m*a



The definition of alpha (angular acceleration)

alpha = a / r



Sub those back in above

I*a/r = m*a*r



The a's cancel

I/r = m*r

I = m*r^2



^This is only good for a point mass where the rotation is not about the COM.



Really, you don't need to integrate dI = r^2 dm. Most of the time this is handled by the parallel axis theorem. It says that the moment of inertia is the moment of inertia about the COM + mr^2... I = I_com + mr^2... It's MOI + It's mass about the axis of rotation.



The only time you would need to actually integrate dI = r^2 dm is when the distance from the axis of rotation is a function of the mass, which I can't really think of an example off the top of my head.



Here is a list of most common I_com's for various configurations:

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

Answer 3

"Is there a way to calculate the inertia of an object when an object is moving linearly or is that just the mass?"



Moment of inertia or just "inertia". If you mean just inertia, avoid that term: it is ambiguous. If you mean moment of inertia, even linearly moving objects have a moment of inertia. No natural center point exists, but one can be defined, and then you use mr^2, same as for a rotating or revolving object.



"If it's just the mass, why can't you just use the mass at the COM to find the moment of inertia?"



Because the parts of the object will not all be at the same distance from the center point.



"Integrate r^2 dm. I would pull dm outside the integral and integrate r^2 as r^3/3, leaving me with dmr^3/3."



You seem to be getting thrown off by integrals with mixed variables. These are tricky and were poorly covered in my calculus class. I learned about them mostly through physics.



You can't pull any arbitrary factor out of an integral, only a constant one. Second, if you are left with integral of r^2, that's meaningless. You have to have a d-something to have a meaningful integral. The integral of r^2 dr is r^3/3, but without the dr it's undefined. Lastly, as I think you recognized, if you do it your way you leave the dm left over and don't know how to determine its value.



In order to do an integral, you need to be able to determine the value of everything inside in terms of the variable that comes after the d. You need to either rewrite r in terms of m, or rewrite dm in terms of dr. Yes, you can write equations for differential factors by themselves. Here is a fairly rich example. Let's find the moment of inertia of a cone around its axis of symmetry. The cone will have radius R, height H, and uniform density p. The moment of inertia is int(r^2 dm), as it is for everything. What's going to change is the value of dm.



The first thing to understand is the physical meaning of dr and dm. dr refers to a thin slice of space between radius r and r + dr. In two dimensions, that's a thin ring. In three dimensions, it's a thin cylindrical shell. dm is the mass enclosed within that space. The volume of that space can be calculated as the area of the cylinder times the thickness, which is dr. We need to know the radius and height of the cylinder. The radius is just the integrand, r. The height is some portion of the total height of the cylinder that decreases as we move away from the center.



(R - r)/R = h/H

h = H(R - r)/R

= H - (H/R)r

V = pi r^2 h dr

= (pi r^2 H - pi r^3 (H/R))dr



To find the mass simply multiply by density. dm = (pi pH r^2 - pi pH/R r^3) dr. Now you can replace dm in the integral.



int(r^2 (pi pH r^2 - pi pH/R r^3) dr)



See that we still have a d-something term, so it's a legitimate integral. Now split it up into two integrals and extract CONSTANT factors.



int(r^2 pi pH r^2 dr) - int(r^2 pi pH/R r^3 dr)

= pi pH int(r^4 dr) - pi pH/R int (r^5 dr)

= pi pH r^5/5 - pi pH/R r^6/6



The range of the integral is 0 to R, and it 0 it evaluates to 0, so just substitute R for r.



pi pH R^5/5 - pi pH/R R^6/6

= pi pH R^5/5 - pi pH R^5/6

= pi pH R^5 (1/5 - 1/6)

= (1/30) pi pH R^5

Answer 4

Your question is not to clear.



If an object is moving linearly, there is no reason for moments, which is the 'rotational' mass or the resistance of a body to circular motion.



Also you cannot take dm or danything 'outside' the integral.

The danything is what you are integrating.