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Moment of inertia and rotational motion

  1. Mar 1, 2015 #1
    Hello,
    I am currently attempting to cover rotational motion using Halliday's Fundamentals of Physics.
    I understand very well the concept of moment of inertia as defined as the sum Σmi*ri2.
    However, the textbook argues that if there are too many particles, the moment of inertia can be defined as an integral. And that is not something that I clearly understand.
    I believe that my problem stems from a lack of understanding of the concept of integrals. I am currently taking a calculus class but the teacher hasn't yet covered this concept (although I believe I have a basic understanding of what it means thanks to some internet research).
    In essence, what I don't understand is the following statement: an infinite sum can be expressed as an integral.
    Could someone explain to me why this is true?
    I faced the same problem when I covered the concept of center of mass (which can also be approached as an infinite sum when there are too many particles). I figured out, however, that if there were a function f(xi)=mi (giving the mass mi of the ith particle, the sum ximi could indeed be expressed as an integral (since it would be the area under the curve).

    However, I have failed to approach the sum related to the moment of inertia in the same way. Indeed, f(m)=r2 and f(r2) are not actual functions since there could be several y's for a single x.

    Could you help me figuring out why integrals can be used to express infinite sums? Was my approach to understand to the sum related to the center of mass incorrect?

    Thank you very much,
    (I don't know if I was right in posting my question in this section since it is more of a calculus question than a physics one; but since it involves physics concepts, I figured out it was legitimate to post it here. My apologies if I was wrong)
     
  2. jcsd
  3. Mar 1, 2015 #2

    sophiecentaur

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    IF you haven't done Integral Calculus at school / college yet then you need to wait before you can fully understand what's going on here. Nevertheless, if you can understand the idea of simple summation of the effect of a small / finite number of masses, linked together then you have the Physics of the situation sorted. Extending this to a Definite Integral is just a bit of Mathematical Wizardry which happens to give the right answer.
    Your Physics Text Book will not have space to explain how Calculus works - it will just give you the results of the operations using Calculus.
    If you want more appropriate information about this then you could probably do best by a Google search to find a website that approaches the topic at a level that suits your particular level of Maths knowledge. You could even search through the Maths Forums on this site to find something to your liking; something that fall within your particular area of knowledge. That may involve a lot of slogging through other stuff though.
     
  4. Mar 1, 2015 #3
    Well, I wouldn't have posted this question hadn't my preliminary Google research been unsuccessful.
    On the internet, functions were always involved; but here, I don't see one, for the reasons I stated above.
    And yes, I know that what is written in the textbook is right, but my question aims at understanding why it is right, which is I believe the best way for one to acquire knowledge...
     
  5. Mar 1, 2015 #4

    sophiecentaur

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    Do you have a maths text book? The way that Integral Calculus is developed is described in all text books of the right level. As I wrote before, this isn't Physics, it's Maths that you need for this. ~What search criteria have you been using? Google is only as good as the questions you ask it.
     
  6. Mar 1, 2015 #5

    PeroK

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    I guess your Google search didn't turn up anything like:

    http://www.mathsisfun.com/calculus/integration-introduction.html
     
  7. Mar 1, 2015 #6
    Actually, I did find this wesbite. However, it is another example of what I said earlier: on the internet, functions are involved. When functions are involved, integration makes sense because the sum is the area under the curve.
    However, here, I don't see a function. Thus, I don't understand why using integration makes sense. Integration is related to the notion of limit; however, here, I don't see why limits are involved.

    I genuinely searched the internet for an answer. I however didn't find something that I could apply here. This is why I thought it'd make sense for me to ask on this forum. I didn't expect sarcasm to be the appropriate answer to a genuine interest in learning. :/
     
  8. Mar 1, 2015 #7

    PeroK

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    Does your physics text mention the "mass density (function)"? You're correct in that in order to use integration, you must define a function to integrate. The mass density function is the limit of a larger and larger number of particles of smaller and smaller mass: in the limit distributed continuously across an area or volume.
     
  9. Mar 1, 2015 #8
    No, it does not. Here is the explanation the book gave the first time it expressed an infinite sum as an integral (when discussing center of mass):
    "An ordinary object, such as a baseball bat, contains so many particles (atoms) that we can best treat it as a continuous distribution of matter. The “particles” then become differential mass elements dm, the sums of Eq. 9-5 (i.e. x(com)=1M ∑ mi*xi) become an integral, and the coordinate of the center of mass is defined as 1/M ∫ x dm"

    I don't quite understand the concept you are defining.
    When I searched Google for "mass density function", I got many websites discussing "probability density function". Is that what you are referring to? I cannot see "mass density function" written as is.

    Thanks for your answer by the way ;)
     
  10. Mar 1, 2015 #9

    PeroK

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    Your book is simply giving a brief summary of the transformation from a finite sum of point masses to an integral of a mass distribution. (It is the same mathematical idea as the transformation of a sum of finite probabilities to the integral of a probability density function.). I suspect you're supposed to accept that this is simply the standard "sum transformed to integral" process. The differential mass element is another way to treat continuous mass density.

    ##dm = \rho(x)dx## where ##\rho(x)## is the (mass) density (function).

    The quickest explanation of what's going on is perhaps to turn this round and use the definition of the integral to see that ##\int x \rho(x)dx## is the limit of ##\sum x_{i}m_{i}## when the object is viewed as a finite set of point masses. The integral needs the point masses to be described (approximately) as a continuous (density) function. This process will turn up in various guises all over physics (and probability theory).

    There's more here:

    http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html#cmc
     
  11. Mar 1, 2015 #10
    I'm getting more and more confused.
    The moment of intertia is the area under a curve, right? Then, of what curve? What kind of function is involved?
    Is there a way to approach this integral as I approached the integral of the center of mass?
    Thanks again for your help ;)
     
  12. Mar 1, 2015 #11

    PeroK

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    Your best bet is to learn the Integral Calculus. Mathematically, one application of integration is to find the area under a curve. Another is to find the volume of a solid of revolution. And a third is to calculate the centres of mass or moment of inertia of an object.

    The simplest case is where you have a bar of varying density along the x-axis. If you have ##y = \rho(x)## where ##\rho## is the density of the bar, then:

    ##\int \rho(x) dx = M## Area under the curve = Mass of the bar

    And:

    ##\frac{1}{M}\int x \rho(x) dx = X = ## Centre of Mass of the bar

    And:

    ##\int x^2 \rho(x) dx = I_x = ## Moment of Inertia of the bar about the origin.
     
    Last edited: Mar 1, 2015
  13. Mar 1, 2015 #12

    sophiecentaur

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    This is not an easy subject. You need approach the ideas involved in a logical order. You can't 'just learn' a small part of Calculus and expect to get the whole picture.
    The use of functions to describe Physical processes needs to be learned in steps. The fact that you are being confused by this (above) means you do not have enough of the Maths to understand it. There is no shame in this - you just need to do a suitable Maths course. There is no simple arm waving explanation and you are wrong to expect it.
     
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