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The angular inertia of a disk

  1. Jun 20, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the angular inertia of a disk (cylinder) with...
    Maximal radius, rmax = 10cm
    Thickness, h = 4cm
    density, d = 3g/cm3
    rotational axis in its centre (like the axis of clock hand in a disk shaped clock)

    Please show and explain your procedure in finding the answer.


    2. Relevant equations
    Angular inertia of point mass = (radius from rotational axis)2 × mass of point mass

    3. The attempt at a solution
    I see that this question obviously involve calculus, but I don't have any real understanding of calculus. I think we need to consider the cylinder as composed of infinitely many concentric tubes each with a negligible volume (the distance between the tubes' inner and outer radius is assumed to be zero) and then add the individual angular inertia of all of the tubes.

    I apologize for any spelling/grammar mistake and any misuse of terminology, please explain any calculus stuff that you would use.
     
    Last edited: Jun 20, 2015
  2. jcsd
  3. Jun 20, 2015 #2

    haruspex

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    First, you need to be clear what axis is to be usd for finding the moment of inertia. I assume the axis is perpendicular to the disk.
    Yes, calculus is necessary for determining it. But are you expected to apply calculus to derive a formula, or are you simply expected to know the formula? Presumably you have been taught one or the other.
     
  4. Jun 20, 2015 #3
    Deriving a formula would be nice.
     
  5. Jun 20, 2015 #4

    haruspex

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    Ok, so follow the plan you outlined. Consider a cylindrical shell, inner radius r, thickness dr. If the density is ##\rho##, what is the mass of the shell? What is its MoI about the axis?
     
  6. Jun 20, 2015 #5
    The mass of the shell is ( (r+dr)2 - r2) × π × ρ × h which, because dr is nearly zero, is also nearly zero.
    and you would have to explain what Mol is.
     
  7. Jun 20, 2015 #6

    SteamKing

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    Just for yucks, how did you come across this problem, if you are calculus-challenged?
     
  8. Jun 20, 2015 #7
    Well, that's not important nor relevant, but just for yucks, it's because I have to do a scientific investigation like that so they will let me go to grade 12

    Oh, by the way, I found the formula: Angular Inertia of a Disk = 0.5 * Mass * Radius^2 (didn't use fancy notations because I'm tired)
    Now just someone explain it.
     
  9. Jun 20, 2015 #8

    Nathanael

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    That's correct but you will need to simplify this expression if you want to calculate the formula.

    (MoI=) Moment of Inertia is a common term for what you called "angular inertia."

    Can you find the moment of inertia of the cylindrical shell?

    This is true, but it is not a problem. The final answer can still be finite because you are going to sum up an infinite amount of infinitesimal terms.

    Do you know about integration? This is the basic idea (as applied to this problem)... You break up the cylinder into cylindrical shells (or "tubes" as you call them) each of thickness Δr. In order for all the Δr's to add up to the total radius R, there must be R/Δr shells. The reason for breaking it up like this is so you can approximate the MoI of each shell and sum them all up. Then you take the limit of this sum as Δr goes to zero. The approximation of the MoI of each shell should "become exact" in this limit. (Note that as Δr goes to zero, the number of shells, R/Δr, goes to infinity.)

    No one actually calculates the discrete sum and then takes the limit, as calculus provides a more graceful and more versatile (yet equivalent) approach (called integration). But if you don't know how to integrate, then this method will give the same answer. [But you will need to know how to calculate discrete sums, things like [itex]\Sigma_{n=1}^{x}(n^3)[/itex]]
     
  10. Jun 20, 2015 #9
    Thank you, may I say that this is the first helpful answer I got from this forum? Anyways, I understand what intergration is I just don't know how it works in this case. I have seen "∫" and that it means "the area under a curve starting from the x-value under the thingy to the x-value above the thingy, but I have no idea how to apply this here.

    Should I visualize a graph of M.o.I.shell over radiusmin.shell and then calculate the area under the curve from zero to radiusdisk? I have a funny feeling that the final answer--0.5*Mass*Radius^2 has something similar to distance=0.5*acceleration*time^2, and that the coefficient--0.5--and the exponent--2 have relations. Just speculating, if that can help you see what's going on in my head.
     
  11. Jun 20, 2015 #10

    Nathanael

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    This is a good idea (except you would want to go from zero to radiusdisk) but if you try to find the shape of the curve you will find that there's no way to find the area (without calculus).

    Start anyway: Find the M.o.I. of a cylindrical shell as a function of it's radius r, thickness dr, and height h.

    Then, if you knew calculus just a little better, I would tell you, "integrate it from 0 to R" which would mean that you take the sum of the M.o.I. of all infinite of these infinitely thin shells and that would be your answer. (And yes it can be interpreted as an area under a curve.)
    Unfortunately you don't know calculus better so it's a little tricky as to how to explain what to do.

    Anyway I still want to see you find that function which you would integrate.

    It's a good guess, but no they are not related. That formula for distance has two assumptions: 1) acceleration is constant (which implies the speed curve is a straight line) and 2) the speed starts out at zero (which means it is a straight line through the origin). You can see from this that the area under the speed curve (a.k.a. the distance) from 0 to some time T can be interpreted as the area of a triangle. The bottom side of the triangle is length T, the height of the triangle is a*T (because that's the velocity at time T) therefore the distance (area) is 0.5*a*T2

    Unfortunately this problem is not so nice as to be the area of a triangle. It is instead the area under some (not straight) curve. It is just a coincidence that these two formulas take a similar form.
     
  12. Jun 20, 2015 #11
    When dr→0, M.o.I.shell = massshell × radiusshell = ( ( (r+dr)2 - r2 ) × π × ρ × h ) × r
     
  13. Jun 20, 2015 #12

    haruspex

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    The trick with integral calculus is that 'first order small quantities' (like dr) will add up to something interesting, so second order small quantities (like dr2) can be ignored in the limit.
    Expand ( (r+dr)2 - r2) to get 2r dr + dr2, then throw away the dr2 to get 2r.dr Summing up these shells we get ##m = \int_0^R \pi\rho h 2r dr##. Can you integrate that? Can you follow the same procedure for the moments of inertia of the shells?
    I wouldn't say it 'means' an area under a curve. It means the limit of a sum as the sum gets taken over finer and finer pieces.
    But if you have an integral ∫f(x).dx and you plot the graph y = f(x) then, yes, the integral will equal an area under the graph.

    Sort of. The 1/2 in s = (1/2) at2 comes from the fact that the integral of xn.dx is ##\frac 1{n+1}x^{n+1}##, so integrating at.dt gives (1/2) at2.
    In the same way, when you do the integral to find the mass of the cylinder the 2r in the integrand turns into r2.
    When you integrate to find the MoI, you start with 2r3 and end with 2r4/4 = r4/2. So it is analogous, but really it's a division by 4 here, not a division by 2.
     
  14. Jun 20, 2015 #13

    Nathanael

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    Compare this to your relevant equation:
    Does it look the same?

    Anyway I think you've made the important realization that all the points of mass of a cylindrical shell (provided it's "infinitely thin") are at the same distance from the rotation axis (therefore your relevant equation 2 for a point mass also applies to a cylindrical shell). This is the whole reason why splitting it up into cylindrical shells is useful.
     
  15. Jun 20, 2015 #14

    SteamKing

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    Thanks for answering my question. It helps to know a little background sometimes.

    I'll give you another useful tip. The correct term is "integration", not "intergration".

    These small differences mean a lot, especially when dealing with calculus. :wink:
     
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