Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Moment of inertia of disc plus point mass at centre

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data

    A uniform disc radius a, mass m
    there is rotation about an axis (z) tangental to the disc and in the plane of the disc.
    a point mass m is placed at the centre of the disc.

    what is the new moment about the axis z

    also show that the period of oscillations will decrease by SQRT(9/10) when the point mass is added

    3. The attempt at a solution

    moment of inertia of just the disc i found to be = (5/4)ma^2
    period before point mass added =

    T=2pi/w = 2pi*sqrt(m/k)
    torque = r x F = mga sin(theta) but we are told small angle approx applies so = mga*theta
    also F=k*theta
    so k = mga

    therefore T = 2pi*sqrt(m/mga) anf because of rotation the m on top = I about z axis
    so T = 2pi*sqrt(I/mga) = 2pi*sqrt(5a/4g)

    I dont know how to get the I when point mass has been added.

    also is the new torque = 2mga * theta so that k = 2mga ?
    Last edited: Apr 5, 2010
  2. jcsd
  3. Apr 5, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi indie452! :smile:

    (have a pi: π and a theta: θ and an omega: ω and an alpha: α and a tau: τ and a square-root: √ :wink:)
    Moment of inertia is additive (about the same axis), so you just add the moment of inertia for the point mass, which of course is ma2, to the moment of inertia of the disc

    (btw, your formula for T is correct, but I'm a bit lost as to how you got there :confused: … simplest is mgaθ = τ = Iα = Iθ'', so T ~ √(I/mga).

    For the new T, use the new total mass, the new total moment of inertia (and the new distance to the overall centre of mass, though of course that's the same). :wink:
  4. Apr 5, 2010 #3

    ok so i got I = (9/8) *ma^2

    and so T = 2[tex]\pi[/tex]*sqrt(a/g)*sqrt(9/8)

    but sqrt(5/4) decreased by a factor of sqrt(9/10) is sqrt(50/36)?
  5. Apr 5, 2010 #4


    User Avatar
    Science Advisor
    Homework Helper

    hey indie452 ! :smile:

    (what happened to that π and √ i gave you? :rolleyes:)
    How did you get 9/8 ma2 ? :confused:
  6. Apr 5, 2010 #5
    sorry meant 9/4 i typed it in wrong
  7. Apr 5, 2010 #6
    btw how do you put the symbols underneath you post like that automatically
  8. Apr 5, 2010 #7


    User Avatar
    Science Advisor
    Homework Helper

    hey indie452! :smile:

    (to get those symbols, if you have a Mac you can type alt-v for √ and alt-p for π, otherwise just copy them from my post above, and paste them in :wink:)
    ah!! :biggrin:

    ok … that's right … so now what do you get for the factor by which the period of oscillation will decrease? :smile:
  9. Apr 5, 2010 #8
    I get a factor of √(10/9) but it wanted a factor of √(9/10)
  10. Apr 5, 2010 #9


    User Avatar
    Science Advisor
    Homework Helper

    hmm :rolleyes: … let's see …

    T ~ √(I/mga) …

    (checking dimensions, that's √(ML2/M(L/T2)L) = T, so that's ok :biggrin:)

    I is multiplied by 9/5, 1/m is multiplied by 1/2, g and a are the same, total 9/10 …

    how did you get 10/9 ? :smile:
  11. Apr 5, 2010 #10

    I of disc = (5/4)Ma^2
    I of point = Ma^2

    so I total = (9/4)Ma^2
    and mga becomes = 2mga

    so T ~ √I/2mga ~ √a/g * √9/8

    and i think the √10/9 is wrong now anyway....how did you get the 9/5? cause the question says its decreased by a factor of √9/10
  12. Apr 5, 2010 #11


    User Avatar
    Science Advisor
    Homework Helper

    (try using the X2 tag just above the Reply box :wink:)

    yes, so new I = 9/5 times old I
    and new m = 2 times old m
    sorry, I can't follow that at all :confused:

    try again. :smile:
  13. Apr 5, 2010 #12
    i see what you're getting at...you're doing it interms of old I...i get it...i was trying to just get the new T and the compare the two which is why i didnt get where the 9/5 was coming from.

    what i tried to do was

    T = 2π√(9/4Ma^2)/(2Mag)
    = 2π√(9a)/(8g)
    = 2π√a/g*√9/8

    and old T = 2π√a/g*√5/4

    old T/new T = √10/9

    but i see what you've done
    newT = 2π√(9/5)I/2Mag where I = old I
    = 2π√9I/10Mag
    = oldT*√9/10
  14. Apr 5, 2010 #13
    I had another question btw that I was wondering if you could answer,

    when calculating the combined I, the I(point) = Ma^2

    why, (when the axis we are rotating around is not through the centre of mass of the point) do we not use the parallel axis theorem and say
    I = I(through cm) + Mr^2

    hang on i think i answered my own qu while typing....is it because the I(through cm) = zero because it is defined as the SUM(mass element*distance to element2) and the distance to the element is zero because it is an infinitesimally small point?
  15. Apr 5, 2010 #14
    Yep, that's right. The moment of inertia of a point mass about its own center of mass is 0! What's there to rotate about that center, exactly? :)
  16. Apr 5, 2010 #15


    User Avatar
    Science Advisor
    Homework Helper

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook