Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    7
    ReputationRep:
    Hi, I keep getting stuck on this seemingly simple question and I cannot figure out why. Firstly, I understand that it's possible to calculate this by finding the moment of inertia of a disc and then integrating across the radius - however I'm unsure why my method fails. If somebody could point out the error I'd be extremely grateful.

    So, "Show that the moment of inertia of a solid sphere of uniform density is  I = \frac{2}{5}MR^2 "

    Taking the upward axis of the sphere as the z-axis we have the integral form of moment of intertia as:  \int^M_0 z^2 dm

    Let the radius of the sphere be  R .
    Create infinitesimally thin discs of mass  dm and width  dz which are at a height  z .

     dm =  \rho \pi r^2dz

    substituting this into the integral gives:
     2\rho\pi \int^R_0 z^2r^2dz The factor of 2 is because we're only integrating across one half of the sphere.
    Using Pythagoras we can get  r^2 = R^2-z^2

    Therefore the integral is  2\rho\pi \int^R_0 z^2(R^2-z^2)dz
    Computing this gives  \frac{4}{15} \rho \pi R^5
    But  \rho=\frac{3M}{4\pi R^3}
    Plugging this in gives
     I=\frac{1}{5}MR^2
    Which obviously isn't the correct answer. If anybody knows what I'm missing I'd be very grateful!
    Offline

    11
    ReputationRep:
    (Original post by Aiden223)
     I=\frac{1}{5}MR^2
    Which obviously isn't the correct answer. If anybody knows what I'm missing I'd be very grateful!
    Your approach seems to be completely wrong, I'm afraid. When you compute \int z^2 dm, you are adding up the contributions due to mass dm that is a distance z from the axis of rotation.

    With your method, the distance from the axis of the mass of the discs is not constant, as far as I can see.
    • Thread Starter
    Offline

    7
    ReputationRep:
    Ooh, so you're not meant to use the distance to the centre of mass? But rather the distance to the centre of mass axis? Which in this case is the z axis?
    Offline

    11
    ReputationRep:
    (Original post by Aiden223)
    Ooh, so you're not meant to use the distance to the centre of mass? But rather the distance to the centre of mass axis? Which in this case is the z axis?
    There are two ways to compute MOI via integration:

    a) You consider an infinitesimal mass dm all of which is the same distance r from the axis of rotation - then you compute \int r^2 dm.

    In your method, your dm is spread out over flat discs, whose centres are on the axis of rotation. This means that some of your dm is closer to the axis than other bits of it. So you wont find the correct MOI as the condition above is not satisfied.

    You can fix this by considering cylinder shells of mass dm and radius r, whose centre line is the axis of rotation - then all of the mass is the same distance from the axis.

    b) You can use an already known result for the MOI of a particular thin body, then integrate over some range that adds up the MOIs of copys of that body that equate to the object that you are considering e.g. you can use the result the MOI of a thin spherical shell radius r, thickness dr - this has infinitesimal MOI

    dI = f(m,r) dr

    where f(m,r) is a function that you can look up. You then compute

    I = \int dI = \int_{r=0}^{r=R} f(m,r) dr
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What's your favourite Christmas sweets?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.