1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: Deriving the moment of inertia of a sphere

  1. Feb 7, 2017 #1
    1. The problem statement, all variables and given/known data
    A uniform spherical ball of mass M and radius R, initially spinning about a horizontal axis with angular speed ω0, is placed gently on the floor. The initial center-of-mass velocity of the ball is zero. Derive that the moment of inertia ICM of the ball about an axis passing through its center of mass is given by ICM = (2/5)MR2 .
    I = ∫ρ2dm

    2. Relevant equations
    I = ∫∫∫(ρ22dρsinθdθdΦ

    3. The attempt at a solution
    I get I = (3/5)MR2. I noticed that this is almost correct apart from a moment of inertia for a point particle, so I subtracted MR2. I reasoned that since the ball is touching the ground and rotating about its axis doesn't it have "two moments"? One touching the ground and the other on its axis. And so to not include the part touching the ground would make sense to get just a sphere. Or is this not right? I saw the "correct" way of adding up disks but does this also work? Also another problem asks me to do the same but for a disc rolling down an incline(except this time the moment of inertia was to be found about the point where the ball touches the ground). I did the same thing as for the sphere but obviously with a different integral and then I added MR2 to MR2/2 to get the correct answer of 3MR2/2.
  2. jcsd
  3. Feb 7, 2017 #2
    Nevermind. Don't pay attention to that. That's just me trying to justify the five hours I spent figuring out why the hell the problem wasn't working out by simply doing the integral. Can you please instead explain the rationale of the correct method? As in why is the integral the same but with a 1/2 out front? Etc...
  4. Feb 7, 2017 #3


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Looks like you might be using the symbol ρ for two different quantities.

    What does ρ stand for in I = ∫ρ2dm?
    What does ρ stand for as a spherical-coordinate variable?
  5. Feb 7, 2017 #4
    Sorry about that. I = ∫∫∫(ρ2)r2drsinθdθdΦ. ρ is the distance from rotation axis and I'm assuming r is the same. Now that I think of it they are not. r is for the radius and ρ is for the distance from center of mass to rotation? In that case if I were to try and derive the correct equation to find moment of inertia, I would first find the center of mass of any object and use that to find ρ (the distance from the CoM to the rotation axis)? To find the moment of inertia of a uniform sphere about an axis through its center implies that the CoM and rotation axis lie along the same line. But it is also rotating about the incline, so I could take the distance ρ and r to be the same?
    Last edited: Feb 7, 2017
  6. Feb 7, 2017 #5


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    r is the distance from the origin to the element of mass dm. ρ is the distance from the axis of rotation to the element of mass. If you take the axis of rotation to be the z-axis of the spherical coordinates and take the origin to be at the center of the sphere, you can express ρ in terms of r and θ.

    You don't need to worry about the location of the center of mass for the evaluation of the integral ∫∫∫(ρ2)r2drsinθdθdΦ. But it should be clear that the center of mass is at the center of the sphere.
  7. Feb 8, 2017 #6
    ρ=rsinθ then. Doing the integral still doesn't work. I don't know what to do about my limits for r. I know it shouldn't just be r but its not rcosθ (which is along the axis of rotation)
  8. Feb 8, 2017 #7
    Solved, thank you
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted