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Moment of Inertia of a rod and two spheres

  1. Nov 21, 2014 #1
    1. The problem statement, all variables and given/known data
    A barbell that consist of a long thin rod of mass M and length L is attached to two uniform spheres on each end. Both spheres have mass M and (1/3)L. The sphere on the left is hallow (spherical shell) and the sphere on the right is solid. What is the moment of inertia for the system as it rotates about an axis about the center of hallow sphere if M=1.0 kg and L = 1.0 m?

    2. Relevant equations
    I assume the parallel axis theorem since it's not rotating about the center of mass.

    3. The attempt at a solution
    What I've set up is: I = Inertia, cm = center of mass

    I = Icm + MD2 = (1/12)ML2 + MM((L/2))2

    This is where I am stuck. This is for the rod but I am assuming I ignore the hallow sphere and add the moment of intertia for the solid sphere. If someone could explain what's going on in this problem or the basics of how to solve moment of inertia problems. Thanks
     
  2. jcsd
  3. Nov 21, 2014 #2

    SteamKing

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    Your OP is not clear. What does (1/3)L represent?
     
  4. Nov 21, 2014 #3
    My apologies. The spheres are 1/3rd the length of the bar. L is the length of the bar. So the length of the spheres are (1/3)L
     
  5. Nov 21, 2014 #4
    I dont understand your attempt...it is for rod I guess...so MD^2 is not [itex]M^2(\frac{L}{2})^2[/itex] , if I understand in good way ( and (1/3)L is diameter of spheres): Just use your theorem for rod and ball on right hand side. Moment of inertia of hollow and solid ball is same. Than [itex]J=J_{leftball}+J_{rod}+J_{rightball}=...[/itex] . For instance: [itex]J_{rod}=\frac{1}{12}ML^2+MD^2[/itex] and your D is not nice [itex]D=\frac{L}{2}+\frac{L}{3*2}[/itex], when the latter term is radius of ball (axis of rotation).
    The result what I tried is terrible number...try and write yours.
    Hope Im right :)
     
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