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A rolling hollow sphere

  1. Dec 6, 2006 #1
    A hollow spherical shell with mass 1.50kg rolls without slipping down a slope that makes an angle of 40.0degree angle with the horizontal.
    -Find the magnitude of the acceleration of the center of mass of the spherical shell

    I am really confused on how to go about this problem. I know it has to do with inertia, but wow, i just don't know where to start.
  2. jcsd
  3. Dec 7, 2006 #2
    U used the term 'rolls without slipping'. I thought of using rotational mechanics to solve it but u are not given the radius of the spherical shell. Thus, i shall treat it like a case of linear motion down the slope. Resolving forces and applying Newton's 2nd law on the shell, we have Normal reaction force = mg cos 40 and mg sin 40 = ma. Thus a = g sin 40 = 9.8 sin 40. If u want to find angular acceleration of the shell, u need to have the shell's radius and use the relation, linear acceleration = spherical radius x angular acceleration.
  4. Oct 22, 2007 #3
    No, you don't need the radius for this one. This is a hollow shell and it's moment of inertia is 2/3mr^2 = I
    Fy = 0
    Fx = mgsin(theta) - f = ma_cm

    I get from Fx=mgsin(theta)-f=ma_cm (1)
    and fR = 2/3mra_cm (2)

    Solve f from (1).
    Substitute that into (2) and then solve for a_cm

    I get a_cm = (3gsin(theta))/5
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