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A spherically symmetric object with radius of .7m and mass of 1.6kg rolls without slipping accross a horizontal surface with velocity of 1.7m/s. It then rolls up an invline with an angle of 28degrees and comes to rest a distance d of 2.3m up the incline, before rolling back down. Find the moment of inertia of this object about an axis through its center of mass.

Isn't I just= mr^2?


    No, not for a solid sphere.

    You can find it from the experiment.

    Conservation of energy:

    final energy= energy at top of incline
    1/2mv^2 + 1/2 I w^2= mgh= 1.6*9.8*2.3sin28

    but w=v*r, so
    1/2 mv^2+1/2 I v^2 r^2= 1.6*9.8*2.3sin28
    Solve for I.

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