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A uniform horizontal rod of mass 2.2 kg and length 0.63 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I =


If an 8.5N force at an angle of 47◦ to the horizontal acts on the rod as shown, what is the magnitude of the resulting angular accel- eration about the pivot point? The accelera- tion of gravity is 9.8 m/s2 . Answer in units of rad/s2

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  1. Angular acceleration = (Torque)/I

    They have given you the formula for I:

    M L^2/12

    I would need to see the figure to tell you what the torque about the spin axis is.

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