Posted by **Latane** on Sunday, February 12, 2012 at 5:48pm.

A uniform horizontal rod of mass 2.4 kg and

length 0.86 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 =

(m*l^2)/12

If a 6.8 N force at an angle of 69 to the hor-

izontal acts on the rod as shown, what is the

magnitude of the resulting angular acceleration about the pivot point? The acceleration

of gravity is 9.8 m/s2 .

Answer in units of rad/s2

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