A uniform rod of mass M = 1.2 kg and length L = 0.80 m,lying on a frictionless horizontal plane,is free to pivot about a vertical axis through one end,as shown.The moment of inertia of the rod about this axis is given by (1/3)ML².If a force (F = 5.0 N,$\theta$ = 40$\degree$)acts as shown,what is the resulting angular acceleration about the pivot point? A)16 rad/s² B)12 rad/s² C)14 rad/s² D)10 rad/s² E)33 rad/s²

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The Answer of A uniform rod of mass M = 1.2 kg and length L = 0.80 m,lying on a frictionless horizontal plane,is free to pivot about a vertical axis through one end,as shown.The moment of inertia of the rod about this axis is given by (1/3)ML².If a force (F = 5.0 N,$\theta$ = 40$\degree$)acts as shown,what is the resulting angular acceleration about the pivot point? A)16 rad/s² B)12 rad/s² C)14 rad/s² D)10 rad/s² E)33 rad/s²

A Uniform Rod of Mass M = 1 θ\thetaθ = 40 °\degree°

A Uniform Rod of Mass M = 1 $\theta$ = 40 $\degree$