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

Question

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

Quizplus · Accepted Answer

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 $\degree$