A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (The moment of inertia of the rod about this axis is ML2/3.) The rod is released when it makes an angle of 37$\degree$ with the horizontal. What is the angular acceleration of the rod at the instant it is released?
A)9.8 rad/s2
B)7.4 rad/s2
C)8.4 rad/s2
D)5.9 rad/s2
E)6.5 rad/s2

Question

Quizplus · Accepted Answer

Explanation:The torque about the pivot is:$$	au = I\alpha = \frac{1}{3}ML^2\alpha$$The net force on the rod is:$$F = mg\sin	heta$$The acceleration of the rod is:$$a = \frac{F}{m} = g\sin	heta$$The angular acceleration of the rod is:$$\alpha = \frac{a}{L} = \frac{g\sin	heta}{L}$$$$\alpha = \frac{(9.8 m/s^2)\sin37\degree}{1.2 m} = 7.4 rad/s^2$$

A Thin Uniform Rod (Length = 1 °\degree° With the Horizontal

A Thin Uniform Rod (Length = 1 $\degree$ With the Horizontal