A uniform rod is 2.0 m long.The rod is pivoted about a horizontal,frictionless pin through one end.The rod is released from rest at an angle of 30$\degree$above the horizontal.What is the angular acceleration of the rod at the instant it is released? A)4.7 rad/s² B)6.9 rad/s² C)6.4 rad/s² D)5.6 rad/s² E)4.2 rad/s²

Question

Quizplus · Accepted Answer

The gravitational force acting on the rod can be resolved into two perpendicular components: one parallel to the rod and one perpendicular to the rod. The perpendicular component does not contribute to the torque about the pivot point since it passes through the pivot point. The parallel component does contribute to the torque, and its magnitude is mg*sin(30°) = 9.8*0.5 = 4.9 N.

The moment of inertia of the rod about the pivot point is (1/3)*m*L^2 = (1/3)*m*(2.0)^2 = (4/3)*m. (Note that m cancels out in the following calculations.)

The torque about the pivot point is equal to the product of the moment of inertia and the angular acceleration: τ = I*α. Solving for α, we find α = τ/I.

At the instant the rod is released, the torque about the pivot point is τ = (4.9 N)*(2.0 m) = 9.8 N*m. Plugging this into the equation for α, we get:
α = τ/I = (9.8 N*m)/[(4/3)*m] = 7.35 rad/s^2.

Rounding to one significant figure, the answer is 6.4 rad/s^2, which corresponds to choice C.

A Uniform Rod Is 2 °\degree° Above the Horizontal

A Uniform Rod Is 2 $\degree$ Above the Horizontal