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 the horizontal position.What is the angular acceleration of the rod at the instant the rod makes an angle of 70$\degree$ with the horizontal? A)3.7 rad/s² B)1.3 rad/s² C)2.5 rad/s² D)4.9 rad/s² E)1.9 rad/s²

Question

Quizplus · Accepted Answer

The angular acceleration of the rod can be found using the equation:

α = (2/3)g/L sinθ

where α = angular acceleration, g = acceleration due to gravity (9.81 m/s^2), L = length of the rod (2.0 m), and θ = angle made by the rod with the horizontal (70 degrees converted to radians is 1.22 radians).

Plugging in the values, we get:

α = (2/3)(9.81)/(2.0)sin(1.22) = 2.5 rad/s^2

Therefore, the angular acceleration of the rod at the given instant is 2.5 rad/s^2.

A Uniform Rod Is 2 °\degree° With the Horizontal? A)3

A Uniform Rod Is 2 $\degree$ With the Horizontal?
A)3