A uniform rod (mass = 2.0 kg, length = 0.60 m) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60$\degree$ below the horizontal?
A)15 rad/s2
B)12 rad/s2
C)18 rad/s2
D)29 rad/s2
E)23 rad/s2

Question

Quizplus · Accepted Answer

The torque due to gravity when the rod is 60° below the horizontal is τ = mg(L/2)sin(60°). The moment of inertia of the rod about the pivot is I = (1/3)mL². Using τ = Iα, we find α = (3g/2L)sin(60°). Substituting the given values, α ≈ 12 rad/s².

A Uniform Rod (Mass = 2 °\degree° Below the Horizontal? A)15 Rad/s2 B)12 Rad/s2 C)18 Rad/s2 D)29

A Uniform Rod (Mass = 2 $\degree$ Below the Horizontal?
A)15 Rad/s2
B)12 Rad/s2
C)18 Rad/s2
D)29