A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The moment of inertia of the rod about this axis is given by (1/3) ML2. The rod is released from rest at an angle of 30° above the horizontal. What is the angular acceleration of the rod at the instant it is released?

Angular acceleration = 6.37rad/sec²

Approximately, Angular acceleration =

6.4 rad/sec²

Explanation:

Length of the rod = 2.0m long

Inclination of the rod (horizontal) = 30°

Mass of the rod is not given so we would refer to it as = M

Rotational Inertia of the Rod (I) = 1/3ML²

Angular Acceleration = ?

There is an equation that shows us the relationship between Torque and Angular acceleration.

The equation is:

Torque (T) = Inertia * Angular Acceleration

Angular acceleration = Torque : Inertia

Where:

Torque = L/2 (MgCosθ)

Where M = Mass

L = Length = 2.0m

θ = Inclination of the rod (horizontal) = 30°

g = Acceleration due to gravity = 9.81m/s²

Inertia = 1/3ML²

Angular Acceleration = (Mass * g * Cos (30°) * (L:2)) : 1/3ML²

Angular Acceleration =

(3 * g * cos 30°) : 2 * L

Angular Acceleration = (3 * 9.81m/s² * cos 30°) : 2 * L

Angular Acceleration = 3 * 9.81m/s² * cos 30°) : 2 * 2.0m

Angular Acceleration = 6.37rad/sec²

Approximately Angular Acceleration =

6.4rad/sec²