Answer:
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
