Write the equations of motion of a pendulum consisting of a thin, 2 kg stick of length suspended from a pivot. How long should the rod be in order for the period to be exactly 1 sec
Question
Question
Write the equations of motion of a pendulum consisting of a thin, 2 kg stick of length suspended from a pivot. How long should the rod be in order for the period to be exactly 1 sec
Answer:
3g/(8π²) ≈ 0.372 m
Explanation:
Draw a free body diagram. There is a weight force at the center of the pendulum.
Sum of the torques about the pivot:
∑τ = Iα
mg (½ L sin θ) = (⅓ mL²) α
3g sin θ = 2L α
α = 3g/(2L) sin θ
For small θ, sin θ ≈ θ.
α = 3g/(2L) θ
θ” = 3g/(2L) θ
The solution of this differential equation is:
θ = θ₀ cos(√(3g/(2L)) t)
So the period is:
T = 2π / √(3g/(2L))
If the period is 1 second:
1 = 2π / √(3g/(2L))
√(3g/(2L)) = 2π
3g/(2L) = 4π²
L = 3g/(8π²)
L ≈ 0.372 m