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Please help me sove this physical pendulum problem?
Problem 49. It is a question in Fundamentals of Physics Extended version 8th edition
I would be able to solve it if the pivot point was at the end of the stick. But I can't, because there is a part of the stick on both sides of the pivot point.
Please show me how you derived the answer
2 Answers
- EMLv 77 years agoFavorite Answer
Good problem.
(a)
The weight of the stick is solely responsible for its torque.
Iθ" = mgx sinθ
Find the moment of inertia of the stick at different pivot points via the parallel axis theorem.
(mL²/12 + mx²)θ" = mgx sinθ
(L²/12 + x²)θ" = gx sinθ
For small oscillations sinθ ≈ θ.
θ" = [gx / (L²/12 + x²)] θ
ω² = gx / (L²/12 + x²)
d(ω²)/dx = [(L²/12 + x²)(g) - (gx)(2x)] / (L²/12 + x²)² = 0
g(L²/12 - x²) = 0
x = L/sqrt(12)
(b)
ω² = [gL/sqrt(12)] / (L²/12 + L²/12)
ω² = [g/sqrt(12)] / (L/6)
(2π/T)² = g sqrt(3) / L
T = 2π sqrt{L / [g sqrt(3)]}
- Anonymous7 years ago
(a)
The weight of the stick is solely responsible for its torque.
Iθ" = mgx sinθ
Find the moment of inertia of the stick at different pivot points via the parallel axis theorem.
(mL²/12 + mx²)θ" = mgx sinθ
(L²/12 + x²)θ" = gx sinθ
For small oscillations sinθ â θ.
θ" = [gx / (L²/12 + x²)] θ
ϲ = gx / (L²/12 + x²)
d(ϲ)/dx = [(L²/12 + x²)(g) - (gx)(2x)] / (L²/12 + x²)² = 0
g(L²/12 - x²) = 0
x = L/sqrt(12)
(b)
ϲ = [gL/sqrt(12)] / (L²/12 + L²/12)
ϲ = [g/sqrt(12)] / (L/6)
(2Ï/T)² = g sqrt(3) / L
T = 2Ï sqrt{L / [g sqrt(3)]}
