University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

182

Pages:

5
Author:

Jabin F.

# M.I. of Thin Hollow Sphere Put R₁ = R and take limit R₁ → R as directly substituting gives 0/0 indeterminate form. I = (M/4πR²) ∫(R² + R₁²)R²sinθ dθ dφ = 8M/5R² ∫(R² + R₁²)R²sinθ dθ dφ = 8MR²/5 Using L'Hospital Rule: I = (8M/5R²) lim(R₁→R) (5R₁⁴) / 3R₁² = 8M/5 * 5/3R² = 8MR²/15 # M.I. of Thin Hollow Sphere Put R₁ = R and take limit R₁ → R as directly substituting gives 0/0 indeterminate form. I = (M/4πR²) ∫(R² + R₁²)R²sinθ dθ dφ = 8M/5R² ∫(R² + R₁²)R²sinθ dθ dφ = 8MR²/5 Using L'Hospital Rule: I = (8M/5R²) lim(R₁→R) (5R₁⁴) / 3R₁² = 8M/5 * 5/3R² = 8MR²/15 ② A thin rod of length l has linear mass density λ = λ₀x where x is measured from one end. Calculate M.I. of rod about an axis through centre and perpendicular to its length. Solution: Linear mass density: λ = dm/dx = λ₀x dm = λ₀x dx Now in polar co-ordinates: Iyy = ∫dm x² Iyy = ∫(σ₀x)x² dx dy Iyy = σ₀ ∫(x³ dx dy) Iyy = σ₀ ∫(x³ √(R² - x²) dx) Iyy = σ₀ ∫(x³ √(R² - x²) dx) from 0 to R Iyy = σ₀R⁴ ∫(sin³θ cosθ dθ) from 0 to π/2 Iyy = σ₀R⁴ ∫(sinθ cosθ - sin³θ cosθ dθ) from 0 to π/2 Iyy = σ₀R⁴ [1/2 sin²θ - 1/4 sin⁴θ]₀^π/2 Iyy = σ₀R⁴ [1/2 - 1/4] Iyy = σ₀R⁴/4 But σ = M/l Iyy = MR²/4