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

2024
Views:

148

Pages:

5
Author:

Jabin F.

# M.I. of thick sphere: (about diameter) Ixx = Iyy = Izz Izz = ∫dm(x²+y²) for uniform object: dm = (M/V)dV Using spherical polar co-ordinates: x = R sinθ cosϕ y = R sinθ sinϕ z = R cosθ dm = (M/4/3π(R₃³-R₁³))R²sinθ dθ dϕ → Izz = ∭(3M/4π(R₃³-R₁³))R²sinθ dθ dϕ (R²sin²θ) = (3M/4π(R₃³-R₁³))∫(R₁)^(R₃)R⁴dR ∫(0)^(π)sin³θ dθ ∫(0)^(2π)dϕ = (3M/4π(R₃³-R₁³))[R⁵/5](R₃)^(R₁) ∫(0)^(π)(3/4sinθ-1/4sin3θ)dθ 2π = (3M/9(R₃³-R₁³))(R₃⁵-R₁⁵)/5 (4/3) Izz = (2/5)M(R₃⁵-R₁⁵)/(R₃³-R₁³) M.I. of solid sphere: R₁=0, R₂=R ⇒ I = 2/5MR² About tangent: using || axes theorem I = I + MR² = 2/5MR² + MR² ⇒ I = 7/5MR² P-122 The moment of inertia of a disc about one of its diameter is Im. The mass per unit area of the disc is proportional to the distance from its centre. If the radius of the disc is R and its mass is M, the value of Im is: Solution: Given that: σ ∝ r Firstly we find the value of k For mass density: dm = krdr ⇒ dm = krdr dθ dA = rdr dθ M = ∫dm = ∫∫kr²drdθ = k∫(0)^(R)r³dr ∫(0)^(2π)dθ = kπR⁴/4 M = k ∫(0)^(R) r³dr dθ = k (R⁴/4)π ⇒ k = 3M/πR³ Now, I_m = ∫dm(x²+y²) = ∫dm r²sin²θ = ∬k r³dr dθ r²sin²θ = k ∫(0)^(R) r⁵dr ∫(0)^(2π)sin²θ dθ = k R⁶/6 π = (3M/πR³) R⁶/6 π = 3MR²/10