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

2024
Views:

408

Pages:

5
Author:

Jabin F.

# M.I. of rectangular plate It is lying on xy-plane : z = 0 For uniform area, mass: dm = M * (dA/A) Ixx = ∫dm(y²+z²) = ∫dm.y² Ixx = (M/ab) ∫∫y²dxdy = (M/ab) ∫(b/a) ∫(a/b) y²dydx = (M/ab) ∫(b/a) [y³/3](b/a) dx Ixx = (M/ab) ∫(b/a) (b³/3a³) dx = (Mb²/12a²) ∫(b/a) dx = (Mb²/12a²) (b/a) = Mb²/12 about x axis Similarly Iyy = Ma²/12 about y axis. Now using parallel axis theorem Izz = Ixx + Iyy = Mb²/12 + Ma²/12 = M(a²+b²)/12 about z axis # M.I. of rectangle about arbitrary axis through centre and lying in its plane A = cosθ + sinθ I = I₁cos²θ + I₂sin²θ As x, y, z are P.A. therefore P.I. are zero I = (Ma²/12)cos²θ + (Mb²/12)sin²θ I = (M/12)(a²cos²θ + b²sin²θ) # M.I. of rectangle about its diagonal cosθ = a / √(a²+b²) sinθ = b / √(a²+b²) I = (M/12)(a²*(a²/(a²+b²)) + b²*(b²/(a²+b²))) I = (M/12)((a^4 + b^4)/(a²+b²)) P-120 Seven uniform discs, each of mass m and radius r, are inscribed inside a regular hexagon, as shown. The moment of inertia of this system of seven discs, about an axis passing through the central disk and perpendicular to the plane of the disks, is: Solution: I = mr²/2 I' = I₁ = I₂ = I₃ = I₄ = I₅ = I₆ = mr²/2 + m(√3r)² = 9mr²/2 I = I₁ + 6I' = mr²/2 + 6(9mr²/2) = 55mr²/2