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

2024
Views:

417

Pages:

5
Author:

Jabin F.

# M.I. of Square a = b in rectangle. I_x = Ma² I_y = Ma² I_zz = Ma² I_d = Ma² + Ma² = 2Ma² ⇒ M.I. of square through center about any axis lying in plane is same (Ma²) P-125 A thin wire of length L and mass M is bent to form a circular loop. M.I. of the loop about a tangent perpendicular to its plane is → Solution: L = 2πR ⇒ R = L/2π I = I_c + MR² = MR² + MR² = 2MR² = 2M(L/2π)² = ML²/2π² Q1 A sphere of mass M and radius R, a cube of maximum possible volume is cut. Calculate M.I. of the cube about axis through body center. Solution: Firstly we have to calculate the parameters of the cube in terms of given quantity. R = (a^2/9 + a^2/4)^1/2 = 3a^2/4 => a = 2R/√3 Mass of cube: (M x V_cube) / V_sphere = (M x a^3) / (4/3πR^3) = M x (3/4πR^3) x (8R^3/3√3) Mass of cube: m' = 9M/π√3 I = m'a^2/6 = (1/6) x (9M/π√3) x (4/3)R^2 = 4MR^2/9π√3 Q2 The system shown in the figure has mass M and uniform density. Calculate M.I. Solution: M is mass of total volume = V_sphere + V_cone (4/3)πR^3 + (1/3)πR^3 = 5/3πR^3 Both are having equal volume, so mass of both are equal = M/2 Now, I = I_cone + I_hemisphere = (3/10)mR^2 + (2/5)mR^2 = (3/10)(M/2)R^2 + (2/5)(M/2)R^2 = 7MR^2/30 P-43 In the figure shown, value of I_xy is, Q-18 Solution: I_xy = Σm_i x_i y_i = - [(m x 0 x 0) + (9m x a x a) + (3m x a√3/2 x a√3/2)] = - 3√3/4 ma²