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

2024
Views:

133

Pages:

6
Author:

Jabin F.

#M.I. of Cuboid: Cuboid is made by combining rectangles. Ixx = M(b² + c²) / 12 Iyy = M(a² + c²) / 12 Izz = M(a² + b²) / 12 #M.I. of Cube: a = b = c = a Ixx = Iyy = Izz = Ma²/12 M.I. of cube about any axis through body center is Ma²/12. So it is called spherical top. P-42: The figure shows a thin square sheet of metal of uniform density along with possible choices for a set of principal axes (indicated by dashed lines) of the moment of inertia, lying in the plane of the sheet. The correct choices for the principal axes would be: (a) (b) (c) (d) If these are P.A, then Ixy = 0, otherwise not. Now Ixy = ∫dmxy ≠ 0 as x, y are in the 1st quadrant where only positive values are found which can't be canceled out. Hence (b) is not P.A. Q. Find principal moment of inertia of system shown in figure: 3 Solution: Method (1) (for objective) Let us first show the P.A. through origin. Then take a symmetry axis through origin and the two more Ir axes. I₁ = Ma²/12 I₂ = Ma²/12 I₃ = Ma²/12 + Ma²/6 = 3Ma²/6 = Ma²/2 Q. Find principal moment of inertia of system shown in figure: 3 Solution: Method (1) (for subjective) Let us first write inertia tensor and find its eigen values: Ixx = Ic + md² = Ma²/12 + Ma² = 4Ma²/3 Iyy = Ic + Md² = Ma²/12 + Ma² = 4Ma²/3 Izz = Ic + Md² = Ma²/12 + Ma² = 4Ma²/3 Ixy = Iyz = Ixz = ∫dmxy = ∫dmxz = 0 II = Ma²/3 * [[4, 0, 0], [0, 4, 0], [0, 0, 4]] Finding eigen values: |I - λI| = 0 |(4Ma²/3 - λ) 0 0 | | 0 (4Ma²/3 - λ) 0 | = 0 | 0 0 (4Ma²/3 - λ) | (4Ma²/3 - λ)³ = 0 λ = 4Ma²/3 λ = 4Ma²/3, 4Ma²/3, 4Ma²/3