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

2024
Views:

279

Pages:

5
Author:

Jabin F.

# M.I. of Hollow Cylinder -> I = MR² thin hollow cylinder. annular cylinder. I = MR² I2 = M(R₁² + R₂²) / 2 I3 = MR² I = I1 + I2 + I3 I = MR² + M(R₁² + R₂²) / 2 + MR² I = 2MR² + M(R₁² + R₂²) / 2 I = (5/2)MR² + (1/2)M(R₁² + R₂²) @ A solid cube has mass M and side 'a'. Calculate its M.I. about an axis consisting with one of its edges/sides. Solution: I = I1 + Md² = Ma²/6 + Ma²/2 I = (3/2)Ma² Two discs are placed as shown in figure. M.I. about given axis lying in plane of discs is: Solution: I1 = Ic + Md² = Ma²/4 + Md² = 5Ma²/4 I2 = Ic + M(d - r)² = Ma²/4 + M(d - r)² = 37Ma²/4 I = I1 + I2 = (5Ma²/4) + (37Ma²/4) = 91Ma²/4 = (91/3)Mr² From a circular disc of radius R and mass M, a circular portion of radius r is cut. Calculate M.I. of remaining portion about axis shown in figure. Solution: Firstly we have to calculate the amount of mass cut. For uniform mass: ρ(disc R) = ρ(disc r) M/πR² = m/πr² m = Mr²/R² = M/4 New I = Im(centre) - Im(end/tangent) = MR²/4 - (5/64)MR² = (16-5)/64 MR² = 11/64 MR² = 11/64 MR²