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

2024
Views:

239

Pages:

5
Author:

Jabin F.

Moment of Inertia (M.I.) of an annular ring about an axis through center and perpendicular to plane: Here, the given axis is the z-axis. Izz = ∫dm(x^2 + y^2) = ∫dmy^2 For a uniform object, dm/dn = A/A ⇒ dm/(πdrdθ) = M/(π(Rg^2 - R1^2)) ⇒ dm = (M/(π(Rg^2 - R1^2)))πdrdθ annular ring with radii Rg and R1, and mass M Izz = ∬(M/(π(Rg^2 - R1^2)))s^2dsdθ = (M/(π(Rg^2 - R1^2)))∫(R1 to Rg)s^2ds∫(0 to 2π)dθ = (Mg/(π(Rg^2 - R1^2)))((Rg^4 - R1^4)/4) [Izz = (1/2)m(Rg^2 + R1^2)] M.I. of a disc about an axis through center and perpendicular to plane: R1 = 0 and Rg = R Izz = (MR^2)/2 M.I. of a thin ring about an axis through center and perpendicular to plane: R1 = Rg = R Izz = MR^2 Moment of Inertia (M.I.) of a disc about some other axis: 1) About diameter: Using 1r axis theorem: Ixx + Iyy = Izz ⇒ ID + ID = MR^2 ⇒ ID = MR^2/2 2) About tangent (in plane): Using II axes theorem: Itam = Ic + MR^2 ⇒ Itam = (MR^2)/4 + MR^2 ⇒ Itam = (5MR^2)/4 3. About an axis through periphery and perpendicular to plane: Using II axes theorem: I = Ic + MR^2 ⇒ I = (MR^2)/2 + MR^2 ⇒ I = (3MR^2)/2