Question

A soda bottle cap wobbles when spinning. Assume that the cap is a thin disk of radius R and a mass M attached to a cylinder of height R, radius R, and mass M, as illustrated below R-height-radius Derive the moment of inertia tensor I around the bottle cap's center of mass. As measured relative to its own center of mass, the MOI for a disk around the principal axis perpendicular to the plane of the disk (ê3) is 1/2MR2, and the MOI around both of the other two (el and ег) is ¼ MR. Similarly, the MOI for a cylinder open at both ends about its own center of mass is 1/2MR2 around ез and 7/12 MR2 about the other two. But keep in mind that the center of mass for the bottle cap isn't the same as the center of mass for either the disk or cylinder alone. Derive an expression for the angular momentum L of the cap if its initial rotation vector a) b) olêut оез. Assume that оз is large and nonzero, that 01 s nonzero but small, and that (02 is initially zero c) Derive an expression for 0, the angle between the principal axis of greatest moment of inertia ês and the angular momentum vector L, as a function of o1 and o3.

Answer 1

The cap is a thin disk of radius R and a mass m attached to a mass mattached to a cylinder of height R, radius R and mass m. The N.O.F for a disk around the principal a proportionalto B perpendicular to the plane of the disk (e_3^^) is 1/2 MR^2 The M.O.I for a bottle cap around the principal axis perpendicular to the plane of cap is According to principal axis theorem. I_bottle cap(e_3^^) = I_disk (p_3^^) + MR^2 = 1/2 MR^2 + MR^2 I_zz(cap) = 3/2 MR^2 The M.O.I around both of the other two (e_1^^and e_2^^) is 1/4 MR^2 I_xx (disk) = I_yy (disk) \r\n I_xx (bottle cap) = I_xx (disk) + I_cylinder = 1/4 MR^2 + 7/2 MR^2 = 5/6 MR^2 I_yy (bottle cap) = I_yy (disk) + I_yy cylinder = 1/4 MR^2 + 7/12 MR^2 = 5/6 MR^2 the moment of Inertia about e^^, for the bottle cap is I_zz (bottle cap)