Question

Homework Topic 8 Due November 28, 2019 1a) The mass moment of inertia (MOI) for a thin ring is the MOI for a thin disk (when masses are the same). Explain why. (3 points) 1b) Why is Izz for a cylinder the same as Izz for a thin disk? (3 points) 1c) Show why Izz' = 3/2 mr for a thin circular disk, start from Izz = 42 m (9 points) 2 Thin Circular disk mln

Answer 1

Mor of Ring about an aw's passing through c.m. and perpendicular to plance of Ring IR = MR? M consider a disc of game mass m and Radius R (thin) GIR consider an elment ring of radius r. and thickness dr. Mass Area = = . of dise TR2 mais of king of radius & in dlm = 24 s dryo MOI of ring of rading is a dhe = (dm) 2 dhe = (2420dr) r² = 240 23 de R Pc = ( dica no sadr - 2006 SIR 2e = 200 x 2 = 24 x M x 2 = 1 MR2 : Mor of the A disc = fc = I MR² = 1 (mol of Ring I22, disc 529 Izle TM Sim thin dise o element disc In cylinder - of mass com) S t dm Rr = £ R som i = £ MR² = Izz, dire we can break the cylinder in seneres am mare disc for which mass distribution about quis will be same. OR. we can consider cylinder as fuck disc.

We know mor of disc about a Taz - Icom Emre for Paлallu quit Неолит I221 = I com of mo [oris the perpendre las distance Iron & mrtmrh of I22 from tomt (com = andre of mass in Izzi 3 mez