Question

Pendulum assembly which is consists of disk with mass 5166 gr, AB and CD thin rods with mass 2.1 (kg/m) is shown in the figure. a) Determine the L length of the DC so that the center of mass of the pendulum is on the O joint. L=? m b) Find the moment of inertia of the system relative to the O joint and the axis perpendicular to the page plane. lo= ? (kgm?) c) What is the radius of gyration of the system relative to 0 ? ko=? (m) I disk= (mR2/2) Trod = (mL2/12) 0.8 m 0.5 m --D 0.2 m A B с

Answer 1

I have been deeply regretted for the inconvinence caused to you due to some technical glitch. Sorry for the trouble
у 0.2m) 0.8m 0.5m Let the origin be at from for the centre ol point 0 mas to be at o mai + m₂ ₂ + m₂ azt mit mat mat my 5.166 kg man of Dinc = mans of horizontal = m = 2.1X 1-3 = 2.13k9 rod - 5-168 (6-8+0-2) - (2 lx 0 -8x9-8 ) +(2.1X0.5X015) + (2.1XLX0.5) mit met my O = (2.1% 0,25) + (2:1XLX0-5) 5-166(1) - 2.140:39 on solving, L : 5.31m.

Izz MR 2 →X k Ikk - 4M ML 12 42 Ik From parallel axis theorem, the above problem be solved Can I = Ic. G + MOOI X-> distance from C.G to the axis where moment of Inertia is required problem For our + M193 Iosperperrlicolar Man + Mix) + + MŽ') Y to pape M 12 where ĐC : ( 6-8 +0.2) : | m 4 10.65 -0.5) = 0.15m 0.5m. Z

5.166x0.2 + 5.166X1 Ig - 2 (2-13) (3) 12 + (2.1X1:3)60-15) (2.1X5-31) (5-31) 12 Io (5.269321+ 0.4459 + 28.9889 JO 34.904 mg k- 347704*10 metery 34.704 kg.my Radius of gyration relative to O. v - mko Io where Hota) mam of pendulumn 5.166 + (2.1x1.3)+(2.1X531) m-> 19.047 kg 34. 704 = 19-047 (ko) O ko 1:35 m 0 0