Question

Review Learning Goal: To be able to use the parallels there to calculate the moment of inertia for an area The paralel-ds theorem can be used to find an area's moment of inertia about any is that is parallel to and that passes through the controid and whose moment of Inertia is known and are the axes that pass through an area controld, the paralel-axis theorem for the moment about the axis, moment about the yaxis. As shown, a rectangle has a base of b 8.00 tanda hage of 4.40 ft. gure 2) The rectangle's bottomis located at a distance - 2.10 it from the and the rectangle's left edge is located at d e 230 ft from the you. What are land e 's moments of inertin about ex and y es, respectively? Express your answers numerically in biquadratic feet feet to the fourth power to three significant figures separated by a comma ► View Available Hints) Figure 1 of 3 LAS i voi ? Submit Part The more shown our has a moment of inertia about the xaxs of it and a moment of about the yaxis of 54.0 Mt. What is the polar momento inertia about point the control? Express your answer numerically in biquadratic feet (eat to the fourth power to three significant figures.

Answer 1

You have not given diagram for c but I tried... Pls rate

-4.4ft Ig= ball x=2. 38 b> 6. 6t y ez 19, = 2. loft distance setween considered axis to parallel controidal axis paralld axis theorem I= Ig de M. I about paralled Age centroidal axis = Igut au = 603 + (orb) 9,+5) = 6.690.487 (6.6*4.)(2.14 in sy 72 7 = 46.8512 +536.9496 In = 583.80 and Iy= Igy + A y ł - hbt (b< 1)(x + 3) = 4.486626)37 (6,6% 4,4) (237 1526) = 1056445 + 910.694 = 1016.1092 fth 1 In, Iy = 583.80, 1816.1052 ty © Giren Ix= 54 ft" , Ig = 56tte for semicirce wchare In = I ay = 1/2 d=6.848 = 3.42 4 ft we have poltar moment of inertia about point'd J = (1 2 3 4 ) e = @ $) (8.4243 = 0.50245( 3.42434 = 69.06 ft