Question

Q2 Manufacturing engineer works to locate a center of gravity as a drilled hole location of a thin composite plate with a density is constant. Figure I shows the shape of a plate consists of a semicircle with a radius of 10 cm and a square. (a) Compute an area of the plate using double integral. (8 marks) (b) Construct a center of gravity using double integral. (10 marks) (C) Determine M, if the density of a plate is 0.5 gem? (7 marks)

Answer 1

Solution at 9) Y-10 X² -110²_yz X=20 X< y=-10 We know theat ley SS da at plante is given SS. dx dx dy By figure R x = -√10²yz to x=20 7-10 to ID

A. $ 500 dx dy -10 -100-42 10 (20 + 1100-y2) dy - 10 25 (20 (20 + 1100-43) dy 10 - 2 2 [200 200 + (stoo-yz dy I (say) 10 10 JE Noo-y2 dy- Krootge Ž 100-y2 + 100 sint By fooman X2 a = x² + asin's te lo -> Soo-y2 + 100 2 Sinty 10 Vio [ot 100 2 100K 4 2 By o A= 400 + 200K 400+ SOK = 50/7+8) Ang

ot So 2 Pap 10 - Too ya 10 5) 0.5 200 (GR-001) -- He 01 - 10 U density given SXD DS. de Me what (mass) m SS Soyda Sorglem da 05 SS dA 0.5 50 (3) - 25(8+) gm 478.5 (A) * 3.14 SS y sunda y (0.5) da 05 / 4 (20 + 100-ye dy C 400-100+ y2 dy -> Fest Meme =) S (300 +42) dy 4 mx -ID 3) --- to face 300 (10 + 10) 4 -flooya => 6600 + www. 3 3 and function - - - 0.5 + Ro y 100- y2) dy 6000 + 2009 3 &0.000 - My 05 - 10 200003 = 1666.67 てい De kwa HAD y = S/ x 84.g) dans olo Rp xp x -R-ool at Comitee dat grauity | Mass Mx 2 لها

(x, y) = 1666.66 278.5 о 278.5 De (5.9844,0) is required centre of gravity C) By cuestion (6) we we have a My = S S « S(x,y) dA = 1666.666 is require ans