Question

04: (12.5 pts) The rotor shown below has a constant angular speed w of 31.46 rad/s. Determine the balancing counter weights which are needed at a radius of 100 mm in balancing planes A and B, and their angular position(s) for dynamic balance of the rotor. All dimensions given are in mm. r1 = 60 mm, 12 = 40 mm, 13 = 90 mm and W1 = 10 kg, w2= 16 kg , W3 = 5 kg. (Note: Problem can be solved by converting dimensions into cm units) PYRIGHT ΠΝΕ SITÉ PLANE 8 PLANEA 1.250 50 20 200 Quyos Bg fy. 135° -80 5","10 kg 90 1200 B them 777777 40 600 Our 16 kg -40 230

Answer 1

Let us Plane. le For dynamic balancing, both wet force and net couple should be 2000 after adding the above masses. Angles are measured from tve x-axis in Counterclockwise direction. put mass MA and mass mo at indination OA and OR at respective

is closed, since couple polygon is closed, since force poly gone and & moro Coso =0 E maro sino; = 0 3 & morili Cosoo =0 E miro li sino =0 be becomes 6 5*60x tos(135) +16% 10x 60 x Cos 30 + 16x40x Cos(300) + 5x80xC05 (135) + MAX 100 X CS@A) + Mpx 100 x Cos (Oc) =0 lo Coele, 556, 77 + 100 MA COSOA + 100 mg cos Op=0 A د 5,5677 + MACOSOAT Mg Cosob

So becomes 3228,041 +25,000 mbsing Tools AM- lly, ean 4 and (D From S 3228.04 ) tando 46557 -> moldo OB = -3.960 ou substitutin in di cosa s prilor dentro del kgebied A and B 3228.0410 -172.922 MB = 0 MON mg = 18,67 kg Lao Snc in Substituting of and me 5.5677 + MACOSOA +18.625 = 0 24,193 + MA COSOA =0 Values and *-1 +masinda =0 masala - satu

solving above tano A = => OA tant La .24,193 24.93 O A = -2,297 substituting for OA -1 + MA sin(-2,297) =0 -24 kg MA=