Question

5. Three particles of masses 1.3 kg, 2.5 kg and 3.7 kg are located at the points ( 3,4) m, ( 16,5)m, and (-1,2) m respectively. Calculate the rotational kinetic energy of this system of particles if they are revolving around the x-axis with an angular velocity of 3 rad/s. (1 point) A. 554.692 J B. 191.216 J C. 441.45 J D. 831.483 J E. 377.97 J 6. Three particles of masses 8.3 kg, 4.5 kg and 9.7 kg are located at the points (-12,5)m,(-11,-5) m, and (8,4) m respectively. Calculate the angular momentum of this system of particles if they are revolving around the x-axis with an angular velocity of 30 rad/s. (1 point) A. 20961.787 Js B. 25317.656 J s C. 8664.091 Js D. 14256 J s E. 3559.751 J s

Answer 1

s. There are three particles of 3.7 kg 2.5 kg are mass t3 kg- and located at the pointa (6.5) mand (-1,2m rexpectively. have to find rotational kinetic energy of this system. Now formula Rotational 3.7 (3,4 m, We of kinetic energy KE ㅗ Totational - where I moment of inertia angular velocity In this quen ation angular uclocity in firstly have to find inertia ayatem. given. moment ane we of the of me 2 2.5kg 5 1,3 kg ma (1,5) 4 (3,4) 3 (-1,2) 2 mg=3.7k + go 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 CS Scanned with Teacher's Signature : CamScanner

formula inertia moment og of the system in Çm; 6? where, mi from the mass of the point particle V Perpendicular distance rotational axir particle. particle m, r + m2 82² + m₂ 7² to the SO auze three There I Given, mo = 1.3kg m₂ = 2.5kg 3.7 kg M3 bi um De 5m 2 m 3 3 rad/sec using egn I = 1-3 kg x (4)2 +2.5*5* + 3.7x2 1.3 x 16 + 9.5x25 + 3-7x4 I = 98.1% CS Scanned with CamScanner

Now Rotational kinetic energy RIKE 12 I w² こ 1 / 4 X98-) X 32 FREE = 441-45 Option C is correet. are three particle of 9.7' 8.3 kg are 6 There masses 4.5 kg, and located at the points (-12,5)m, (-11,5)m, and (0,4m reppectively. We find angular momentuh system. Now formula of angular momentum in L = I w have to thin where T moment of past inertia of tha System angular velocity Teacher's Signature : CS Scanned with CamScanner

C (321) 28.8=1m Ms = S 4 9.71g (84) 3 2 1 1 1 1 - 12 +1 -10 -9 -8 -7 -6 -5 _4 -3 2 1 3 ។ 5 6 -2 -1 -1 8 7 9 11 12 141,-5) -3 -4 m₂ = 4.5kg use equo I = more +more +moris Given! a'u こ 8.3 kg 4.5 kg m2 = m₂ = 9.7 kg Y 5m =-5m ste Ym w= 8.3x5? 30 rad rad/ pee +4.5*(-5) + 9.7x42 I = 0:3X25+ 4.5 x 95 + 9.7*16 475.2 kg / m2 CS Scanned with CamScanner

Now, angular momentum the oyaten in Angular momentum (L) I w = 47.5-2 X 30 L 14256 J-see Option (1) jo correet. CS Scanned with CamScanner