Question

To apply the equations of motion to a system that involves rotation about a fixed axis and to use this information to determine key characteristics.

The slender rod ABAB shown has a mass of m=51.0m=51.0 kgkg and is being supported by a rope and pulley system stationed at CC. Starting from rest (in the position shown), the rope and pulley system tug on the rod causing it to rotate about AA. The torque applied to the pulley is T=2.85T=2.85 kN⋅mkN⋅m and has an effective moment arm of r=0.160r=0.160 mm. The dimensions shown in the figure are l=2.30l=2.30 mm and h=1.40h=1.40 mm. Assume the pulley is frictionless and massless.

MasteringEngineering Masterir x C Get Homework Help With Cher X + • session.masteringengineering.com/myct/itemView?assignment ProblemiD=11226176 O* H MAE 242 Summer 2020 Hader <EC10 Equations of Motion: Rotation about a Fixed Axis < 10 of 15 1 Review Part A - Angular Acceleration of the Rod Learning Goal: To apply the equations of motion to a system that involves rolation about a fixed axis and to use this information to determine key characteristics. The slender rod AB shown has a mass of m=51.0 kg and is being supported by a rope and pulley system stationed at C. Starting from rest in the position shown), the rope and pulley system tug on the rod causing it to rotate about A. The torque applied to the pulley is T = 2.85 kN-m and has an effective moment arm of r = 0.160 m. The dimensions shown in the figure are l = 2.30 m and h = 1.40 m. Assume the pulley is frictionless and massless. Determine the angular acceleration of the rod the instant the rope and pulley systerrı have pulled the rod through an angle of 8 = 2.70°. Express your answer to three significant figures and include the appropriate units. View Available Hint(s) TH пA th ? (Figure 1) a= Value Units Submit Figure < 1 of 1 > Part B - Normal Component of the Reaction at A Determine the normal component of the reaction the rod exerts on the pin at A. Use the coordinate system set up by the free-body diagram below. B B )

Answer 1

Given dater: mass(m) = 51.0 kg Torque applied (1) = gol 2.85 kN-m C k h A B r=0. lbom da I 2.30m ; halihom Solutioni "A k l alby applying Shine vile, ܀ l Som Q0+2:46 - ) Got 250 p 90-2.75 K入 T cus 2 Sin (3) 1.10 Son ) Sin 92.90-1) 270. W Six (4.2.10 )of) ws 92.No-Bosh B 9.30 liko Sins layer() + 0.0471 1.643 3) 1B = 39.07 T= 28500? Frig XV - kg xo.160 ho Sin (32.01) TER 17812,5 N MA=Oi Ipd = pain (a) xl - wws@) W ={17810.5) Sn (32.67) (2:30) All wes (216) - (510) Acos (216) x 2.30 AT A Ghi) Led = 21752.861 -88.58589-81) Id=21177.843 ) rall go d235.493 CAM) W sn (25) kg 80g (32.07) 3, 2 IA=mea 54428 3 3 89.93

1) Fro; Aga wsino) fros (B] + c (57*9.81) Sim (2.76) +17612.52.00>(32019 At a 15117.880 Aga 15. 118 kN زد) W caso c) .An HF 8MCB] 17812.5 Ant to sin (32.09) - 6189.85 x [email protected]) An = 93.78967, 881 N (Ana-8.958 kN ( 2 ) Arswens & 22 236.493 va rad's Ana -8.968 KN Q Aga Isill8 kN - स c)