Question

Problem II (34 marks) VA The homogeneous, Skg bar AB is released from rest In the position shown(Neglect the friction/between wheel A and ground. 『こ3437. 5 kg t7.as ack ect J.9-01 a6y 12 m NES.4.81 aA 1.5 m NA 1. Draw the free body diagram and kinetic diagram, and define appropriate coordinate system (10 marks). 2. Determine the following parameters (magnitude and direction) immediately after release (24 marks). (1) The angular acceleration of the bar. (2) The tension in the cord BC ax-cos3487. 0.1)5 ay sin3h70. a1500 + Sf masr= T. SP=MabywtNA VENA IA 5.25 0. -w.0.t2T. toperteter. -10.x2a36.73-10 -37r0=48054NA r0.K20=36.78+2T Tatd

（Pls tell me why i cannot use IA=1/3（ml^2）❗️❗️❗️

The homogencous, Skg bar AB is releayed from rest n the position shown(Neglect the friction between wheel A and ground. 1. Draw the free body diagram and kinetic diagram, and define appropriate coordinate system (10 marks). 2. Determine the following parameters (magnitude and direction) immediately after release (24 marks). (1) The angular acceleration of the bar.

Answer 1

2m 1-sm MEskg D'Atembert pineipe EMA TX Mgx 3 2 SX 2.S Sax 18bxS 3 31 25 2. 36. 7815- 2T 31 25 36. 787 T

Accelenstion ait A tve X -0.75j-0.5ai -o-Sa anx D'ALembeppe XO Q.Sd - TE 31. 25 2.386 iad s Sb STL

Answer 2

YOU SHOULD USE IA=(ML^2)/3 BUT if you take moment about A, then on the right side of your moment equation you should also consider the moment caused by inertia vectors m(a)x and m(a)y. So on the rhs of moment equation you should have (1/3)m(l)^2 + moment produced by m(a)x + moment produced by m(a)y