GENERAL RIGID BODY MOTIONA disk is originally rotating at w = -8 rad/s. It is given a constant angular acceleration a = 1 rad/s?.Consider the pin to be the origin. At time t = 0, point A is then located at (2,0) ft as shown in thepicture.• Where is point A after 1 second (x, y)?• What x and y components of the velocity and acceleration of point A after 1 second?wo = 8 rad/sB1.3 ft- 2 ft-

Given: initial angular velocity of rotor, ωo=8 rad/sconstant angular acceleration of rotor, α=1 rad/s2Newton's first law for angular motion,ω=ωo+αtωo=8 rad/secα=1 rad/s2t=1 secω=ωo+αtω=8+1×1ω=9 rad/sω=θtθ=ω×t=9×1=9 rad1 radian=57.2958 Degreeθ=515.66o=360o+155.66o360o means 1 full rotation of disk and it will be in the original positionso position of point A from initial point is 155.66o clockwise =90o+65.66oTo calculate the coordinate of point A, Refer the figure for coordinate x, sin θ=ABOAsin 65.66o=x2x=-1.82 ft (negative sign used because the direction of x is left from origin )for coordinate y, sin θ=ACOAsin 24.34o=y2y=-0.824 ft (negative sign used because the direction of y is downward from origin )Point A after 1 second= A(-1.82,-0.824) Part b:After 1 sec tangential velocity of point A is,VA=r×ω=2×9=18 ft/sVAx=VAcos 65.66oVAx=7.42 ft/s ,this is the x component of the velocity of point AVAy=VAsin 65.66oVAy=16.4 ft/s ,this is the y component of the velocity of point Atangential Acceleration of point A isat=r×α=2×1=2 ft/s2atx=at cos 65.66oatx=0.8243 ft/s2 , this is the x component of the tangential acceleration of point Aaty=at sin 65.66oaty=1.82 ft/s2 , this is the y component of the tangential acceleration of point A