show complete solution help is very much appreciated thank y

show complete solution. help is very much appreciated. thank you! :)

circle of as rod anoched A and out the For the an velocity olay a Angular speed of the s ink der votive Aareleration or A as the crank AC re wks

given Radius = 90mm;

OC = 300mm;

Angular velocity = 60 rad/sec;

Velocity of A is =wr= 60 * 90 mm/sec; = 5400 mm/sec = 5.4m/sec;

now,

the point A about O is (300 - r cos(theta) ,r sin(theta));

where theta = wt = 60t;

(300 - r cos(60t) ,r sin(60t));

Velocity of A along radial axis;

velocity = V (cos(90- (theta+ rho));

5.4 sin(theta+ rho);

rho = 30 degrees;

OC = 300mm,

AC = 90mm;

theta = 11.456 degrees;

velocity = 5.4 sin(11.456 + 30) = 3.58 m/sec;

now, angular speed about O is 5.4 cos(11.456 + 30) = 4.05 m/sec / r = 4.05/0.226571 = 17.875 rad/sec;

3) double derivative of r;

r = sqrt ((300-90cos(60t))^ + 90 sin(60t)^2) = {300^2 + 90^2 - 2*300*90*cos(60t)}^1/2;

2r dr/dt = 2*300*90*60*sin(60t)

2r * d^2r /dt^2 + 2 (dr/dt)^2 = 2*300*90*60*60*cos(60t);

r\'\' = (2*300*90*60*60*cos(60t) - 2*(300*90*60*sin(60t))^2/(300^2 + 90^2 - 2*300*90*cos(60t)) )/(300^2 + 90^2 - 2*300*90*cos(60t))^1/2;

Double derivative of Theta;

sin (theta )/90 = sin(180 - theta - rho) /300 =>

sin(theta) / 90 = sin(theta + rho) /300;

sin(theta) /90 = (sin(theta)cos(wt) + sin(wt)cos(theta))/300;

differentiate wrt t;