In this example, we will examine a rigid body problem involving a linkage.
In the linkage shown here, link AO rotates with a constant angular velocity, omegaAO, 
of 10 radians per second in the counter clockwise direction.
Link AO is 0.10 meters in length. 
Link AB is 0.26 meters in length. 
Link BC is 0.18 meters in length. 
The distance between points O and C is 0.18 meters.
When point A is located 0.06 meters to the left and 0.08 meters above point O, link BC is vertical.
At this instant, what is alphaAB, which is the angular acceleration of link AB?
Identify your coordinate system and label the positive direction of your coordinates. 
In this case a rectangular coordinate system is more convenient.
Indicate the positive direction of rotation.
The angular acceleration of link AB appears in the equation relating the acceleration of points A and B. 
The acceleration of point A is equal to the acceleration of point B, plus the acceleration of point A relative to B. This is equation 1.
The relative acceleration term is decomposed into tangential and normal components. 
The tangential component is alphaAB cross rA/B, the position vector pointing from point B toward point A. 
The normal component is the angular velocity of link AB, called omegaAB, cross the quantity omegaAB cross rA/B.
The angular acceleration vector for link AB has the magnitude of alphaAB and points in the z direction.
The angular velocity vector for link AB has the magnitude of omegaAB and points in the z direction.
rA/B starts at point B and terminates at point A. 
Point A is located 0.24 meters to the left and 0.10 meters downward from point B.
In order to find alphaAB, we need to obtain omegaAB, the acceleration vector for point A, and the acceleration vector for point B.
OmegaAB is obtained by relating the velocity of points A and B. 
The velocity of point A is equal to the velocity of point B plus the velocity of point A relative to point B. 
We do not know the velocity of points A and B, but we can find expressions for them by relating  
those velocities to the velocity at points O and C.
The velocity of point A is equal to the velocity of point O plus vA/O, the velocity of point A relative to point O.
The velocity of point O is zero,
and the relative velocity is rewritten as omegaAO cross rA/O, the position vector pointing from point O toward point A.
The angular velocity vector for link AO has a magnitude of 10 radians per second and points in the positive z direction.
rA/O starts at point O and terminates at point A. 
Point A is located 0.06 meters to the left and 0.08 meters upward from point O.
Take the cross product of omegaAO and rA/O to obtain the velocity of point A.
The velocity of point B is equal to the velocity of point C plus vB/C, the velocity of point B relative to point C.
The velocity of point C is zero and the relative velocity is rewritten as omegaBC cross rB/C,  
the position vector pointing from point C toward point B.
The angular velocity vector for link BC is omegaBC and points in the z direction.
rB/C starts at point C and terminates at point B. 
Point B is located 0.18 meters upward from point C.
Take the cross product of omegaBC and rB/C to obtain the velocity of point B. 
Now we turn our attention to the last term in equation 2, vA/B.
vA/B is rewritten as omegaAB cross rA/B.
We already found expressions for omegaAB and rA/B and can evaluate their cross product.
Plug in all the velocities into equation 2.
Equating the x-terms and the y-terms, we obtain two equations and two unknowns.
Link AB rotates in the counter clockwise direction at a rate of 2.5 radians per second. 
Link BC also rotates in the counter clockwise direction, but at a rate of 5.833 radians per second.
The acceleration of point A is obtained by relating the acceleration of points A and O. 
The acceleration of point A is equal to the acceleration of point O plus the acceleration of point A relative to point O.
The acceleration of point O is zero since it is fixed. 
The relative acceleration term is decomposed into tangential and normal components. 
The tangential component is alphaAO cross rA/O. 
The normal component is omegaAO cross the quantity, omegaAO cross rA/O.
alphaAO is zero because link AO rotates with constant angular velocity.
We already found expressions for omegaAO and rA/O.
Evaluating the cross products gives us the acceleration of point A.
The acceleration of point B is obtained by relating the acceleration of points B and C. 
The acceleration of point B is equal to the acceleration of point C plus the acceleration of point B relative to point C.
The acceleration of point C is zero since it is fixed. 
The relative acceleration term is decomposed into tangential and normal components. 
The tangential component is the angular acceleration of link BC cross rB/C. 
The normal component is omegaBC cross the quantity, omegaBC cross rB/C.
The angular acceleration vector for link BC has the magnitude of alphaBC and points in the z direction.
We already obtained expressions for omegaBC and rB/C.
Evaluating the cross products gives us the acceleration of point B.
Plugging in all the expressions into equation 1 and evaluating the cross products gives us the following equation.
Equating the x terms and the y terms, 
we obtain two equations and two unknowns and can solve for the angular acceleration of link AB and link BC. 
The angular acceleration of link AB is 10.42 radians per second squared in the counterclockwise direction. 
The angular acceleration of link BC is 19.21 radians per second squared in the clockwise direction.
