Problem 5/163  Page 403, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: A link rotating about a fixed pivot point C at a constant counterclockwise rate N.  At the instant of interest the link is vertical.  Point A of the link is a distance H directly below the link pivot point.  A slotted member is also rotating about a fixed point O.  The point O is a distance H below point C and a distance L to the left of point C.  The point A on the link is constrained to slide in the slot of the slotted member.  At the instant of interest the point A is distance L to the right of point C.  At the instant of interest the slot makes an angle q with the horizontal.

Find: The angular velocity and angular acceleration of the slotted member.

H = 0.12 m
L = 0.12 m
q = 45 degrees
N = 4 rad/s

The motion of point A can be considered from two perspectives.  It is attached to the link and can be analyzed using the two point formula for the link.  It is moving in the slot and be considered using the relative motion two point formula for the slotted member.  Thus (L for link):
vA = vC + wL X r_A/C
vA = 0 + N k X -H j
vA = N H i

From the perspective of the slotted member (S for slotted member) :
vA = vO + wS X r_A/O + vREL_A/S

Noting that the point A is constrained to move in the slot, then any relative velocity must be parallel to the slot:
vA = 0 + wS k X L i + vREL ( cos q  i  +  sin q  j )
vA =  wS L j +  vREL ( cos q  i  +  sin q  j )

 Equating the components of the two expressions for the velocity of point A:
N H = vREL  cos q
0 =  wS L + vREL sin q 

The top equation can be solved for velocity of point A relative to the slot.  Then the second equation can be solved for the angular velocity of the slotted member.
vREL = N H / cos q   =  0.679 m/s
wS = - N H tan q / L = -4 rad/s

The acceleration of point A can be considered from the same two perspectives.  Noting that the point A is attached to the link and that the angular acceleration of the link is zero (constant angular velocity):
aA = aC + wL X ( wL X r_A/C  )  + aL X r_A/C
aA = 0 + N k X ( N k X -H j ) + 0
aA = N k X  N H i
aA = N² H j

From the perspective of the slotted member:
aA = aO +  wS X ( wS X r_A/O  )  + aS X r_A/O + 2 wS X vREL_A/S + aREL_A/S

Noting that the point A is constrained to move in the straight slot, then any relative acceleration must be strictly tangent to the path and must be parallel to the slot (the relative motion is in a straight line and thus the relative normal acceleration is zero):
aA = 0 + wS k X ( wS k X L i  ) + aS k X L i + 2 wS k X  vREL ( cos q  i  +  sin q  j ) + aREL  ( cos q  i  +  sin q  j )
aA = wS k X wS L j +  aS L j + 2 wS vREL ( cos q  j  -  sin q  i ) + aREL  ( cos q  i  +  sin q  j )
aA = -wS² L i +  aS L j + 2 wS vREL ( cos q  j  -  sin q  i ) + aREL  ( cos q  i  +  sin q  j )

Equating the components of the two expressions for the acceleration of point A:
0 = -wS² L - 2 wS vREL sin q   + aREL  cos q
N² H =   aS L + 2 wS vREL cos q + aREL  sin q

Noting that the angular velocity of the slotted member and the velocity of point A relative to the slot have already been determined, the above can be seen to be two equations in the two unknowns, the angular acceleration of the slotted member and the acceleration of point A relative to the slot.  The top equation can be solved for the relative acceleration and the second equation can then be solved for the angular acceleration:
aREL = { wS² L + 2 wS vREL sin q } /  cos q = -2.72 m/s²
aS = { N² H -  2 wS vREL cos q - aREL  sin q } / L = 64 rad/s²