Kinematics: How Do You Solve for Velocity in General Plane Motion?

[JaWiB]

This is my first post, so I apologize if any formatting is off or if I'm not clear enough.

Homework Statement

A ride in an amusement park consists of a rotating arm AB having a constant angular velocity $$\omega_{AB}$$=2rad/s about point A and a car mounted at the end of the arm which has a constant angular velocity $$\omega'$$={-.5k}rad/s, measured relative to the arm. At the instant shown, determine the velocity and acceleration of the passenger at C.

Since I don't have a scanner handy, I'm attaching my paint replication of the figure.

For now, I'm just looking at solving for velocity, and I think I'll be able to find acceleration if I can figure that part out.

Homework Equations

I'm using the kinematic equations derived in the book for a rotating and translating reference frame x,y. Using the point B in the figure as the origin for these axes, and the point A in the figure for a fixed reference frame X,Y, we have:

$$\vec{V_{C}} = \vec{V_{B}} + \vec{\omega_{BC}}\times\vec{r_{C/B}}+(\vec{V_{C/B}})_{xy}$$

The Attempt at a Solution

This didn't seem too complicated to me, but I can't get the same answer that the book does. First of all, I think that the $$(\vec{V_{C/B}})_{xy}$$ term should be zero since C is fixed to B (it's not moving with respect to the x,y axes).

$$\vec{V_{B}}$$ should be equal to the angular velocity of the AB arm crossed with it's radius (2 rad/s * 10 ft tangent to the arm) which ends up as { -20icos(60) + 20jsin(60) } ft/s = { -10i + 10$$\sqrt{3}$$j } ft/s

Then $${\omega_{BC}}\times\vec{r_{C/B}}$$ is completely in the negative X direction at .5 rad/s * 2 ft which is just { -1i } ft/s

Finally, my answer is the sum of those two vectors: { -11i + 10$$\sqrt{3}$$j } ft/s

The answer the book gives, however, is { -7i + 17.3j } ft/s. So the Y component of my answer is correct, but somehow my X component's magnitude is too great. What am I missing?


Hopefully my figure is clear enough. C is attached 2ft from the center of the clockwise rotating disk B, which is in turn attached to a 10ft arm which is rotating anti-clockwise about A. The X and Y axes given are horizontal and vertical.

 

[JaWiB]

If anyone's interested, the trick is that $$\omega'$$ is measured relative to the arm, so the correct value of $$\omega_{BC}$$ is $$(\omega_{AB}-\omega')$$