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JaWiB

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

A ride in an amusement park consists of a rotating arm AB having a constant angular velocity [tex]\omega_{AB}[/tex]=2rad/s about point A and a car mounted at the end of the arm which has a constant angular velocity [tex]\omega'[/tex]={-.5

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.

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:

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

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 [tex](\vec{V_{C/B}})_{xy}[/tex] term should be zero since C is fixed to B (it's not moving with respect to the x,y axes).

[tex]\vec{V_{B}}[/tex] 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 { -20

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

Finally, my answer is the sum of those two vectors: { -11

The answer the book gives, however, is { -7

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.

## Homework Statement

A ride in an amusement park consists of a rotating arm AB having a constant angular velocity [tex]\omega_{AB}[/tex]=2rad/s about point A and a car mounted at the end of the arm which has a constant angular velocity [tex]\omega'[/tex]={-.5

**k**}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:

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

## 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 [tex](\vec{V_{C/B}})_{xy}[/tex] term should be zero since C is fixed to B (it's not moving with respect to the x,y axes).

[tex]\vec{V_{B}}[/tex] 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 { -20

**i**cos(60) + 20**j**sin(60) } ft/s = { -10**i**+ 10[tex]\sqrt{3}[/tex]**j**} ft/sThen [tex]{\omega_{BC}}\times\vec{r_{C/B}}[/tex] is completely in the negative X direction at .5 rad/s * 2 ft which is just { -1

**i**} ft/sFinally, my answer is the sum of those two vectors: { -11

**i**+ 10[tex]\sqrt{3}[/tex]**j**} ft/sThe answer the book gives, however, is { -7

**i**+ 17.3**j**} 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.

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