Kinematics of Rigid Body Plane Motion

1. Jul 18, 2016

Auburn2017

1. The problem statement, all variables and given/known data
The wheel shown rotates about point O.
Point A has a velocity=-7j in/sec.
Point B has a tangential velocity=-4i in/sec^2
Determine the absolute acceleration of Point C located on the circle.
You will have to look at the figure for more clarification please.
2. Relevant equations
rA=rB+rA/B
vA=vB+vA/B
aA=aB+aA/B
vA/B=ω×rA/B
aA/B=(aA/B)n+(aA/B)t
(aA/B)n=ω×(ω×rA/B)
(aA/B)t=α×rA/B
3. The attempt at a solution
I tried finding the velocity of C relative to A. I feel like there is an easy way to find the angular acceleration and angular velocity of the wheel but I dont really know how. Thanks for your help.

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2. Jul 18, 2016

TSny

The wheel is considered to be a rigid body rotating about a fixed axis. How does the absolute speed of point C compare to the absolute speed of A? How does the magnitude of the absolute tangential acceleration of C compare to that of B?

Last edited: Jul 18, 2016
3. Jul 18, 2016

Auburn2017

Since all three points are on the edge of the wheel then they should all have the same absolute acceleration and velocity.

4. Jul 18, 2016

TSny

The magnitudes of the velocity and tangential acceleration will be the same. Directions?

5. Jul 18, 2016

Auburn2017

velocity CW
acceleration CCW

6. Jul 18, 2016

TSny

OK. But more specifically, can you express the tangential acceleration vector of point C in terms of the unit vectors $\hat{i}$ and $\hat{j}$?

7. Jul 18, 2016

Auburn2017

No I can't. Isn't that what the problem is asking?

8. Jul 18, 2016

TSny

No, there's more to the problem than just doing that. Does point C have any other component of acceleration besides tangential?

9. Jul 18, 2016

Auburn2017

Yes it has a normal acceleration toward the center of the circle. I guess the tangential acceleration would be atcos60° i and atsin60° j

10. Jul 18, 2016

TSny

You need to draw the tangential and normal acceleration vectors for point C on your diagram. The tangential acceleration of C is not vertically upward (i.e., not +j) and it's not "to the right" (i.e., not i).

11. Jul 18, 2016

Auburn2017

12. Jul 18, 2016

TSny

That's getting close. But you need to check if 60o is correct here.

13. Jul 18, 2016

Auburn2017

yes it is correct

14. Jul 18, 2016

TSny

Did you actually draw the tangential acceleration vector at C on your diagram? What angle does it make to the horizontal?

15. Jul 18, 2016

Auburn2017

Yes I drew it. I'm not sure how to tell the angle though...

16. Jul 18, 2016

TSny

You'll need to use a little geometry.

17. Jul 18, 2016

Auburn2017

I'm going with 30 degrees

18. Jul 18, 2016

TSny

Did you work it out, or is that a guess?

Look at the figure below. What is the angle $\phi$? Form $\phi$ do you see how to get $\theta$?

19. Jul 18, 2016

Auburn2017

Yes theta would equal 30 degrees

20. Jul 18, 2016

TSny

Good. Repeat for the normal acceleration.