Rotating reference frame help

  • Thread starter natalie
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  • #1
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Homework Statement




I have attached the problem as a picture on this post, am really really unsure on how to start!

so far the only thing i can think of doing is using this equation

[tex] (\frac{d^{2}r}{dt^{2}})_{s} [/tex] = [tex]( \frac{d^{2}r}{dt^{2}})_{s'} + 2ω \times (\frac{dr}{dt})_{s'} + \dot{ω} \times r + ω \times [ω \times r] [/tex]


And now just solve for r, but in the s reference frame ?

any help appreciated really stuck.
 

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Answers and Replies

  • #2
6,054
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Not even part (i)?
 
  • #3
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Not even part (i)?

No, I dont see how it works?
 
  • #4
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Assume that the rod is pointing momentarily north.

What forces acting on the bead, in the inertial frame, can you think of? What are their directions?
 
  • #5
vanhees71
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Hm, I must admit that I always have trouble using forces to derive the equations of motion. If you have Hamilton's principle at hand, it's way simpler to use the Euler-Lagrange equations and then work out the forces at the very end ;-)).
 
  • #6
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then work out the forces at the very end ;-)).

That's cheeeeeating! :)
 
  • #7
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Assume that the rod is pointing momentarily north.

What forces acting on the bead, in the inertial frame, can you think of? What are their directions?

so basically, if i understand correctly, the rod is lying in the x' axis. we have a weight force, and a normal force. thats in the intertial reference frame? the weight force cancels out the normal....
 
  • #9
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so basically, if i understand correctly, the rod is lying in the x' axis. we have a weight force, and a normal force. thats in the intertial reference frame? the weight force cancels out the normal....

If the weight cancelled the normal force, the bead would have zero resultant force acting on it. What it is the motion under zero resultant force? Does that seem plausible in the situation at hand?
 

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