Circular motion problem.

  • #1
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Poster has been reminded to show their work in their OP, instead of having to be reminded of that by other respondents

Homework Statement


A particle A moves along a circle of radius ##R = 50 cm## so that its radius vector ##r## relative to the point O (Fig. 1.5) rotates with the constant angular velocity ω = 0.40 . Find the modulus of the velocity of the particle, and the modulus and direction of its total acceleration.
6.PNG

Homework Equations


Circular movement equations.

The Attempt at a Solution


I tried to do it bye setting the coordinate system on the point O, but after that I really don't know what to do next.
 

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Answers and Replies

  • #2
kuruman
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Please show what you did and how far you got.
 
  • #3
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Please show what you did and how far you got.
I just set the coordinate system on O. Then I ended up with a triangle and used the sine law, ending up with ##\frac{R}{\sin \theta}=\frac{r}{\sin(\pi - 2\theta)}##. After that I don't know what to do.
 

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  • #4
kuruman
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Your drawing will lead you to trouble because you show two items of length R. If you consider them as vectors, one of them from O to the center of the circle is fixed while the other varies with time because its direction changes. I would call them different names, then I would write the vector from O to A in terms of the angles you specified and take time derivatives. Note that in your drawing ##\frac{d\theta}{dt}=\omega=const.##
 
  • #5
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Your drawing will lead you to trouble because you show two items of length R. If you consider them as vectors, one of them from O to the center of the circle is fixed while the other varies with time because its direction changes. I would call them different names, then I would write the vector from O to A in terms of the angles you specified and take time derivatives. Note that in your drawing ##\frac{d\theta}{dt}=\omega=const.##
Ok, I see. I continued with what I had before, ended up with this: ##r=2R\cos\theta##. The system is in polar coordinates, right? So ##x=r\cos\theta## and ##y=r\sin\theta##, so I replaced r and ended up with ##\vec r=2R\cos^{2}(\theta)\vec i + 2R\cos(\theta)\sin(\theta)\vec j##. How do I take the derivative with respec to fime of that?
 
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  • #6
kuruman
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Have you done all that? Do the same for the other vector to A shown in the original drawing. That also goes around. What is its angular velocity relative to ω? Think about all that, do what what you have to do and maybe you will see where to go next. Be sure to show what you have done up to that point if you need to ask for more help.
 
  • #7
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Have you done all that? Do the same for the other vector to A shown in the original drawing. That also goes around. What is its angular velocity relative to ω? Think about all that, do what what you have to do and maybe you will see where to go next. Be sure to show what you have done up to that point if you need to ask for more help.
Sorry What other vector? I did it for the r vector that is the drawing.
 
  • #8
haruspex
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I just set the coordinate system on O. Then I ended up with a triangle and used the sine law, ending up with ##\frac{R}{\sin \theta}=\frac{r}{\sin(\pi - 2\theta)}##. After that I don't know what to do.
Figuring out that the R radius shown makes twice the angle to the horizontal that the r vector makes to it was a good start. You can now write down the locus of the particle in Cartesian coordinates centred on the centre of the circle. Adjusting to make the origin at O is trivial.
 
  • #9
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Figuring out that the R radius shown makes twice the angle to the horizontal that the r vector makes to it was a good start. You can now write down the locus of the particle in Cartesian coordinates centred on the centre of the circle. Adjusting to make the origin at O is trivial.
How do I do that? I tried just taking the derivative with respect to time of ##\vec r=2R\cos^{2}(\theta)\vec i + 2R\cos(\theta)\sin(\theta)\vec j##.
 
  • #10
haruspex
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How do I do that? I tried just taking the derivative with respect to time of ##\vec r=2R\cos^{2}(\theta)\vec i + 2R\cos(\theta)\sin(\theta)\vec j##.
I wouid think it is easier to leave it in terms of 2θ. What do you get when you differentiate?
 
  • #11
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I wouid think it is easier to leave it in terms of 2θ. What do you get when you differentiate?
I get ##\vec v=-2R\sin(2\theta)\omega \vec i + 2R\cos(2\theta)\omega \vec j##. I got ##\omega## because when I differentiate it it was like differentiating it with recpect to ##\theta##, so then by the chain rule I ended up with ##\frac{d\theta}{dt}=\omega## is it right?
 
  • #12
kuruman
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Yes. So what is an expression for the modulus of the velocity?
 
  • #13
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Yes. So what is an expression for the modulus of the velocity?
It is the square root of the component on x squared and the component on y squared. I got ##2R\omega##.
 
  • #14
kuruman
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It is the square root of the component on x squared and the component on y squared. I got ##2R\omega##.
Right. Now do the acceleration by taking another derivative.
 
  • #15
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Right. Now do the acceleration by taking another derivative.
Exactly. I did the same process and got ##4R\omega^{2}##.
 
  • #17
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Then you're done.
Thanks. I made another question, and I'm having trouble with it, can I post the link here?
 
  • #18
kuruman
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Thanks. I made another question, and I'm having trouble with it, can I post the link here?
No. Please create a separate thread.
 
  • #19
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No. Please create a separate thread.
Ok, thanks I created another one.
 

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