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Particle moving in a circle

  • Thread starter Jahnavi
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  • #1
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Homework Statement



circle.jpg

Homework Equations




The Attempt at a Solution



Since the force is always directed towards C , angular momentum about C should be conserved . But that doesn't seem to help as we need the relation at any general angle .

How should I proceed ?
 

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  • #2
ehild
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Homework Statement



View attachment 214888

Homework Equations




The Attempt at a Solution



Since the force is always directed towards C , angular momentum about C should be conserved . But that doesn't seem to help as we need the relation at any general angle .

How should I proceed ?
Point P moves along a circle centered at point O. Knowing that the force exerted on it points towards C, what are the radial and tangential accelerations in terms of theta?
 
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  • #3
Abhishek kumar
Point P moves along a circle centered at point O. Knowing that the force exerted on it points towards C, what are the radial and tangential accelerations in terms of theta?
In question already mentioned that point is moving only under the force acting towards C then from where tangential acceleration comes.
 
  • #4
cnh1995
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In question already mentioned that point is moving only under the force acting towards C then from where tangential acceleration comes.
If it were moving only under the force acting towards C, it wouldn't be moving in a circle as shown. It has both radial and tangential accelerations.
 
  • #5
Abhishek kumar
If it were moving only under the force acting towards C, it wouldn't be moving in a circle as shown. It has both radial and tangential accelerations.
Read question
 
  • #6
cnh1995
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Read question
I did. It's the 'net force' that acts towards C. It has both radial and tangential components such that the particle moves on a circle as shown.
 
  • #7
Abhishek kumar
I did. It's the 'net force' that acts towards C. It has both radial and tangential components such that the particle moves on a circle as shown.
You can broke the component of F along x and y direction.point is moving in circle but force not directed towards centre and as point move distance between point c and point p changes.i think it moves under some kind of restoring force
 
  • #8
cnh1995
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i think it moves under some kind of restoring force
Fortunately, we don't have to analyse the nature of force in the question. You can arrive at the answer with elementary trig and vector algebra.
 
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  • #9
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Point P moves along a circle centered at point O. Knowing that the force exerted on it points towards C, what are the radial and tangential accelerations in terms of theta?
Thanks !

Is this type of circular motion actually possible with the single force as shown in the picture ?

Is this some sort of planetary motion around the sun with perigee coinciding the focus ?
 
  • #10
cnh1995
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Is this type of circular motion actually possible with the single force as shown in the picture ?
Yes, and the condition under which it is possible is what the question is asking for.
 
  • #11
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Yes, and the condition under which it is possible is what the question is asking for.
I think it's the other way round :smile:
 
  • #12
cnh1995
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I think it's the other way round :smile:
Possibly, but did you get the answer? If you did, I can stop posting here since it's beyond my expertise to discuss the nature of this force (and the planetary motion you mentioned):-p.
 
  • #13
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Possibly, but did you get the answer?
I get option a) . Do you get the same ?
 
  • #14
cnh1995
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  • #15
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Sorry . I am actually getting option b) .

@ehild , please reply to post#9 .
 
  • #16
cnh1995
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Sorry . I am actually getting option b) .

@ehild , please reply to post#9 .
Unless I made some stupid mistake, it's d).
But I suppose now you know how to do it.
 
  • #17
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Unless I made some stupid mistake, it's d).
Let me check my work .

Edit : I am still getting option b) .
 
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  • #18
cnh1995
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Let me check my work .

Edit : I am still getting option b) .
You calculated (d2θ/dt2)/ (dθ/dt)2 right? It comes out to be 2cotθ.

I think you calculated the reciprocal.
 
  • #19
haruspex
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You calculated (d2θ/dt2)/ (dθ/dt)2 right? It comes out to be 2cotθ.

I think you calculated the reciprocal.
I also get (B). Consider θ=0. Would it not make sense that ##\ddot\theta=0## but ##\dot\theta<0##?
 
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  • #20
cnh1995
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I also get (B). Consider θ=0. Would it not make sense that ##\ddot\theta=0## but ##\dot\theta<0##?
You're right!
In my working, I had drawn the 2 radii perpendicular to each other in the diagram, so my θ was 45 degrees there. And using the alternate angles, I concluded that tanθ=Ft/Fr. Actually, tanθ=cotθ at θ=45°, so my assumption worked for θ=45° but it's actually cotθ=Ft/Fr.
Tricked by my own drawing!

My apologies!
 
  • #21
ehild
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Thanks !

Is this type of circular motion actually possible with the single force as shown in the picture ?
Yes.
Is this some sort of planetary motion around the sun with perigee coinciding the focus ?
I don't think so. The force points to C, but its magnitude changes along the circle. It is zero at C and highest at the opposite point.
upload_2017-11-12_21-17-42.png

I also got B.
 

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  • #22
ehild
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In question already mentioned that point is moving only under the force acting towards C then from where tangential acceleration comes.
See Post #21
 
  • #23
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but its magnitude changes along the circle. It is zero at C and highest at the opposite point.
Please explain how do we infer this from the question .
 
  • #24
Abhishek kumar
If we look in polar form then r and Θ both varies with respect to time.we can find expression for acceleration using r and Θ
 

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