Bead sliding on a Vertical Circular Loop versus in Free Fall

  • #1
Rubberduck2005
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Homework Statement:
A circle with a radius R lies in the vertical plane. A bead falling under gravity moves from A( the top point) to B ( an antipodal point to A), Moving a long the diameter. Now consider a separate bead, such a bead leaves A at an angle W from the vertical and moves a long a cord to a point C. Prove that both beads take the same time to complete their respective journeys.
Relevant Equations:
g=9.81ms^-2 j is the y unit vector , i the x unit vector
I can evaluate the first beads motion easily A to B is -2Rj considering the point B as y=0 the motion of the bead will be -gt^2/2+2R=0 which implies t=2√(R/G) , this is ok but what I am struggling with is A to C I can see that the angle between the beads weight and it's negative normal force is W-90 as (180=180-W + 90 + (the new angle=L), given this I can derivate that the acceleration moving down the plane is gsin(L)=gsin(W-90)=-gcos(W) now the motion of this bead need be -1/2(gcos(w))t^2+2R=C(J) I can also see that A to C= A to O + O to C= -R + O to C, I can't find the Y coordinated of C. If you see a flaw in my reason please point it out.

 

Answers and Replies

  • #2
Rubberduck2005
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Untitled.png
 
  • #3
haruspex
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the angle between the beads weight and it's negative normal force is W-90
I get 270-W.
can't find the Y coordinated of C
What is angle BOC in terms of BAC? (Well known theorem.)
 
  • #4
Rubberduck2005
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haruspex With regards to your first comment I create a triangle using (180 -W) as my top angle and the bottom left being 90 from this I deduced that the angle in the bottom left need be 90-W thus the angle between the weight and the negative normal be that angle?
 
  • #5
Rubberduck2005
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@haruspex I am also not sure what theorem you are talking about
 
  • #6
haruspex
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the angle in the bottom left
I assume you mean bottom right, and specifically the angle that AC makes to the horizontal. That is not what I thought you meant by the angle between the beads weight and it's negative normal force. I took that to mean the angle between a vertical line down from the particle and the upward normal to AC from the particle.
@haruspex I am also not sure what theorem you are talking about
Angles subtended by a chord at centre and circumference.
Google that if you still don't know.
 
  • #7
kuruman
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angle BOC = 2BAC-180?@haruspex
Let's see if that works when OC is horizontal, in which case BOC = 90o and BAC = 45o. If your formula is correct, it should be true that 90o = 2×45o - 180o. Is it true? If not, what does this simple test suggest about the correct formula?
 
  • #8
Rubberduck2005
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kuruman hi thank you for dismissing my claim can you link me to a video on this theorem or simply hint at how to achieve insight into what the idea could be
 
  • #9
Rubberduck2005
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angles of the same base cord are equal? but it doesn't connect with the circumforance
 
  • #10
haruspex
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angles of the same base cord are equal? but it doesn't connect with the circumforance

The first that came up when I googled was https://www.cuemath.com/geometry/arcs-and-subtended-angles, though it refers to angles subtended by arcs rather than chords. Same deal.
A little further on I hit https://amsi.org.au/teacher_modules/Circle_Geometry.html#Angles_at_the_centre_and_circumference.
Immediately following, https://www.scienceandmathsrevision.co.uk/topic/circle-theorems/.
How did you not find a relevant one?
 
  • #11
Delta2
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Your schematic is a bit off regarding how W angle is defined: The problem statement clearly says that W is the angle from the vertical while in your schematic you show it to be the angle from the horizontal.

If we define W as above then the component of acceleration parallel to AC is simply ##g\cos W##.
Then I used cosine law to find AC as a function of the radius R and the angle AOC. Then used a bit of trigonometry and geometry (namely that the angles OAC(=W by definition) and OCA are equal (why?), hence AOC=180-2W) to transform the expression for AC. Then simply used $$\frac{1}{2}g\cos W t^2=AC$$ and I got the desired result.
 
  • #12
haruspex
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while in your schematic you show it to be the angle from the horizontal.
As I read the drawing, it is the angle to the upward vertical, so obtuse.
 
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  • #13
Delta2
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As I read the drawing, it is the angle to the upward vertical, so obtuse.
Ah yes, I see that now. Well , anyway, in my opinion it works better if we define W=OAC.
 
  • #14
Delta2
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I think this problem is more of a problem in trigonometry and geometry, from physics the only equation we use is the equation of uniform accelerating motion in a line, that is##s=\frac{1}{2}at^2##.
 
  • #15
kuruman
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I think this problem is more of a problem in trigonometry and geometry, from physics the only equation we use is the equation of uniform accelerating motion in a line, that is##s=\frac{1}{2}at^2##.
I agree. One can write two equations
##AC=\frac{1}{2}at_1^2##
and
##AB=\frac{1}{2}gt_2^2##

If it is true that the times are equal (##t_1=t_2##), then it should also be true that
##\dfrac{AC}{AB}=\dfrac{a}{g}##.
 

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