Kids on a merry-go-round & Relative Motion

[Shreya]

Homework Statement

Three schoolboys, Sam, John, and Nick, are on merry-go-round. Sam and John occupy diametrically opposite points on a merry-go-round of radius R. The position of the boys at the initial instant are shown in fig
Considering that the merry-go-round touch each other and rotate in the same direction at the same angular velocity ω determine the nature of motion of Nick from John's point of view.

I have understood the argument given in the solution provided by the textbook which is

Let the merry-go-round turn through a certain angle φ Fig. We construct a point O(OA=R) such that points O,S,A and J lie on the same straight line. Then it is clear that ON=R+r at any instant of time. Besides, Point Ois at rest relative to John (Sam is always opposite to John). Therefore, from John's point of view, NIck translates in a circle of radius R+r with the centre at point O which moves relative to the ground in a circle of radius R with the centre at pointA.

But, I can't seem to convince myself that Nick moves in a circle. I tried to reason using relative velocity follows:

Relevant Equations

$$v = \omega r$$
$$v_{AB} = v_A - v_B$$

(Please refer the image below. ) The velocity of Nick $v_N=\omega r$ and the velocity of John $v_S=\omega R$ is depicted. The relative velocity of Nick with respect to John will be $v_{NJ} = \omega (R-r)$. The velocity is along the tangent to the circle centred at B. If Nick were to move in a circle centred at O from Sam's point of view, i would expect his relative velocity to be perpendicular to ON (which it is not).
Similarly, the velocity at instant of time of Nick with respect to point O is 0.

Both of these confuse me a lot. I would really appreciate any help. Thank You!
(Edit: Sorry for the wrong formatting of the equations. I have edited them.)

 

[kuruman]

But, I can't seem to convince myself that Nick moves in a circle.

Then start by writing an equation giving Nick's position vector as a function of time relative to the center of Nick's merry-go-round:
$$\vec R_N=-R_N\cos\omega t~\hat x+R_N\sin\omega t~\hat y.$$ A similar-looking equation applies for Sam's position vector $\vec R_S$ and John's position vector $\vec R_J.$ Write them down. Now you want to write new position vectors for Sam and John with reference to Nick's origin. How can you do that? Once you have the position vectors, you can easily find the velocities. That should be sufficient to get you going.

 

[Shreya]

Let me try that.
Nick:
$$\vec R_N = -R cos (\omega t) \hat {x} + R sin (\omega t) \hat y$$
$$\vec V_N = R\omega sin (\omega t) \hat x + R\omega cos (\omega t ) \hat y$$
Sam:
$$\vec R_S = -r cos (\omega t) \hat x + r sin (\omega t) \hat y $$
$$\vec V_S = r\omega sin (\omega t) \hat x + r\omega cos (\omega t) \hat y $$
John:
$$\vec R_J = r cos (\omega t) \hat x - r sin (\omega t) \hat y $$
$$\vec V_J = -r \omega sin (\omega t) \hat x - r \omega cos (\omega t) \hat y$$
Nick with respect to John:
$$\vec R_{NJ} = - (R+r) cos (\omega t) \hat x + (R+r) sin (\omega t) \hat y$$
$$\vec V_{NJ} = (R+r) \omega cos (\omega t) \hat x + (R+r) \omega cos (\omega t) \hat y $$

nice! I can now see why Nick moves in a circle of radius $R+r$ with respect to John.
Could you please help me understand why the circular motion is around O through these equations?

 

[kuruman]

You missed my point. I will use double subscripts with the understanding that Nick rotates about point B and the other two rotate about point A as shown in the drawing.$$\vec R_{\text{NB}} = -R cos (\omega t) \hat {x} + R sin (\omega t) \hat y.$$ For the other two, $$\begin{align} & \vec R_{\text{SA}} = -r cos (\omega t) \hat {x} + r sin (\omega t) \hat y \nonumber \\ & \vec R_{\text{JA}} = r cos (\omega t) \hat {x} - r sin (\omega t) \hat y \nonumber \end{align}$$ You need $\vec R_{\text{NJ}}.$ How do you get there? Hint: What is $\vec R_{\text{AB}}$?

 

[Shreya]

Sorry i missed that.

$\vec R_{AB} = (R+r) \hat x$
$\vec R_{NJ} = \vec R_{NB} +\vec R_{BA}+ \vec R_{AJ}$
$\vec R_{NJ} =[ - (R+r) cos \omega t + (R+r)] \hat x + (R+r) sin \omega t \hat y$​

Is this better?
(Ps: Sorry for the late reply, it was close to midnight in my timezone)

 

[haruspex]

Sorry i missed that.

$\vec R_{AB} = (R+r) \hat x$
$\vec R_{NJ} = \vec R_{NB} +\vec R_{BA}+ \vec R_{AJ}$
$\vec R_{NJ} =[ - (R+r) cos \omega t + (R+r)] \hat x + (R+r) sin \omega t \hat y$​

Is this better?
(Ps: Sorry for the late reply, it was close to midnight in my timezone)

That would all be correct if you just change the first line to $\vec R_{BA} = (R+r) \hat x$.
Can you see that the final expression describes a circular trajectory?

 

[Shreya]

That would all be correct if you just change the first line to R→BA=(R+r)x^.

but isnt $\vec R_{BA} = - (R+r) \hat x$ as AB is the direction of positive x.

Can you see that the final expression describes a circular trajectory?

Yes! I do. Nick seems to move in a circular trajectory with respect to John with a shifted centre by R+r.
But, I can't understand why the shifted centre has to be O using these equations.

 

[kuruman]

Drawing a diagram helps. Starting at J, you can get to N directly along the red vector or indirectly via the sum of the black vectors. That sets up the equation for $\vec R_{\text{NJ}}.$

 

[Shreya]

Thanks for all the help. I got to go to school now. I will think about all that you have said regarding the problem and reply back as soon as i get back.

 

[haruspex]

but isnt $\vec R_{BA} = - (R+r) \hat x$ as AB is the direction of positive x.

Sorry, wrong correction.
I should have written, change the sign on the $(R+r) \hat x$ term.

 

[kuruman]

but isnt $\vec R_{BA} = - (R+r) \hat x$ as AB is the direction of positive x.

The double subscript convention "AB" normally stands for "B relative to A", which mathematically is written as $\vec R_{\text{B}}-\vec R_{\text{A}}$. The direction implied in "B relative to A" is "from A to B."

 

[Shreya]

The double subscript convention "AB" normally stands for "B relative to A", which mathematically is written as $\vec R_{\text{B}}-\vec R_{\text{A}}$. The direction implied in "B relative to A" is "from A to B."

You were right, @haruspex, It was my fault that I hadn't thought much before replying.

I realise that $\vec R_{NJ} =[ - (R+r) \cos \omega t + (R+r)] \hat x + (R+r) \sin \omega t \hat y$ means that Nick moves in a circle with respect to John with a radius of $(R+r)$
But, I interpret the orange-coloured term as shifting of the origin from John to a point $(R+r)$ away from John around which Nick moves. I also can see that the point O is $(R+r)$ from John always.
But, Isnt point O & $(R+r) \hat x$ different?
Please be kind to help

 

[kuruman]

Imagine being John. From your point of view Nick is describing a circle. You have correctly deduced that the radius of the circle is $(R+r)$. Where in relation to this circle do you, John, who are at rest do you stand? Clearly you are not at its center.

 

[Shreya]

Well, The centre of the circle is $(R+r)$ term away from me due to the $(R+r) \hat x$ term.
I also tried plotting it (by taking $\omega = \pi$ and $(R+r)=5$:

But this centre isn't O. What am I missing out ?

Wait a min, It is O, isnt it. I gotta draw more diagrams.

I tried to represent how I see Nick while depicting myself in ground frame. The solid line is how I move and the dotted line is how I see Nick at every instant.
Hmm, there's just one thing missing - the centre of nick's motion should be R away from me (not R+r, as in the diagrams), why is it so?
Could you please help me find it?

 

[kuruman]

You have answered the question posed in the problem. The circle that you show in post #14 is Nick's position relative to John at different times. John stands on the circumference of a circle of radius $(R+r)$ and sees Nick move on the circumference of this circle. I think this is what the statement of the problem meant with "determine the nature of motion of Nick from John's point of view."

As for point O, imagine Oscar standing at point O and moving with it. What is John's motion from Oscar's point of view? Hint: Find an expression for $\vec R_{\text{JO}}$ and see what it says to you.

 

[TSny]

$\vec R_{NJ} =[ - (R+r) \cos \omega t + (R+r)] \hat x + (R+r) \sin \omega t \hat y$

This expresses $\vec R_{NJ}$ in terms of the x-y coordinate system attached to the earth. That is, you have $$x_{ _{NJ} }= -(R+r) \cos \omega t +(R+r) \;\;\;\;\;\;\;\; y_{ _{NJ} }= (R+r) \sin \omega t$$Relative to this frame, J, O, and N are all in motion.

J's frame of reference is the frame attached to disk $A$ and rotating with the disk. Call this the x'-y' frame.

Relative to this frame, J and O are at rest and only N is in motion. The motion of N relative to J in this frame is the motion of N as seen by J. Can you express $\vec R_{NJ}$ in the primed frame? That is, find $x'_{ _{NJ} }$ and $y'_{ _{NJ} }$ from the known $x_{ _{NJ} }$ and $y_{ _{NJ} }$. See this for a review of transforming between rotated frames.

 

[Shreya]

Relative to this frame, J and O are at rest and only N is in motion. The motion of N relative to J in this frame is the motion of N as seen by J. Can you express R→NJ in the primed frame? Equivalently, find xNJ′ and yNJ′ from the known xNJ and yNJ. See this for a review of transforming between rotated frames.

(I am not quite familiar with rotating reference frames, but I have tried to do it. Please do check for any errors.)
As $$x'=x \cos \theta + y sin \theta $$
and $$y'=-x \cos \theta + y \cos \theta $$
Here, $\theta = - \omega t$,
$$x'_{NJ}= - (R+r) + (R+r) \cos \omega t$$
$$y'_{NJ} = (R+r) \sin \omega t$$

So I redrew my diagram, to account for these equations

Could you please check if this is right?

 

[kuruman]

Could you please check if this is right?

What question are you trying to answer now? Please state it clearly. I think that you have already answered the question posed in the problem "determine the nature of motion of Nick from John's point of view."

 

[TSny]

(I am not quite familiar with rotating reference frames, but I have tried to do it. Please do check for any errors.)
As $$x'=x \cos \theta + y sin \theta $$
and $$y'=-x \cos \theta + y \cos \theta $$
Here, $\theta = - \omega t$,
$$x'_{NJ}= - (R+r) + (R+r) \cos \omega t$$
$$y'_{NJ} = (R+r) \sin \omega t$$

Yes. These expressions look correct.

So I redrew my diagram, to account for these equations
View attachment 320121

This looks good for the motion of Nick in the primed frame. John and point O are fixed in place in this frame. N moves counterclockwise on a circular path of radius R+r as shown in your diagram. O is the center of the path.

View attachment 320122

I'm not sure what this picture represents.

 

[kuruman]

J's frame of reference is the frame attached to disk A and rotating with the disk.

My interpretation is that J's frame of reference is one in which J is at rest and everything else rotates about him. That includes points A and O. Point N is not at fixed distance from J. I still am not sure what specific question the posted diagrams and equations are intended to answer.

For any of pair of points of interest, it seems to me that it would be sufficient to write down a relative position vector, e.g. $\vec R_{\text{NJ}}$ and then generate a parametric plot. This was done for NJ and illustrated in post #14. Exactly what points are we concerned with now? I still don't know.

 

[TSny]

My interpretation is that J's frame of reference is one in which J is at rest and everything else rotates about him. That includes points A and O. Point N is not at fixed distance from J.

I imagine John and Sam as always facing each other. John thinks of himself as at rest. He sees Sam as always in front of him at a fixed distance 2r. So, Sam is at rest from the point of view of John. From John's point of view, point O is always located behind Sam at a constant distance R+r from John. So, John takes O to be at rest relative to John. At the initial time, John and Nick are coincident. As time increases, John continues to face Sam, and Nick comes into John's peripheral view on John's right. John sees Nick move out in front of him along a circle with a radius R+r and center at O. At the instant that Nick has made half an orbit on this circle, John would see Nick located directly behind Sam at a distance of 2(R+r) from John. Nick returns to John's location after one full orbit.

I still am not sure what specific question the posted diagrams and equations are intended to answer.

For any of pair of points of interest, it seems to me that it would be sufficient to write down a relative position vector, e.g. $\vec R_{\text{NJ}}$ and then generate a parametric plot. This was done for NJ and illustrated in post #14. Exactly what points are we concerned with now? I still don't know.

The circle plotted in post #14 can be thought of as representing the motion of N relative to J if J always faces in one direction relative to the earth frame. So, if J is facing toward the left at the initial position, then J remains facing toward the left relative to the earth as time goes on. If we thought of John as a rigid body, we would say that John is in pure translation relative to the earth as he rides the merry-go-round. He does not rotate relative to the earth as he did when he was always facing Sam. In this scenario, John would now consider Sam to be orbiting around John.

John would now say that N moves along a fixed circle of radius R+r that lies behind him rather than in front of him. This is the circle plotted in post #14. John is at the origin and always facing left along the negative x-axis. John would say that point O also moves in a circle of radius R+r with the center of the circle located at John's position. So, O's orbit is identical to N's orbit except O's orbit is shifted a distance R+r in the negative x direction. John would say that point O is located, at each instant, at a distance R+r in the negative x direction relative to N. So, point O is not at the center of the circle of post #14 except at the instant when N is directly to the rear of John at maximum distance from John.

But I don't think this scenario is what was intended in the problem. I think we are to assume that John is always facing in a fixed direction relative to disk A, say towards Sam. So, John's frame of reference is a frame that is rotating relative to the earth. So, we transform to the rotating frame to get John's viewpoint.

 

[kuruman]

I imagine John and Sam as always facing each other. John $\dots$

I agree with your clear explanation. My interpretation was that John and Nick and any other observer are fixed relative to the surface they stand on but they can swivel their heads panoramically to get the big picture.

 

[Shreya]

Exactly what points are we concerned with now? I still don't know.

I am sorry. I was just trying to do it in the rotating frame and understand it better.

but they can swivel their heads panoramically to get the big picture

well that mustn't be hard as they are schoolboys :)

I'm not sure what this picture represents.

That awfully drawn diagram was meant to represent how Nick's motion was perceived by John at different instants of time. I wanted to just confirm that the circle made by O was at a distance of $R$ from A (as was mentioned in the textbook explanation).

Could you tell me which software do you use to draw images like the ones you posted in #8 & #16? I would love to draw to like that.

I imagine John and Sam as always facing each other. John thinks of himself as at rest. He sees Sam as always in front of him at a fixed distance 2r. So, Sam is at rest from the point of view of John. From John's point of view, point O is always located behind Sam at a constant distance R+r from John. So, John takes O to be at rest relative to John. At the initial time, John and Nick are coincident. As time increases, John continues to face Sam, and Nick comes into John's peripheral view on John's right. John sees Nick move out in front of him along a circle with a radius R+r and center at O. At the instant that Nick has made half an orbit on this circle, John would see Nick located directly behind Sam at a distance of 2(R+r) from John. Nick returns to John's location after one full orbit.

Thanks a lot for this, it really nailed down my understanding on how rotating reference frames work.

Thank you so much @kuruman & @TSny. I was struggling with this problem, until you both helped me. I appreciate that you helped me uncover the solution without straight away giving me the solution. I am not sure how I can express my gratitude for all the time & effort you took,except by this message. So, Thank you!

 

[TSny]

Could you tell me which software do you use to draw images like the ones you posted in #8 & #16?

I use Microsoft Paint. If I need to add mathematical expressions, I usually create them using the equation editor in PowerPoint and then copy-and-paste into Paint.

 

[kuruman]

I use Microsoft Paint. If I need to add mathematical expressions, I usually create them using the equation editor in PowerPoint and then copy-and-paste into Paint.

For #8 I used the drawing tools in Microsoft Powerpoint.

 

[Shreya]

Thanks! I'll try them when I draw diagrams :)