Recently I saw the following interesting problem

Ann is sitting on the edge of a carousel that has a radius of 6 m and is rotating steadily. Bob is standing still on the ground at a point that is 12 m from the centre of the carousel. At a particular instant, Bob observes Ann moving directly towards him with a speed of $1m/s$ With what speed does Ann observe Bob to be moving at that same moment?

The solution to this problem has also been given . However , there is a small sentence in the solution that I find quite confusing. the solution of the problem is very clear up to this point:

enter image description here

Since Ann is moving directly towards Bob, his position, B in the figure, must lie on the tangent to the carousel at Ann’s position A. Thus A, B and C, the centre of the carousel, must form a right-angled triangle. Using the given geometrical data, it follows that the distance between Ann and Bob at the given moment is $6\sqrt{3}$m. It also follows that the tangential speed of the carousel is $1m/s$ and that its angular velocity is therefore $ω = \frac{1}{6}rad/s$ If Ann were sitting at the centre of the carousel, she would see the whole world around her rotating with the same angular speed ω, but in the opposite direction. That means she would observe Bob standing 12 m away from the centre of the carousel, but moving with a speed of $ \frac{1}{6}$x12 = 2 m/s in a direction perpendicular to the line joining him to the centre of the carousel.

However in the next line , they’ve stated that :

Although Ann is not sitting at the centre of the carousel, but at its edge, the same conclusion applies – namely that, according to Ann, Bob’s speed is 2 m /s.

Initially , I was very confused about this statement. If Ann sits on the centre of the carousel , she will see her friend Bob moving with a velocity 2m/s. However , the same kind of motion cannot be observed from the frame of reference of the edge of the carousel. This is because the edge of the carousel also has a linear velocity which has to be accounted for, in order to truly bring Bob into Ann’s frame of reference . However , the diagram on the right indeed implies that they have considered the same thing and have taken into account the linear velocity of Ann. My question is : Is it correct to assume that the velocity of 2m/s was given to Bob in order to account solely for the rotational motion of the carousel , and that the resultant of the velocities $2m/s$ and $1m/s$ gives the true velocity of Bob as observed by Ann?

This problem is taken from 200 More Puzzling Problems in Physics

  • $\begingroup$ Let's assume that Bob and Ann each has a radar gun. Bob points his radar gun at Ann, and Ann points her radar gun at Bob. When Ann gets to the indicated position, they both take a reading. In my opinion, both radar guns will read 1 m/s at that instant. $\endgroup$ – David White Jun 9 '19 at 16:26
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    $\begingroup$ @David thank you , but the hint to the question says that the frames of reference are not inertial and hence it would be wrong to suggest Galilean Transformation in this . Thus the velocity with which Ann sees Bob would not be 1m/s $\endgroup$ – Aditi Jun 9 '19 at 16:37
  • $\begingroup$ I'd still be interested in testing your assertion in the real world. $\endgroup$ – David White Jun 9 '19 at 17:44
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    $\begingroup$ @David I’d be thankful to you for that ! I’d really like to know where I went wrong :) $\endgroup$ – Aditi Jun 9 '19 at 17:51

$\let\om=\omega \def\v#1#2{\vec{#1}_{\rm #2}} \def\ora#1{\overrightarrow{#1}}$ Your error (and not only yours) is in saying a thing like "Ann's frame of reference". By this way you don't specify a reference frame. It's not enough to say that in that frame Ann is steady - you also have to tell if and how that frame is rotating.

Actually there are only two frames relevant to the problem:

  • K, fixed to ground
  • K$'$, rotating with the carousel.

In K, B is steady (velocity $\v vB=0$) and A's velocity is $$\v vA = \v\om{}\times\v rA$$ always tangent to the circle. There are two positions of A where $\v vA$ directly points to B: the one drawn, where A is approaching B, and that symmetric wrt CB, where A is receding from B.

In K$'$, $\v vA'=0$ whereas $$\v vB' = -\v\om{} \times \v rB.$$

This is on a straight line from B orthogonal to CB and never touches the circumference. Then B's velocity is never directed towards A. The right drawing shows that: B's velocity has a component towards A, of $1\,\rm m\,s^{-1}$, but this is not the full velocity of B as seen from K$'$ frame. It's drawn as the downward $2\,\rm m\,s^{-1}$ vector. This has nothing to do with A's location.


In order to better understand the meaning of both components of $\v vB'$ let's decompose $\v rB$ in eq. (1): $$\v rB = \ora{\rm CA} + \ora{\rm AB}.$$ Then $$\v vB' = -\v\om{} \times \ora{CA} - \v\om{} \times \ora{AB}.\tag2$$ Eq. (2) shows that $\v vB'$ is the (vector) sum of two contributions. The former is the one pointing towards A, the latter is perpendicular. The latter arises because what you called "A's frame" is rotating with angular velocity $\v\om{}$ so that B looks rotating around A in the opposite direction.

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    $\begingroup$ Thank you , but could you please explain why $\sqrt{3}$m/s is also relevant to this answer ? I was wondering whether that was the actual velocity of B as seen by A when she sits on the edge of the carousel . Thanks again ! $\endgroup$ – Aditi Jun 10 '19 at 3:36
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    $\begingroup$ @Aditi I edited my post to clarify that point. $\endgroup$ – Elio Fabri Jun 10 '19 at 8:32

I can't see the relevance of the non-inertial frame. The velocity of B relative to A is equal and opposite to that of A relative to B.

But, you say, when A is at the centre, in her frame of reference, B's velocity relative to her is 2 m/s, but in B's frame, her velocity relative to B is zero.

But I'd still have thought that when A is at the edge, the 1 m/s fully takes account of the rotation.

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    $\begingroup$ Thank you , but I was wondering if the principal of velocities being equal and opposite was true only if both the objects were in the same reference frame or atleast their different reference frames were not accelerating with respect to each other ? $\endgroup$ – Aditi Jun 9 '19 at 17:35
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    $\begingroup$ I also think that when A is at the centre , she would see B moving past her and hence would not say that B’s velocity is zero , that is if we’re analyzing from the reference frame of A. $\endgroup$ – Aditi Jun 9 '19 at 17:49
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    $\begingroup$ From our reference frame we can say that B is not being displaced , but I think that may not be the case with A’s reference frame fixed to the centre right ? A would see B being displaced . $\endgroup$ – Aditi Jun 9 '19 at 17:54
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    $\begingroup$ I'm sorry; I thought I'd removed this answer! In A's frame, even when A is at the centre, B's displacement IS changing. But when A is at the edge, I'd still have thought that the 1 m/s velocity of A relative to B is equal and opposite to that of B relative to A $\endgroup$ – Philip Wood Jun 9 '19 at 18:52
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    $\begingroup$ Suppose we stand at opposite ends of a room. Clearly our relative velocities are zero. But if I start spinning in place, then I see you moving, and hence assign you a nonzero velocity in my noninertial frame, despite the fact that you still assign me a zero velocity in your frame. That's what this question is getting at. $\endgroup$ – knzhou Mar 29 '20 at 18:11

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