Relative velocities in Circular Motion

Question

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:

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

Answer 1

$\let\om=\omega \def\v#1#2{\vec{#1}_{\rm #2}} \def\ora#1{\overrightarrow{#1}}$ 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.

Edit

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.

Answer 2

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.