# A professor's ride (Don't worry! You would be safe) This situation is very interesting -

A professor placed two of his deserving students on a special ride. The special ride consists of a disc which spins with angular speed $ω$. The two students sit on chairs which are aligned along one of the radii of the disc at distance $S1$ and $S2$ respectively from the center. Spinning the disc, the professor asks them to find each other's relative velocity.

Opinion 1: The relative velocity among the students is zero as they don't appear to move with respect to each other.

Opinion 2: The relative velocity of the students is not zero as the velocity of one student is $ωS1$ and the other's velocity is $ωS2$. Hence the relative velocity among them is $ω(S1-S2)$.

[By relative velocity, I mean velocity of A with respect to B or vice-versa, where A and B are the two students.]

Which one of them do you think is right and why? Note by Lokesh Sharma
6 years, 1 month ago

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Opinion one is correct. Person A is stationary in the frame of reference of person B. Therefore the relative velocity must surely be 0. But more interestingly is to argue why then, opinion 2 is incorrect? The relative velocity seems to indeed be $\omega(S1 - S2)$. So we would be inclined to say that both are correct, yet the two opinions are mutually exclusive? This conclusion is absurd, thus one of our assumptions or intermediary conclusions must be wrong. The frame of reference! It has no uniform velocity! When we take the frame of reference of person B to include person A, we must realize that this frame of reference itself has different velocities! Thus, it would actually mean that if A had the same speed as B, then A would not appear stationary to B. Reversely, if their relative velocity in a frame of reference that is moving linearly that for a moment includes persons A and B (e.g., a crow that ascends from the ground to the top of the disc), would be exactly $\omega(S1 - S2)$, then the frame of reference that appears stationary to B will have a relative velocity between persons A and B of 0.

- 6 years, 1 month ago

A is not stationary in rest frame of B. At the top and bottom of the wheel, the positions of A and B relative to each other are interchanged.

- 6 years, 1 month ago

Let the two students be A & B. We define relative velocity of B with respect to A as the velocity of A in the rest frame of B. In this case the rest frame of B is moving with an angular velocity equal to that of A. So in the frame of B, A is always at the same point P irrespective of how much time has passed. Therefore, their relative velocity is ZERO. So, according to me, Opinion 1 is more accurate. Hope this helps.

- 6 years, 1 month ago

I think Opinion 2 is much more sensible.

- 6 years, 1 month ago

But what explanation would you give for invalidation of Opinion 1?

- 6 years, 1 month ago

opinion 2. consider it as a plane motion..relative velocity of any point on a rigid body w.r.t. any other point on the same body is =distance between points *angular velocity=ω(S1−S2)...

- 6 years, 1 month ago

Option 2...As their velocities are undoubtedly different.Its just that S2 covers different distance then S1 does..such that for S1 it apparently looks as if S2 is at rest..Hope this helps.. OR after going around half circle,the relative positions are reversed..in fact their relative position is changing every instant.See it as a vector joining S1 to S2.

- 6 years, 1 month ago

What about the speed? Is relative speed among them is also non zero? If yes, then why won't distance among them changes?

- 6 years, 1 month ago

Because the student at S2 has to move further to move the same angle as S1

- 6 years, 1 month ago

Actually, both happens. The relative velocity is zero. But the angular velocity will be different. right?

- 6 years, 1 month ago

Relative velocity is not zero. Relative angular velocity, if such a thing exists, is again $\omega$, I think.

- 6 years, 1 month ago

look, when you draw a circle, the radius is same at all points. Now, draw another circle with a common centre. Now join a line from the centre to the big circle. Again do the same thing but at a different angle. What will happen? the distances will always be same! so, velocity is displacement by time. so relative velocity is relative displacement by time. when displacement don't change, relative velocity will be zero!

- 6 years, 1 month ago

Okay. Now draw a line from the centre of the small circle to the small circle. Now do the same thing, but at a different angle. The distances will be the same. By this logic, Neither of the students are moving at all. Do you see the flaw?

You are correct in saying that they remain at constant distance from each other. Yet any form of circular motion remains at constant distance from the centre, which would imply that objects in circular motion are at rest!

The problem is that the angle at which they are located relative to each other is different despite the distance between them being equal.

- 6 years, 1 month ago

Velocity is a vector.!!!

- 6 years, 1 month ago

If the angular velocity is different,then S1 would have moved in front of S2 which is clearly not happening because they are on the same disk.

- 6 years, 1 month ago

Don't worry... y'all are right... There will be different answers but it points to a basic concept in physics. It is due to the frame of reference concept. That means what we see wouldn't be same at all views.

- 6 years, 1 month ago

Yeah I agree with you that what we see wouldn't be same at all the views but here, in this case, we have to see only the view of either of the students which follows from the definition of relative velocity. So relative velocity will be zero as to a student, other would be appear to be stationary.

- 6 years, 1 month ago

Opinion 1. If I stood on the wheel and I look at any one student,I would always be looking at that student,no matter how fast the wheel is spinning. Because neither student is moving relative to me,they are not moving relative to each other.

- 6 years, 1 month ago

Opinion 2 because $S2>S1$ so the distance which student 2 has to travel is more than student 1.

- 6 years, 1 month ago

Opinion 2. If we consider student 1 as the frame of reference, we can see that student 2 is moving in a circular path around student 1. Pretty sure that if $S_{1} < S_{2}$ 2's angular momentum with respect to 1 will still be $\omega$, and $-\omega$ otherwise.

- 6 years, 1 month ago

Invalidation of opinion 1 - although the distance between them is constant, the direction of displacement changes continuously, so relative velocity isn't zero.

- 6 years, 1 month ago

The problem is due to the frame of reference concept. Actually all are right. But different perceives differently at different views... lol!!!

- 6 years, 1 month ago

How come you can say that student 2 is moving in a circular path around student 1 even when student1 is moving with the same angular velocity as student 2? Because of same angular velocities about the same axis, it would seem to student 1 that student 2 is at rest. Which gives their relative velocity to be zero. So opinion 1 is apt according to me. What are your views against my statement?

- 6 years, 1 month ago

Because the magnitude of the velocity of student 2 is greater. When we say relative velocity, we compare only the magnitudes and directions of linear velocity, not angular velocity.

Note how at different positions of student 1, student 2 is at the same distance from student 1, but is in a different direction from him.

Suppose you they at the bottom, initially, student 2 is below student 1. At the top, student 2 is above student 1. So the position of student 2 relative to student 1 keeps changing.

Edit:

$v_{21} = v_{2} - v_{1} = \omega S_{2} - \omega S_{1} = \omega(S_{2} - S_{1})$

Now student two is always at distance $S_{2} - S_{1}$ from student 1, but at different angles. This is circular motion, with $r = S_{2} - S_{1}$ and $v = v_{21}$

And the angular velocity of this circular motion of student 2 with respect to student 1 is

$\frac{v_{21}}{r} = \frac{\omega(S_{2} - S_{1})}{S_{2} - S_{1}} = \omega$

That's how I see it, anyway.

- 6 years, 1 month ago

When we say that at a given position student 1 is above 2 and vice-versa, we are telling what we would see if we were to look at the disc from top. But in order to find the relative velocities of 1 and 2 , we have to consider what we would see if we were to be at their positions. So when I am at the place of either of the students, the other would seem to be at rest to me. Hence I am able to say that the other one is NOT moving according to me irrespective of what the magnitude of his linear velocity is. If the are moving around the same axis with same angular velocity then in the rest frame of 1 (which is also moving with same angular velocity about same axis), 2 always appears to be at the same given point. Hence no displacement occurs between them in this frame (neither in direction nor in magnitude). Hence relative velocity is zero.

- 6 years, 1 month ago

The same logic applies for both horizontal and vertical circles.

Suppose they start at the southern part of the circle. 2 is south of 1. When they reach the eastern most point, 2 is east of 1. When they reach the northern most point, 2 is north of 1.

Now consider if there was only one student on a horizontal circle. At the south, he is south of the centre. At the east, he is east of the centre and at the north, he is north of the centre. Do the two cases look similar?

- 6 years, 1 month ago

Opinion 1 is the right one. Pretending the relatice velocity of the students equals 0 this implicates w(S1-S1) equals 0 and because the radial velocity is greater than 0 : S1-S2=0 which is true because they're aligned along one of the radii of the disc from the two different sides.

- 6 years, 1 month ago

Define your terms properly. By "relative velocity", do you mean "the velocity of A in the reference frame of B" or "the apparent difference in their velocities as observed by the professor"?

- 6 years, 1 month ago

The first one; velocity of A in the reference frame of B, where A and B are the two students.

- 6 years, 1 month ago

I think both are one and the same..!!!

- 6 years, 1 month ago

any person would feel that the disk is not rotating about center but is rotating about himself so there would be a relative velocity between the two so, option 2 is correct

option 1 is wrong as it does not see the persons from there reference frames, instead it sees the situation from the reference frame of a person situated at the center of the disk(so in this, w=0)

- 6 years, 1 month ago

opinion 2 is not the correct option... it is just the difference of the velocities of the speeds of the 2 students relative to a watcher.

- 6 years, 1 month ago

Angular velocity of an object does not depend upon its 'speed', it only sees what 'angle' the body sweeps about the center in unit time. What it means is that at whatever the distance the two bodies are from the center respectively, if they subtend the same angle at the center in unit time of their circular motion, they are said to have the same 'angular velocity'. Let a body lie 1km away from a point while another lie 1m away; suppose both have the same angular velocity about the point, the body lying 1km away clearly has a much greater speed compared to the one lying 1m away, but both take the same time to sweep a particular angle. This hypothesis supports the first opinion. However, the second opinion also seems convincing. The problem here is that we can give two different explanations depending upon the frame of reference specified. We can observe the motions taking the center as the point of reference or from any other frame lying independent of the circular motion of the bodies. If we take the center as the point of reference, then the first opinion makes more sense from our hypothesis. But if we take an external frame of reference, then we can say that the 'linear' velocities of the two bodies vary at 'every' instant in such a way that the relative velocity between them at 'that' instant is zero. And yes, if we take only linear velocities into consideration, then relative velocity is not zero; but here the bodies do not continue to move linearly, but in a circular path, and their velocities vary in accordance to two-dimensional circular motion. Thus, even if the instantaneous linear relative velocity is not zero; in circular motion, the relative velocity between them remains zero. So, I think opinion 1 is correct.

- 6 years, 1 month ago