# Homework Help: Remaining in the same spot on a spinning disk

1. Dec 1, 2017

### Fibo112

1. The problem statement, all variables and given/known data
The problem is to figure out what acceleration is needed relative to a rotating disk to remain in the same spot(meaning to not rotate with the disk).

2. Relevant equations
a=wr^2

3. The attempt at a solution
I thought that since the Person remaining stationary would be rotating in the opposite direction from the viewpoint of a rotating system the acceleration needed would be w^r2 pointing radially inwards.

2. Dec 1, 2017

### Staff: Mentor

Relative to the non-spinning surroundings, what would be the person's acceleration?

3. Dec 1, 2017

### Fibo112

When the Person is stationary the net acceleration has to be zero.

4. Dec 1, 2017

### Staff: Mentor

Relative to all inertial frames of reference, acceleration is invariant.

5. Dec 1, 2017

### Staff: Mentor

I may have misinterpreted this question. It seems to me that your original answer is correct.

6. Dec 1, 2017

### Fibo112

I may have phrased it poorly..Let's say I am on a carousel and I am holding hands with someone standing on the ground next to the carousel. What does my acceleration have to be relative to the carousel so that we can remain holding hands

7. Dec 1, 2017

### Fibo112

Heres what confuses me also. In the question it is phrased: with what acceleration should the person beginn walking in y direction, where the y direction is perpendicular to the vector pointing from the person to the center.

8. Dec 1, 2017

### Staff: Mentor

Yes. That would be correct.

9. Dec 1, 2017

### Fibo112

?

10. Dec 1, 2017

### haruspex

If A has acceleration a, in an inertial frame, and B has acceleration b, what is A's acceleration relative to B? What equation does that allow you to write in this context?

11. Dec 1, 2017

### haruspex

That is rather a different question. In fact, I am not sure it makes sense. Please quote the entire question word for word.

12. Dec 1, 2017

### Staff: Mentor

From the framework of the rotating table, if you are standing still on the rotating table (i.e., rotating with the table), you would be experiencing an inwardly directed force from the table on the sole of your shoe and an outwardly directed body (pseudo) force (centrifugal force, equivalent to an outwardly directed gravitational force). These two forces would be judged to be equal, and you would think you were in equilibrium.

If you were stationary in actual space (moving tangent to the radial lines of the table as reckoned from the table frame of reference at a velocity $\omega r$ opposite to the actual direction of rotation), as reckoned from the frame of reference of the rotating table, you would be experiencing a radial acceleration inward toward the center of the table. There would be no inward or outward force of the table on the soles of your feet. So you would have to conclude that there is some kind of body force drawing you in toward the center of the table. It would be as if you were in free fall (i.e., in orbit).

13. Dec 1, 2017

### haruspex

If I were to let go a coin whilst in a rotating drum I think I would see it accelerate outwards.

14. Dec 1, 2017

### Staff: Mentor

This refers to the case where you are not rotating with the table. It is rotating underneath you. In that inertial frame of reference, the coin would fall straight downward.

15. Dec 1, 2017

### haruspex

No. In the below, substitute a coin for the person and have me rotating with the table. Instead of dropping the coin, I'll throw it with velocity opposite and equal to my instantaneous velocity in the ground frame. Ignore gravity.
As reckoned in my rotating frame of reference I would see the coin accelerate radially outward.

The simpler approach is to apply the usual rules of relative motion. Acceleration of point on disc, adisc+ acceleration of person relative to disc, aperson:disc = acceleration of person in ground frame, aperson. The question asks for the value of aperson:disc such that aperson=0.

Anyway, Fibo112's puzzlement is, at least in part, that the question specifies a tangential acceleration. That makes no sense to me either. Seems it would need an infinite acceleration until reaching the disc's tangential velocity.

16. Dec 1, 2017

### Staff: Mentor

Well, certainly, this is a deceptively complicated situation. The case I was referring to is where the person is hovering above the table (levitating) and, in the laboratory frame of reference, not rotating along with the table. If he dropped a coin in this situation, it would land at the same radial distance from the center of rotation as when it was released.

17. Dec 1, 2017

### haruspex

You have not performed the substitution I described.
You wrote:
Performing the subsititution that becomes:
If a coin were stationary in actual space (moving tangent to the radial lines of the table as reckoned from your frame of reference in the rotating drum, at a velocity ωr opposite to your actual direction of rotation), then as reckoned from your frame of reference in the rotating drum, the coin would be experiencing a radial acceleration inward toward the center of the table.

I am saying no, you would observe it accelerating outwards, i.e. centrifugally.
(How does one unindent nowadays? There used to be an unindent button.)​

18. Dec 1, 2017

### Fibo112

Heres the question.

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19. Dec 1, 2017

### Fibo112

The part I'm referring to is d...I'm afraid its in german.

20. Dec 1, 2017