Remaining in the same spot on a spinning disk

[Fibo112]

Homework Statement

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).

Homework Equations

a=wr^2

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.

 

[Chestermiller]

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

 

[Fibo112]

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

 

[Chestermiller]

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

 

[Chestermiller]

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

 

[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

 

[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.

 

[Chestermiller]

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.

Yes. That would be correct.

 

[Fibo112]

?

 

[haruspex]

w^r2 pointing radially inwards.

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?

 

[haruspex]

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.

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

 

[Chestermiller]

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).

 

[haruspex]

from the frame of reference of the rotating table, you would be experiencing a radial acceleration inward toward the center of the table.

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

 

[Chestermiller]

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

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.

 

[haruspex]

This refers to the case where you are not rotating with the table.

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.

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 ωr 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.

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, a_disc+ acceleration of person relative to disc, a_person:disc = acceleration of person in ground frame, a_person. The question asks for the value of a_person:disc such that a_person=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.

 

[Chestermiller]

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, a_disc+ acceleration of person relative to disc, a_person:disc = acceleration of person in ground frame, a_person. The question asks for the value of a_person:disc such that a_person=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.

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.

 

[haruspex]

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.

You have not performed the substitution I described.
You wrote:

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 ωrωr\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.

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.)​

 

[Fibo112]

Heres the question.

 

[Fibo112]

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

 

[Fibo112]

Can you read it? The upload quality is pretty bad..

 

[Fibo112]

This might be better

 

[haruspex]

This might be better

Yes, I can now read the lettering, but I cannot find an OCR app that can. If you can be bothered to type it all in, I can try various translation packages.

 

[Chestermiller]

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.)​

Sorry. I still don't get what you're saying. If I suddenly increment the velocity of the body so that it is permanently zero in the laboratory frame of reverence, it will stay at a constant radius, and it will be rotating in the opposite direction or the original rotation as reckoned from the frame of the rotating turntable. So its acceleration will be radially inward as reckoned from the rotating turntable.

 

[Fibo112]

I missread the question myself and missed some crucial details, apologies. Here is the question.

Inclined rotating disc:
We consider a frictionless, rotatably mounted disc which is inclined at a constant angle β relative to the Earth's surface, see Figure 4. The disc has a radius of R = 2 m, a thickness of 5 cm and a mass density of ρ = 1 g / cm3.

On the plate is a child with a mass of m = 30 kg. The finite extent of the child, as well as friction due to air resistance and storage of the disc can be neglected. We choose a spatially fixed coordinate system so that the disk surface rotates in the xy plane and the origin of the coordinates coincides with the center of the disk. The system is under the influence of gravitational force Fg. Note: Numerical values are to be calculated only in part a) and b). d) We assume that the disc is initially at rest and that his father puts the child at the point {R / 2, 0, 0} as shown in the figure on the right. The child now tries to stay at this point in the laboratory system by starting to move in positive y-direction with constant acceleration a relative to the disk. How big should this acceleration be, so that the child remains in the same place in the laboratory system? Which angular acceleration of the disk ˙ω corresponds to this? How do you interpret the dependence on the angle of inclination β? What happens in the limit case β = 0?

So in this case the weight of the child will cause a torque which will beginn to accelerate the disc. The childs acceleration would have to be (d/dt w)*R/2, is that correct? On a side note would it be correct to say that if the kid was dropped on the disk after the disk was already rotating the kids acceleration would also have to have an x component for him to remain in place?

 

[haruspex]

Sorry. I still don't get what you're saying. If I suddenly increment the velocity of the body so that it is permanently zero in the laboratory frame of reverence, it will stay at a constant radius, and it will be rotating in the opposite direction or the original rotation as reckoned from the frame of the rotating turntable. So its acceleration will be radially inward as reckoned from the rotating turntable.

Ok, I get it at last! My mistake was to interpret "acceleration relative to the rotating disc" as acceleration relative to a neighbouring point on the disc. That is not the same as acceleration in a frame of reference rotating with the disc.
If A is a fixed point next to a disc centred at O and B is a point on the edge of the disc currently passing A, the actual acceleration of B is towards O. So the acceleration of stationary A relative to B must be away from O. Yet an observer at B sees A as accelerating towards O.

 

[haruspex]

the weight of the child will cause a torque which will beginn to accelerate the disc.

Yes, but don't forget that the child's attempt to stay in place will also affect the torque.

if the kid was dropped on the disk after the disk was already rotating the kids acceleration would also have to have an x component for him to remain in place?

In order to stay in place in that context the child would need to have her feet moving to match the disc speed on landing. Thereafter, it is the same problem as before - the right tangential acceleration will keep the child in place.

 

[Fibo112]

(R/2*30*(sin(b)+a)/Moment of Inertia)*R/2 = a. Is this equation for determining a correct?

 

[Fibo112]

I am having trouble picturing what is even meant by this childs acceleration. I guess what is meant is what is the childs acceleration in the rotating frame of reference which is the disk? From the point of view of the disks frame of reference the child is rotating with an increasing velocity isn't it? If something is rotating with an increasing velocity doesn't it have a tangential and a radial component of acceleration? Am I making any sense?

 

[haruspex]

(R/2*30*(sin(b)+a)/Moment of Inertia)*R/2 = a. Is this equation for determining a correct?

No.
Let's take this in steps. What forces act on the child? What equation does that give?
What forces act on the disc?

 

[Fibo112]

The way I thought was that the childs weight creates a torque of R/2*30kg * sin(b). I also thought that the Force for the childs acceleration has to come from friction so this will create a torque of R/2*30kg*a. So then I thought since T=d/dt(MOI*w)=MOI*d/dt w I would get d/dt w = T/MOI. Lastly I thought that acceleration is equal to R/2 d/dt w...

 

[haruspex]

the childs weight creates a torque of R/2*30kg * sin(b)

That is dimensionally wrong. You forgot something.
But please, let us do it the safe way, with free body diagrams, ΣF=ma etc.

 

[Fibo112]

That is dimensionally wrong. You forgot something.

yes I forgot g

 

[haruspex]

yes I forgot g

Right... and please see my edit above.

 

[Fibo112]

Ok. So since the child remains motionless in an Inertial system we have a=0. The forces on the child are the Gravitational force, the normal force of the disc and the friction force of the disc. The component of the Gravitational force which is perpendicular to the disk will cancel out with the discs normal force. The remaining forces on the child are 30kg*g*sin(b) for the remaining gravitational force and a frictional force of equal magnitude and opposite direction to compensate the gravitational force. By Newtons third law the child enacts an equal and opposite force on the disk. The normal component causes no torque. The frictional component causes a torque of T= R/2 30kg*g*sin(b). This causes an angular acceleration of T/Moment of Intertia. The acceleration of the disk at the point of the child is R/2*T/Moment of inertia.?

 

[haruspex]

Ok. So since the child remains motionless in an Inertial system we have a=0. The forces on the child are the Gravitational force, the normal force of the disc and the friction force of the disc. The component of the Gravitational force which is perpendicular to the disk will cancel out with the discs normal force. The remaining forces on the child are 30kg*g*sin(b) for the remaining gravitational force and a frictional force of equal magnitude and opposite direction to compensate the gravitational force. By Newtons third law the child enacts an equal and opposite force on the disk. The normal component causes no torque. The frictional component causes a torque of T= R/2 30kg*g*sin(b). This causes an angular acceleration of T/Moment of Intertia. The acceleration of the disk at the point of the child is R/2*T/Moment of inertia.?

Yes.
Note that this differs from what your equation in post #27 would have given, even after restoring the g factor.

Out of interest, note that your analysis interprets this:

positive y-direction with constant acceleration a relative to the disk

as meaning relative to the tangential acceleration of the disc, not an acceleration of magnitude a relative to the reference frame of the disc. I think you have the right reading.