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

1. Dec 1, 2017

### Fibo112

This might be better

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2. Dec 1, 2017

### haruspex

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.

3. Dec 1, 2017

### Staff: Mentor

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.

4. Dec 1, 2017

### 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?

5. Dec 1, 2017

### haruspex

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.

6. Dec 1, 2017

### haruspex

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

7. Dec 1, 2017

### Fibo112

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

8. Dec 1, 2017

### 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?

9. Dec 1, 2017

### haruspex

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

10. Dec 1, 2017

### 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...

11. Dec 1, 2017

### haruspex

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

12. Dec 1, 2017

### Fibo112

yes I forgot g

13. Dec 1, 2017

### haruspex

Right... and please see my edit above.

14. Dec 1, 2017

### 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.?

15. Dec 1, 2017

### haruspex

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:
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.

16. Dec 1, 2017

### Fibo112

I am still bothered by this questions notion of acceleration, it just doesn't make any sense to me to be talking about acceleration without a clear frame of reference as to where this acceleration is taking place...

17. Dec 1, 2017

### haruspex

The question states that the child is to stay in the same place in the laboratory frame. That is enough to determine the force the child must exert on the disc, and the disc's consequent acceleration.
The slightly awkward part is that it asks for "the child's acceleration relative to the disk." I now consider that ambiguous. I think what they want is relative to the tangential acceleration of the disc. But if it means relative to the reference frame of the rotating disc then you would have to add the apparent radial acceleration, $-\omega^2\frac R2\hat r$.

18. Dec 1, 2017

### Fibo112

Ok. Thanks a lot for your help.