# Rotating disk problem

1. Aug 20, 2014

### Satvik Pandey

1. The problem statement, all variables and given/known data
We have a disk of mass M and radius R placed on a horizontal plane. A cylindrical groove of radius r is made on a diameter.Now a sphere of mass m and radius r is placed in the groove at the circumference of the disk.

At t=0 the whole system is rotated with an angular velocity 1rad/s as shown in the figure.Also at the same time the sphere was projected with a velocity v towards the center of the disk.Find the minimum velocity of the sphere such that it is able to reach the other end of the groove.

Details and Assumptions

M=2kg
m=3kg
R=2m
r<<R
https://d18l82el6cdm1i.cloudfront.net/solvable/2676130fa9.119ef294cc.dlI6so.png
The disk is fixed about it's centre that is it can only rotate

2. Relevant equations

3. The attempt at a solution
I think if the ball crosses the center of the disk the centrifugal force will be enough to move that ball to the other side of the groove.
I think the minimum velocity in this case is the velocity which is enough to move the ball to the center of the disk.
Here the angular velocity of ball keeps changing, I think.
At distance x from the center of the disk let the angular velocity be ω_x.
I$_{i}$ ω$_{i}$=I$_{x}$ω$_{x}$

ω$_{x}$=MR2+2mR2)/(MR2+2mx2)

ω$_{x}$=16/4+3x2

I think the total energy of the ball will be entirely converted to the potential energy.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 20, 2014

### Satvik Pandey

1. The problem statement, all variables and given/known data
We have a disk of mass M and radius R placed on a horizontal plane. A cylindrical groove of radius r is made on a diameter.Now a sphere of mass m and radius r is placed in the groove at the circumference of the disk.

At t=0 the whole system is rotated with an angular velocity 1rad/s as shown in the figure.Also at the same time the sphere was projected with a velocity v towards the center of the disk.Find the minimum velocity of the sphere such that it is able to reach the other end of the groove.

Details and Assumptions

M=2kg
m=3kg
R=2m
r<<R
https://d18l82el6cdm1i.cloudfront.net/solvable/2676130fa9.119ef294cc.dlI6so.png
The disk is fixed about it's centre that is it can only rotate

2. Relevant equations

3. The attempt at a solution
I think if the ball crosses the center of the disk the centrifugal force will be enough to move that ball to the other side of the groove.
I think the minimum velocity in this case is the velocity which is enough to move the ball to the center of the disk.
Here the angular velocity of ball keeps changing, I think.
At distance x from the center of the disk let the angular velocity be ω_x.
I$_{i}$ ω$_{i}$=I$_{x}$ω$_{x}$

ω$_{x}$=(MR2+2mR2)/(MR2+2mx2)

ω$_{x}$=16/4+3x2

I think the total energy of the ball(rotational+kinetic) will be entirely converted to the potential energy.

Potential energy=$\int mω2xdx$
Limits are from R to 0
=256m$\int xdx/16+9x4$
Substituting x2/t=2 and xdx=dt
256m$\int \frac{dt}{4(4+t2)}$

=$\frac{256m}{4}$$\frac{1}{2}$tan$^{-1}$$\frac{t}{2}$
=256mπ/32
=1/2 mV2+1/2 Iω2 =256mπ/32
=1/2 mV2+1/2 mR2/2 =256mπ/32
on solving we get
V2=16π-4
V=6.80

3. Aug 20, 2014

### ehild

Where are the parentheses???? Because you wrote ω=4+3x2. Do you think it is right?
What frame of reference you use? If it is a rotating frame, the angular velocity is zero and so is the rotational energy. If it is a rest frame of reference, there is no potential energy.

And comb up your formulae. Tex does not understand

ehild

4. Aug 20, 2014

### haruspex

Those numbers were not given in the problem statement. Anyway, what's interesting is to get the right formula, so the numbers don't matter.
Yes
What happened to $\omega_i$?
What potential energy? It's all in a horizontal plane.

5. Aug 20, 2014

### haruspex

6. Aug 20, 2014

### Orodruin

Staff Emeritus
I have some doubts regarding the problem formulation, the answers to which may either make the problem relatively simple or significantly complicate life.

First of all, is there an external constraint keeping the disk rotating at the same angular velocity? If not its angular velocity is going to change with time, making it more difficult to go to a frame rotating with constant angular velocity.

Second, is the ball rolling or sliding through the groove? If it is rolling without slipping you will also have to take that kinetic energy into account.

7. Aug 20, 2014

### haruspex

Since the mass of the disk is given, I think it is safe to say this is an isolated system.
Rolling without slipping would be extremely tough to handle. Since the ball exactly fits the groove, you'd have to take it as slipping, frictionlessly, everywhere except where the normal force acts. That point will change, resulting in complex gyrations of the ball. So I'm pretty certain it should be treated as a frictionless particle.

8. Aug 20, 2014

### Satvik Pandey

Ep=$\int mω2xdx$

ω2=256/(4+3x)2

On putting values I got

Ep=-$\int \frac{xdx}{(4+3x^{2})^{2}}$

Putting (4+3x)2=a So da=6xdx.
Substituting the values

Ep=-$\frac{256m}{6}$$\int \frac{da}{(a)^{2}}$

On putting m=3 and after simplifying

=128-$\frac{1}{a}$
Putting a=(4+3x)2

=-$\left (128-\frac{1}{(4-3x^{2})}\right)$$^{0}_{2}$

Om putting the values I got it 24

As 1/2 Mv2=E$_{p}$

So 3/2 V2=24
So V=√16
=4 m/s.

I did it by choosing rotating frame of reference.

9. Aug 20, 2014

### Satvik Pandey

Could you please explain why there is no potential energy in a rest frame of reference?
I searched for it in senior secondary school physics book but didn't find.

I used rotating frame of reference and I got correct answer.

10. Aug 21, 2014

### Nathanael

I'm wondering about a different way to solve this. Can anyone please explain why what I did does not work?

$ω=\frac{16}{4+3x^2}$

(x is the distance from the center)

centrifugal acceleration $=\frac{dv}{dt}=\frac{1}{2}\frac{dx}{dt}\frac{dv}{dx}=\frac{1}{2}\frac{d(v^2)}{dx}=Rω=\frac{32}{4+3x^2}$

$\int_R^0d(v^2)=\int_R^0\frac{64}{4+3x^2}dx=-v^2$

This gives me $v\approx 4.4 m/s$

Is this method even close? If it is, where did I go wrong?

11. Aug 21, 2014

### ehild

The centrifugal acceleration is not Rω

ehild

Last edited: Aug 21, 2014
12. Aug 21, 2014

### haruspex

dv/dt will be negative, no?
I don't understand how you get to equate it to Rω. It's dimensionally wrong. On the left you have an acceleration, on the right a velocity.

13. Aug 21, 2014

### Nathanael

Ahhh of course, what was I thinking thank you for noticing that, it should be $Rω^2$

But now I get:
$v^2=\int_0^R2(\frac{32}{3+4x^2})^2dx\approx 150$

14. Aug 21, 2014

### haruspex

Yes, that all looks good, and quite neat.

15. Aug 21, 2014

### ehild

The rest frame of reference is fixed to the outside world. To the table, for example. The only potential energy could be that of the gravity. But it does not change as the whole motion is horizontal. In the rotating frame of reference, the centrifugal force has potential. You did it well, congratulation. But next time indicate what frame of reference you use.

I used the rest frame of reference and conservation of energy. Got the same result.

Check your formulae, there are typos.

m is missing

Are not a few parentheses missing?

It should be 1/2 mv2

Use itex and /itex at the beginning and at the end of a formula. For power, use ^ . For lower index, use _.

ehild

16. Aug 21, 2014

### ehild

$dv/dt = -x ω^2$ : The radius changes during the motion from R to 0. It was denoted by x.

ehild

17. Aug 21, 2014

### Nathanael

Thank you once again ehild, and sorry for making such silly mistakes, I guess I was not paying very much attention.

Anyway thank you for your help; I now get the correct answer

18. Aug 21, 2014

### ehild

You are welcome, but check your post before sending it

ehild

19. Aug 21, 2014

### Orodruin

Staff Emeritus
I would like to highlight this and suggest that Satvik tries doing it this way too. Of course the end result is the same, but I think the solution using the non-rotating frame is a bit more straightforward as there is no need to introduce an effective potential or integrate.

20. Aug 22, 2014

### Satvik Pandey

Sorry for the late response.

I have studied somewhere that centrifugal force is a pseudo force.
And I know that we have to take pseudo force in consideration if we are watching an object from an accelerating frame of reference.
Here I have used a rotating frame of reference which was moving with the rotating disc.
So, I have used an accelerating frame of reference hence I have to consider pseudo force( i.e. centrifugal force in his case) and work done by this force get stored in the form of potential energy.
If I observe the motion of the ball from a rest frame of reference(like from the Earth) then in that case in we don't need to consider pseudo force as it is an inertial frame.
So if we are watching ball from the Earth then which force is decelerating the ball.
As ball is also following a circular motion so there must be centripetal which is acting towards center so ball should accelerate while going towards the center.