What Is the Minimum Velocity for a Sphere to Traverse a Grooved Disk?

[Satvik Pandey]

Homework Statement

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

Homework Equations

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.

 

[Satvik Pandey]

Homework Statement

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

Homework Equations

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

=$\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

 

[ehild]

Homework Statement

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

Homework Equations

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

Where are the parentheses? Because you wrote ω=4+3x². Do you think it is right?

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

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

 

[haruspex]

M=2kg
m=3kg
R=2m

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.

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.

Yes

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.
ω$_{x}$=MR²+2mR²)/(MR²+2mx²)

What happened to $\omega_i$?

I think the total energy of the ball will be entirely converted to the potential energy.

What potential energy? It's all in a horizontal plane.

 

[haruspex]

Duplicate of https://www.physicsforums.com/showthread.php?t=766787

 

[Orodruin]

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.

 

[haruspex]

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.

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.

 

[Satvik Pandey]

E_p=[itex]\int mω²xdx[/itex]

ω²=256/(4+3x)²

On putting values I got

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

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

E_p=-$\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.

Got correct answer.
I did it by choosing rotating frame of reference.

 

[Satvik Pandey]

If it is a rest frame of reference, there is no potential energy.

ehild

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.

 

[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?

 

[ehild]

centrifugal acceleration [itex]=\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}[/itex]

The centrifugal acceleration is not Rω

ehild

 

[haruspex]

centrifugal acceleration $=\frac{dv}{dt}=\frac{1}{2}\frac{d(v^2)}{dx}=Rω$

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.

 

[Nathanael]

also wrong.

Ahhh of course, what was I thinking thank you for noticing that, it should be $Rω^2$But now I get:
$v^2=\int_0^R2(\frac{32}{3+4x^2})^2dx\approx 150$

 

[haruspex]

E_p=[itex]\int mω²xdx[/itex]

ω²=256/(4+3x)²

On putting values I got

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

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

E_p=-$\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.

Got correct answer.
I did it by choosing rotating frame of reference.

Yes, that all looks good, and quite neat.

 

[ehild]

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.

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.

E_p=[itex]\int mω²xdx[/itex]



ω²=256/(4+3x)²

On putting values I got

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

m is missing

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

E_p=-$\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}$

Are not a few parentheses missing?

Om putting the values I got it 24

As 1/2 Mv²=E$_{p}$

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

 

[ehild]

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


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

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

ehild

 

[Nathanael]

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

ehild

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

 

[ehild]

You are welcome, but check your post before sending it

ehild

 

[Orodruin]

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

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.

 

[Satvik Pandey]

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.
ehild

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.
I know that this is an odd confusion but could you please help?

 

[ehild]

There is no centripetal force in the inertial frame of reference to keep the ball on circular track, so it would move along a straight line if it was free. But the groove forces it to change direction. Using polar coordinate system, it has tangential component of acceleration. And the radial acceleration is outward without the centripetal force, which would force it to travel along a circle.

ehild

 

[Satvik Pandey]

Using polar coordinate system, it has tangential component of acceleration. And the radial acceleration is outward without the centripetal force, which would force it to travel along a circle.

ehild

As 'v' velocity is initially given to ball so it it tends to move towards the center of disk due to inertia of motion. I think a reactionary force is acting on it by the walls of groove is responsible for tangential acceleration.
How do you found that the radial acceleration is outward?

 

[ehild]

I was not exact enough. The ball motion towards the centre slows down, it means "outward acceleration" but I meant $\ddot r$. As the centripetal force is zero, the whole radial component of the acceleration $\ddot r -r\omega^2=0$.ehild

 

[Orodruin]

As 'v' velocity is initially given to ball so it it tends to move towards the center of disk due to inertia of motion. I think a reactionary force is acting on it by the walls of groove is responsible for tangential acceleration.
How do you found that the radial acceleration is outward?

You don't need to. It is enough with conservation of energy and angular momentum.

 

[ehild]

You can also follow Nathanael's hint. $\ddot {r }=0.5 \frac{d(v^2)}{dr} =r \omega^2$ . You denoted r by x, so $\ddot x = 0.5\frac{dv^2}{dx} =x \omega^2$ and $\omega=\frac {16}{4+3x^2}$ Integrate.

ehild

 

[Satvik Pandey]

You don't need to. It is enough with conservation of energy and angular momentum.

Initial energy of the system=$\frac{mv^{2}}{2}$+$\frac{I_{B} ω^{2}}{2}$+$\frac{I_{D} ω^{2}}{2}$

Final energy of the system will be in the form of rotational energy of the disc.

So
E_F=$\frac{I ω^{2}_{f}}{2}$

By law of conservation of angular momentum
I$_{i}$ω_i=I_fω_f

ω_f=$\frac{MR^{2} +2mR^{2}}{MR^{2}}$

ω_f=4

On equating initial and final energies

$\frac{3v^{2}}{2}+\frac{mR^{2}}{2}+\frac{MR^{2}}{4}$=$\frac{16 MR^{2} }{4}$

$\frac{3v^{2}}{2}+8$=32

So v=$\sqrt{16}$=4m/s.

Got correct answer.Thanks Orodruin.It is very much easy to solve this question through with conservation of energy and angular momentum.