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Force on a particle constrained to move on the surface of a sphere

by silmaril89
Tags: classical dynamics, force, marion, thornton
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silmaril89
#1
May26-12, 04:34 PM
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Problem 2-2 from "Classical Dynamics of Particles and Systems" By Thornton and Marion is stated as follows:

A particle of mass m is constrained to move on the surface of a sphere of radius R by an applied force [itex]\textbf{F}[/itex](θ, [itex]\phi[/itex]). Write the equation of motion.

Initially I felt that the force should only point in the [itex]\hat{r}[/itex] direction. It seems obvious to me that [itex]\textbf{F} = \frac{m v^2}{R} \hat{r}[/itex].

However, looking at the solution manual here, they claim the solution should be of the form, [itex]\textbf{F} = F_\theta \hat{\theta} + F_\phi \hat{\phi}[/itex]. If there were a force in any direction except toward the center, then the particle would would not stay on the surface of the sphere.

Can someone explain what is going on here?
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jfy4
#2
May26-12, 04:57 PM
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When it says "constrained" to the sphere I don't think you have to imply [itex]v\cdot a=0[/itex]. I think it's just saying this thing lives on a closed 2d surface called a sphere. So the only forces that can act have to be theta and phi direction forces.
silmaril89
#3
May26-12, 05:04 PM
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Quote Quote by jfy4 View Post
When it says "constrained" to the sphere I don't think you have to imply [itex]v\cdot a=0[/itex]. I think it's just saying this thing lives on a closed 2d surface called a sphere. So the only forces that can act have to be theta and phi direction forces.
That certainly helps me understand how [itex]F_\theta[/itex] and [itex]F_\phi[/itex] could be non-zero, but how could [itex]F_r[/itex] be zero?

K^2
#4
May26-12, 05:07 PM
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Force on a particle constrained to move on the surface of a sphere

To clarify on jfy4's point, yes, there is a radial force, but we can happily ignore it, as all the motion it provides is already covered by the constraint. So you might as well pretend that the entire space is 2-dimensional, and radial direction simply does not exist.
silmaril89
#5
May26-12, 05:49 PM
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Quote Quote by K^2 View Post
To clarify on jfy4's point, yes, there is a radial force, but we can happily ignore it, as all the motion it provides is already covered by the constraint. So you might as well pretend that the entire space is 2-dimensional, and radial direction simply does not exist.
If you look at the solution manual, it doesn't say we simply ignore it since it's irrelevant. It actually says that the total force in the [itex]\hat{r}[/itex] direction sums to zero. This doesn't make sense.
Rap
#6
May26-12, 05:55 PM
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Quote Quote by silmaril89 View Post
That certainly helps me understand how [itex]F_\theta[/itex] and [itex]F_\phi[/itex] could be non-zero, but how could [itex]F_r[/itex] be zero?
[itex]F_r[/itex] is not generally zero. A planet in a circular orbit is confined to a sphere (a circle, actually) yet there is a net radial force inward, due to gravity.
Vanadium 50
#7
May26-12, 05:57 PM
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The acceleration in the r-hat direction is zero, right? So the force in that direction also needs to be zero, right?
silmaril89
#8
May26-12, 06:00 PM
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Quote Quote by Vanadium 50 View Post
The acceleration in the r-hat direction is zero, right? So the force in that direction also needs to be zero, right?
No, if it's moving on a sphere, then it's acceleration should be [itex]\frac{v^2}{r} \hat{r}[/itex].
ParticleGrl
#9
May26-12, 08:30 PM
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Quote Quote by silmaril89 View Post
No, if it's moving on a sphere, then it's acceleration should be [itex]\frac{v^2}{r} \hat{r}[/itex].
Only for constant velocity along a great circle of the sphere- lots of other things could be happening that complicate matters. How do you modify things if v is changing, or the plane of the circle great circle is changing? Or if its moving along a smaller circle of the sphere?

You are right in that you need a force of constraint in the r direction to keep the particle on the surface of the sphere (try taking the second derivative of [itex]\hat{r}[/itex], you should find it has non-zero components in the [itex]\hat{r}[/itex] direction, even if the magnitude of r is constant. If a solution says otherwise, its wrong.
silmaril89
#10
May26-12, 10:59 PM
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Quote Quote by ParticleGrl View Post
Only for constant velocity along a great circle of the sphere- lots of other things could be happening that complicate matters. How do you modify things if v is changing, or the plane of the circle great circle is changing? Or if its moving along a smaller circle of the sphere?

You are right in that you need a force of constraint in the r direction to keep the particle on the surface of the sphere (try taking the second derivative of [itex]\hat{r}[/itex], you should find it has non-zero components in the [itex]\hat{r}[/itex] direction, even if the magnitude of r is constant. If a solution says otherwise, its wrong.
Right, I guess I was just trying to prove a point that it would be non-zero. Correct me if I'm wrong, but I think the acceleration in the [itex] \hat{r} [/itex] direction for constant [itex]r = R[/itex] should be,

[tex] \textbf{a} = -(R \dot{\theta}^2 + R \dot{\phi}^2\sin^2{\theta}) \hat{r} [/tex]

It's too bad the solution manual is wrong. I thought the book would have accurate solutions since it is quite popular.
haruspex
#11
May27-12, 12:36 AM
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It's not clear whether the particle is constrained to the surface of the sphere by F (as stated at top of thread) or by some other force. Looking at the solution I would say it's the second. That being so, Fr is irrelevant - the other force will always neutralise this, as well as providing the force necessary for spherical motion. I think that's what the text in the manual is trying to say, but it's rather unclear.
I checked the manual answer for the simple cases of constant longitude and constant latitude (but variable speeds). They look right.
K^2
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May27-12, 03:24 AM
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Quote Quote by Vanadium 50 View Post
The acceleration in the r-hat direction is zero, right? So the force in that direction also needs to be zero, right?
Coordinate acceleration in r-hat direction is zero. Actual acceleration in r-hat is non-zero. There is acceleration due to metric of coordinate system as well.

Quote Quote by silmaril89 View Post
Correct me if I'm wrong, but I think the acceleration in the [itex] \hat{r} [/itex] direction for constant [itex]r = R[/itex] should be,

[tex] \textbf{a} = -(R \dot{\theta}^2 + R \dot{\phi}^2\sin^2{\theta}) \hat{r} [/tex]
Yes, that actually is correct, somewhat surprisingly. I was expecting more terms to come out from that sinθ, but none of them are in radial direction.
Vanadium 50
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May27-12, 05:41 AM
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The whole point of generalized coordinates is that they let you turn a 3-dimensional problem into a two-dimensional problem. (x, y, z) --> (r = constant, theta, phi). Using this technique, because r is constant, you can ignore all forces in that direction. That's how this technique works.

Note that this does not require one to figure out the constraining force to keep r = constant, which is what some of you are trying to do. Again, that's part of the simplification.
K^2
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May27-12, 06:30 AM
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That doesn't make your previous statement any less false. Yes, you don't need to account for force in radial direction. In fact, you can solve this problem in a 2d manifold with metric of a 2d spherical shell. In that case, the question of radial force goes away, because there is no such thing as radial direction. But none of this has anything to do with r'' being zero.
Vanadium 50
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May27-12, 08:09 AM
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PF has a problem where people try and make things more complicated than they have to be. This is Chapter 2 in Marion, for heaven's sake, and it should be answered at a level appropriate to Chapter 2 in Marion.

In this formulation, the constraining force is not part of the problem, and there is no net force in the r-direction, because r is constant. Yes, one can formulate this problem in other ways, some of which can provide additional answers - but the question is about understanding things at the level of the beginning of Marion.
K^2
#16
May27-12, 08:16 AM
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Again, that's not what you said.

The acceleration in the r-hat direction is zero, right? So the force in that direction also needs to be zero, right?
Is not even close to saying the force in radial direction can be ignored because it's just a constraint force. The logic of that statement is faulty and would lead to a completely wrong conclusion in some problems. Orbital mechanics, for example, where movement in radial direction is not constrained, but [itex]\partial^2 r/\partial t^2=0[/itex] is a special case, where ar is non-zero nonetheless.
Vanadium 50
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May27-12, 10:18 AM
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But this is not orbital mechanics.

This is Chapter 2 in Marion and it should be answered at a level appropriate to Chapter 2 in Marion. We should try and make this concept clear, and not drown the OP in advanced "yeahbuts".
Rap
#18
May27-12, 11:13 AM
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Quote Quote by Vanadium 50 View Post
PF has a problem where people try and make things more complicated than they have to be. This is Chapter 2 in Marion, for heaven's sake, and it should be answered at a level appropriate to Chapter 2 in Marion.

In this formulation, the constraining force is not part of the problem, and there is no net force in the r-direction, because r is constant. Yes, one can formulate this problem in other ways, some of which can provide additional answers - but the question is about understanding things at the level of the beginning of Marion.
Agree with the first paragraph, disagree with the second. As noted before, there is a radial component of the acceleration vector, equal to [itex]-r \left(\theta '^2+\sin ^2(\theta ) \phi '^2\right)[/itex] and so there is a radial component to the net force on the particle. Motion on the sphere implies a radial component of the force on the particle.

So what is the answer to the OP? The force on the particle is:[tex]F_r(\theta,\phi)=-mr \left(\theta '^2+\sin ^2(\theta ) \phi '^2\right)[/tex][tex]F_\theta(\theta,\phi)=mr(-\cos(\theta)\sin(\theta)\phi'^2+\theta'')[/tex][tex]F_\phi(\theta,\phi)=mr(2\cos(\theta)\theta'\phi'+\sin(\theta)\phi'')[/tex]This is three equations in three unknowns ([itex]F_r,\theta,\phi[/itex]), so solve the last two for [itex]\theta[/itex] and [itex]\phi[/itex] (given initial conditions) to get the equations of motion, then solve the first for [itex]F_r[/itex], if you feel like it.

That's a big job, but the above is brute force. Maybe there is a more elegant way? I think solving it by the Lagrangian method will be tricky, since [itex]F_\theta[/itex] and [itex]F_\phi[/itex] might not be described by a potential function.

Maybe the bottom line is that it's a much more complicated question than Thornton and Marion thought.


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