Force on a particle constrained to move on the surface of a sphereby silmaril89 Tags: classical dynamics, force, marion, thornton 

#1
May2612, 04:34 PM

P: 84

Problem 22 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? 



#2
May2612, 04:57 PM

P: 647

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.




#3
May2612, 05:04 PM

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#4
May2612, 05:07 PM

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P: 2,470

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 2dimensional, and radial direction simply does not exist.




#5
May2612, 05:49 PM

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#6
May2612, 05:55 PM

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#7
May2612, 05:57 PM

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The acceleration in the rhat direction is zero, right? So the force in that direction also needs to be zero, right?




#8
May2612, 06:00 PM

P: 84





#9
May2612, 08:30 PM

P: 669

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 nonzero components in the [itex]\hat{r}[/itex] direction, even if the magnitude of r is constant. If a solution says otherwise, its wrong. 



#10
May2612, 10:59 PM

P: 84

[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. 



#11
May2712, 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. 



#12
May2712, 03:24 AM

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#13
May2712, 05:41 AM

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The whole point of generalized coordinates is that they let you turn a 3dimensional problem into a twodimensional 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. 



#14
May2712, 06:30 AM

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P: 2,470

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.




#15
May2712, 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 rdirection, 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. 



#16
May2712, 08:16 AM

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Again, that's not what you said.




#17
May2712, 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". 



#18
May2712, 11:13 AM

P: 789

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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