Interesting kinematics problem.

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Main Question or Discussion Point

I want to find a function describing the motion of the particle undergoing the type of motion depicted.

Basically it is a particle on a rotating vane (constant angular speed) and at some time it is released. The particle will move along the edge to the tip of the vane (and eventually fall off). This is rotation on a horizontal plane, so gravity is pointing down into the page (or monitor screen).

[PLAIN]http://img203.imageshack.us/img203/5241/rotation.gif [Broken]

Unfortunately, as you all know, I'm not good at math, so I got stuck. I tried this:

*Put a coordinate system on the ball. One direction is the r direction (along the vane) and the second is the theta direction (normal to r).
*I said the acceleration in the r direction is r*w^2, and in the theta direction is dr/dt*w
Anything I do from here becomes a mess.

Can anyone get a function for the motion of this particle? I know it must look like a snail shell type curve. Thanks.
 
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Answers and Replies

  • #2
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First define the planes as XYZ. The Z plane is "down" and that velocity vector is determined solely by acceleration due to gravity. The velocity vector resulting from the rotational acceleration is in solely in the XY plane. The easiest thing for me to do would be to use two equations to determine the particle's velocity in the three planes. The velocity due to gravity will occur independently of the velocity in the XY plane and is described by:

Vz = g * T

Assuming the particle "slides" rather than "rolls", and there is no friction, the particle will exit the vane with a speed equal to:

V = r * w and the direction will be determined by the actual rotational speed of the vane, length of the vane and relative angular position of the vane at T = 0, but will be between 0 and 2 * Pi Radians. Once the particle ceases to be accelerated on by the vane, its velocity in the XY plane and its direction will remain unchanged, thus the only acceleration will be due to gravity.

The problem is no different than a ballistic solution except during the period of time the vane is accelerating the particle; no "snail shell" after it reaches then end of the vain.

Just my thoughts.

Fish
 
  • #3
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Double post removed.
 
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  • #4
LeonhardEuler
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This seems like one of those situations where knowing the Lagrangian formulation of mechanics is useful:
http://en.wikipedia.org/wiki/Lagrangian_mechanics

In this case, as I understand it, we have a particle with polar coordinates [itex](r,\theta)[/itex], and the particle is constrained so that
[tex]\theta=\gamma t[/tex]
for some constant [itex]\gamma[/itex]. Considering only planar motion (if there is free motion allowed in the z-direction then adding it is trivial: there will just be uniform acceleration downward in the z-direction), then there is no potential energy and the lagrangian is just given by:

[tex]\mathcal{L}=T=\frac{1}{2}m(r^2\gamma^2+\dot{r}^2)[/tex]
The Euler-Lagrange equation for the one degree of freedom, 'r', is:
[tex]\frac{\partial\mathcal{L}}{\partial r}=\frac{d}{dt}(\frac{\partial\mathcal{L}}{\partial\dot{r}})[/tex]
[tex]m\gamma^2r=m\ddot{r}[/tex]
[tex]\ddot{r}=\gamma^2 r[/tex]
[tex]\rightarrow r(t) = C_{0}e^{\gamma t} + C_{1}e^{-\gamma t}[/tex]
If the initial velocity of the particle is 0, then we get C0=C1. If the initial position is r0, then the final form of the solution is
[tex]r(t)=r_0\cosh{(\gamma t)}[/tex]
[tex]\theta (t) = \gamma t[/tex]
The curve swept out is:
[tex]r(\theta)=r_0\cosh{\theta}[/tex]
so it is approximately a logarithmic spiral for large t or [itex]\theta[/itex].
 
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  • #5
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LeonhardEuler,

The math and the case you present is interesting, and beyond what I remember about calculus, so I will not argue it for the case of the particle traveling from one radius point to another, nor will I argue that the acceleration of the particle might cause the particle to exceed the absolute value of the vane's outside radial velocity, though I have serious reservations on this point, but, I cannot understand how the velocity vector of the particle could change once it has left the influence of the vane (except in the Z direction), or as influenced by friction.

SO I assume you are only defining the path of the particle while it is in contact with the vane? And, if this is correct, am I correct in my assumption that the vane cannot accelerate the particle to a velocity higher than its outer radial velocity, or am I in fact mistaken in this assumption and that is the result of the equations I have failed to grasp?

I am not challenging your math, I am simply attempting to grasp the implications. The concept of the particle attaining a higher velocity than the outer radius of the vane intrigues me, and if this is the case I would be willing to devote time to reviewing my calculus until I could follow along. If, in fact, the direction of the particle continues to rotate around the original radius after it has left the vane's influence then I would be forced to reconsider everything I thought I knew about physics.

Thanks in Advance,

Fish
 
  • #6
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Why is LeonhardEuler such a genius?

And how can you modify the Lagrangian to include a friction force?

SO I assume you are only defining the path of the particle while it is in contact with the vane?
Obviously yes, he is.

The concept of the particle attaining a higher velocity than the outer radius of the vane intrigues me, and if this is the case I would be willing to devote time to reviewing my calculus until I could follow along.
That can never happen. Draw it on paper or imagine it. If it leaves the vane, it will travel at constant velocity. But then the vane will hit it again (at a higher point). You can think of the motion as the limit of this type of motion with infinite frequency (hit hit hit hit etc).
 
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  • #7
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Curl,

Thanks. I thought I understood what was going on, obviously I was wrong. Your OP was about the particle while in contact with the vane, I assumed you were talking about the mundane period after it left the vane. LeonhardEuler understood your OP, and answered it very eloquently in the mathematical terms you were thinking of, but in the process confused me, LOL.

All is good, thank you for helping me regain my sanity, hehe, and apologies for misunderstanding your OP.

Fish
 
  • #8
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double post removed.
 
  • #9
Rap
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Umm - I thought gravity was involved, so I got

[tex]r=r_0 \cosh(\theta)+(mg/2\omega^2)(\sin(\theta)-\sinh(\theta))[/tex]

where [tex]\omega[/tex] is the angular velocity.
 
  • #10
LeonhardEuler
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Why is LeonhardEuler such a genius?
Practice :wink:
And how can you modify the Lagrangian to include a friction force?
There are two ways of doing that result in the same equation in the end. One way is to define a velocity dependent pseudo-potential where the friction force is obtained by taking the "gradient" of the potential, but the gradient takes derivatives with respect to the velocities instead of the normal coordinates and putting that in the lagrangian. A simpler way is to put the friction force directly into the Euler-Lagrange equation:
[tex]\frac{\partial\mathcal{L}}{\partial r}=\frac{d}{dt}(\frac{\partial\mathcal{L}}{\partial \dot{r}}) - Q_{r}[/tex]
with Qr being the friction force.
 

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