Motion of particle changing forces

  • Thread starter Pcmath
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  • #1
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I've got a question and can't find any answer on google.

So the centripetal force for a particle in uniform circular motion is mv^2/r. This also means that if F > mv^2/r than the particle will get closer to the center and if F < mv^2/r than it will travel further from the center.

Say a 5 kg object at point (0,5) on cartesian plane with initial speed of 10 ms-1 and direction parallel to x-axis, a force of 35 N directed towards point (0,0) acts on the object. It no longer follows a circular path. So is it possible to get a equation that shows the motion of the object? Note the force is towards specific point (0,0) no matter where the object is.
 

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  • #2
Ibix
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Yes.

Do you know how to integrate?

If the mass has mass m and is at point (x,y), can you write down the force in the x direction and force in the y direction? As a vector if you know about vectors, or two equations if you don't?

Note that I haven't worked through this. We may end up with an integral we can't do.
 
  • #3
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I tried this method before.

However the problem is that the path is not a circle so it is difficult when working with the angular displacement about the point (0,0) because the force in x and y depends on it.

I think we should end up with a differential equation that can't be solved in terms of elementary functions but I fail to even derive it.
 
  • #4
Dale
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Say a 5 kg object at point (0,5) on cartesian plane with initial speed of 10 ms-1 and direction parallel to x-axis, a force of 35 N directed towards point (0,0) acts on the object. It no longer follows a circular path. So is it possible to get a equation that shows the motion of the object?
You will want to approach this problem with Lagrangian mechanics. Simply write down expressions for the kinetic and potential energies in polar coordinates. Then use the standard methods of Lagrangian mechanics to get the equations of motion

The potential energy is ##V=fr ## and the kinetic energy is ##T=\frac{1}{2}m(\dot r^2 + r^2 \dot \theta^2 )##. Everything else is entirely algorithmic to get the equations of motion
 
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