Finding motion where the acceleration depends on position and time

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Summary:

Finding motion by making a differential equation using initial values and an acceleration that depends on position and time.

Main Question or Discussion Point

I have computed that the acceleration in my problem is
a(t) = -gj - k/m(|r(t)| - L_0) * r(t)/|r(t)|

Where a(t) is the acceleration vector, g is the gravitational acceleration, j is the unit vector in y-direction, k is the spring constant, m is the mass, r(t) is the position vector, |r(t)| is the length of r(t) and L_0 is the equilibrium length of a rope. I am asked in my problem to make a differential equation by using some initial values and the acceleration to compute the motion.

I was told by the teacher not to use the equations of motion because those require a constant acceleration, while the one we have here is dependent on the position. Is there a way for me to compute the motion given an initial velocity and position?
 
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Answers and Replies

  • #2
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Are you familiar with setting up and solving differential equations?
 
  • #3
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I'm somewhat familiar with it. I was thinking that since we know r(0), v(0) and a(t), I could simply compute
r(t) = r(0) + v(0)*t + ∫∫a(t)dtdt, but I do not know how to integrate a(t) when it depends on r(t)
 
  • #4
PeroK
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Summary:: Finding motion by making a differential equation using initial values and an acceleration that depends on position and time.

I have computed that the acceleration in my problem is
a(t) = -gj - k/m(|r(t)| - L_0) * r(t)/|r(t)|

Where a(t) is the acceleration vector, g is the gravitational acceleration, j is the unit vector in y-direction, k is the spring constant, m is the mass, r(t) is the position vector, |r(t)| is the length of r(t) and L_0 is the equilibrium length of a rope. I am asked in my problem to make a differential equation by using some initial values and the acceleration to compute the motion.

I was told by the teacher not to use the equations of motion because those require a constant acceleration, while the one we have here is dependent on the position. Is there a way for me to compute the motion given an initial velocity and position?
What physical scenario does this represent?
 
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It is a pendulum where we think of the rope as a spring
 
  • #6
PeroK
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It is a pendulum where we think of the rope as a spring
Ah, ##r(t)## is a vector! You must write ##\vec r##. And the thing on the end is ##\hat r##. I guess you mean the bob is attached by a spring. It can't be a rope.

So, how are you going to resolve your vectors? ##\hat x, \hat y## or along the line of the spring and tangential to it?
 
  • #7
vanhees71
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Once more I'd rather suggest to use Hamilton's principle (which is adequate for an I-level thread, I hope).
 
  • #8
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Well, I have also computed the components of the acceleration, a_x and a_y, so I guess I could do it component-wise
 
  • #9
PeroK
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Well, I have also computed the components of the acceleration, a_x and a_y, so I guess I could do it component-wise
At this level, you may struggle without using Latex to render your equations:

https://www.physicsforums.com/help/latexhelp/

Another observation: this seems quite an advanced problem, so the comment about "not using equations of motion", by which you mean "not using SUVAT" looks strange. If you are looking at this problem, then the simplicity of SUVAT should be long ago! Have you not studied simple harmonic motion or other variable force problems before this?

In this case, perhaps resolving along the spring and tangential to it seems more natural. I haven't got time to analyse this myself, but my guess is that's a more promising approach.
 
  • #10
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2nd derivative of distance with respect to time= acceleration. your title seems to have pointed to the definition of velocity, which is the 1st derivative of distance with respect to time. acceleration is the change of velocity with respect to time
 

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