Cart Striking Spring: Velocity & Equations

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    Cart Spring
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Discussion Overview

The discussion revolves around a physics problem involving a cart that descends from a height and moves through a loop before striking a spring. Participants explore the equations of motion and energy conservation principles to determine how far the spring is depressed. The scope includes theoretical reasoning and mathematical modeling.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes that the velocity at the top of the loop can be expressed as \(v^2 = gr\) and questions the equation of motion for the cart when it strikes the spring.
  • Another participant suggests using the energy conservation approach, stating that the energy of the compressed spring must equal the initial gravitational energy, expressed as \(\frac{1}{2}kx^2 = mgh\).
  • There is a clarification that the height \(h\) is a point before the loop and that \(h\) must be at least \(\frac{5}{2}r\) for the cart to successfully navigate the loop.
  • Participants discuss the implications of the height \(h\) in relation to the spring's position, with one asserting that \(h\) is the height above the spring.
  • There is a derived expression for the spring's depression, \(x = \sqrt{\frac{2mgh}{k}}\), and a condition that \(x\) must be greater than or equal to \(\sqrt{\frac{5mgr}{k}}\) based on the height constraint.
  • One participant notes that the assumption about the cart staying in contact with the loop while going through it is not explicitly stated in the problem.

Areas of Agreement / Disagreement

Participants generally agree on the use of energy conservation principles and the derived equations, but there is uncertainty regarding the assumptions about the cart's motion through the loop and the initial conditions. Multiple competing views remain regarding the specifics of the problem setup.

Contextual Notes

There are limitations regarding the assumptions made about the cart's trajectory and the conditions under which it strikes the spring. The discussion highlights the need for clarity on whether the cart remains in contact with the loop throughout its motion.

Dustinsfl
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A cart starts from a height h and goes through a loop of radius r. After going through the loop the cart stikes a spring with spring constant k. How for is the spring depressed.

I have solved for the velocity at the top of loop, \(v^2 = gr\).
I am stuck with coming up with the equation of motion for the cart strike.
\[
m\ddot{x} = -kx +\text{veloctiy?}
\]
do I have \(\sqrt{gr}\) as well? If not, are there any other forces to consider?
 
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dwsmith said:
A cart starts from a height h and goes through a loop of radius r. After going through the loop the cart stikes a spring with spring constant k. How for is the spring depressed.

I have solved for the velocity at the top of loop, \(v^2 = gr\).
I am stuck with coming up with the equation of motion for the cart strike.
\[
m\ddot{x} = -kx +\text{veloctiy?}
\]
do I have \(\sqrt{gr}\) as well? If not, are there any other forces to consider?

The equation of motion is:
\[
m\ddot{x} = -kx
\]
And no, there are no other forces to consider once the cart is coming out of the loop.

But rather than using the equation of motion, I would use that the energy of a compressed spring is $\frac 1 2 k x^2$, which must be equal to the initial gravitational energy $mgh$.

Btw, note that it is not given that the cart starts from the top of the loop. It is only given that it starts at height $h$.
 
I like Serena said:
The equation of motion is:
\[
m\ddot{x} = -kx
\]
And no, there are no other forces to consider.

But rather than using the equation of motion, I would use that the energy of a compressed spring is $\frac 1 2 k x^2$, which must be equal to the initial gravitational energy $mgh$.

Btw, note that it is not given that the cart starts from the top of the loop. It is only given that it starts at height $h$.

The height h is a point before the loop. \(h\geq \frac{5}{2}r\) in order to make it through the loop.

So you are saying use the conversation of energy equation.
\[
mgh_i+ 0\cdot KE = mgh_f + \frac{1}{2}kx^2
\]
What would \(mgh_f\) be though?
 
dwsmith said:
The height h is a point before the loop. \(h\geq \frac{5}{2}r\) in order to make it through the loop.

So you are saying use the conversation of energy equation.
\[
mgh_i+ 0\cdot KE = mgh_f + \frac{1}{2}kx^2
\]
What would \(mgh_f\) be though?

I assume $h$ is the height above the spring.
 
I like Serena said:
I assume $h$ is the height above the spring.

Yes. The spring would be h = 0 after the cart goes through the loop.

So is it
\[
mgh = \frac{1}{2}kx^2\Rightarrow x = \sqrt{\frac{2mgh}{k}}
\]
but using the fact that \(h\geq \frac{5}{2}r\), we have
\[
x\geq\sqrt{\frac{5mgr}{k}},
\]
correct?
 
Last edited:
dwsmith said:
Yes. The spring would be h = 0 after the cart goes through the loop.

So is it
\[
mgh = \frac{1}{2}kx^2\Rightarrow x = \sqrt{\frac{2mgh}{k}}
\]
but using the fact that \(h\geq \frac{5}{2}r\), we have
\[
x\geq\sqrt{\frac{5mgr}{k}},
\]
correct?

Yes.
That is assuming that the cart stays in contact with the loop while going through it.
It seems that that piece of information is not given.
 

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