Particle constrained on a curve

In summary: Looking at ##\dot x## it seems to be monotonically increasing for any choice of ##\beta## (between 0 and 1).
  • #1
randy
7
3
Homework Statement
A particle of mass m slides down a curve ##y = - kx^2, (k > 0)## under the influence of
gravity. There is no friction, and the particle is constrained to stay on the curve.
It starts from the top with negligible velocity.

1. Find the velocity v as a function of x
2. Next, assume that the particle initially slides down the curve under gravity, but this
time is not constrained to the curve. Does it leave the curve after it has fallen a
certain distance? Prove your answer.
Relevant Equations
T = 1/2mv^2; U = mgh; L = T-U
I tried 1. using the Lagrangian method:
From ##y=-kx^2## I got ##\dot y = -2kx \dot x## and ##\ddot y = -2k \dot x^2 - 2 kx \dot x##.
(Can I use ##\dot y = g## here due to gravity?)

This gives for kinetic energy:
$$T = \frac{1}{2} mv^2 = \frac{1}{2} m (\dot x^2 + \dot y^2) = \frac{1}{2} m (\dot x^2 + 4k^2x^2 \dot x^2)$$
and for potential energy just ##U = mgy = -mgkx^2##.
Thus the Lagrangian should be
$$L = \frac{1}{2} m (\dot x^2 + 4k^2x^2 \dot x^2) + mgkx^2 $$
$$\Rightarrow \frac{\partial L}{\partial x} = 4mk^2x^2\dot x^2 + 2mgkx $$
$$\Rightarrow \frac{d}{dt}(\frac{\partial L}{\partial \dot x}) = m\ddot x + 4mk^2(2x \dot x^2 + x^2 \ddot x)$$
From which I can derive the equation of motion of the particle with respect to x:
$$ \ddot x + 4k^2x^2\ddot x^2 + 4k^2x \dot x^2 -2gkx = 0$$
This is where I'm stuck. I don't know how this could give me the velocity as function of x.
So I tried it again, but this time I set ##\dot y = g##, and (using the same approach) ended up with
$$2mgkx = m \ddot x$$
I then tried to solve this differential equation by setting ##x = e^{-\beta t}##, what gave me ##\beta = \sqrt{2gk}##. However, this doesn't seem to fit the problem, since this would lead to ##x(0) = 1 \neq 0##, which is obviously wrong.

Concerning part 2:
I was going to use Lagrange multipliers in order to get the constraining force using the relation ## F_c = \lambda \frac{\partial}{\partial x} (y-kx^2)## and setting it 0 to find the point, where the particle would leave the curve. Given that this is the correct idea, I'll probably be able to solve this once I have the correct Lagrangian.Thanks for any help.
 
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  • #2
What about using energy?
 
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  • #3
Hello Randy,
:welcome: !​
randy said:
Can I use ##\dot y=g## here due to gravity?
No. Nor ##\ \dot y=-g\ ## (just dimension wise). But not even ##\ \ddot y=-g## .
Reason: the force of constraint is not limited to the x-direction.

randy said:
$$L = \frac{1}{2} m (\dot x^2 + 4k^2x^2 \dot x^2) + mgkx^2 \Rightarrow \frac{\partial L}{\partial x} = 4mk^2x^2\dot x^2 + 2mgkx $$
isn't right.
 
  • #4
PeroK said:
What about using energy?
Thanks a lot! This approach is so much easier! Ended up with ##v(x) = \sqrt{2gk}x## for part 1, which seems appropriate. However, now I am kinda lost for part 2... If I'm not mistaken, the normal force on the particle must vanish in order for it to leave the curve. So the normal vector at a point x on the curve should be parallel to the negative inverse of the derivative of y (since ##\dot y## gives the slope of the curve) and the normal force needs to counteract ##mg \cos{\alpha}## in order to constrain it to the curve. But how do I now get alpha dependent on x?
 
  • #5
randy said:
Thanks a lot! This approach is so much easier! Ended up with ##v(x) = \sqrt{2gk}x## for part 1,
What about calculating the components ##v_x, v_y##? It did ask for velocity, rather than speed, so it might want the components in any case.

Then, take a close look at ##v_x##.
 
  • #6
PeroK said:
What about calculating the components ##v_x, v_y##? It did ask for velocity, rather than speed, so it might want the components in any case.

Then, take a close look at ##v_x##.
Okay, so from conversation of energy I have ## E = T + U = \frac{1}{2}m(\dot x^2 + \dot y^2) -mgkx^2= 0## (due to kinetic and potential energy being 0 at the start) and from ##y=-kx^2## I get ##\dot y= -2kx\dot x##.
Now solving for ##\dot x##:
\begin{align*}
&\frac{1}{2}m(\dot x^2 + \dot y^2) =mgkx^2 \\
&\Rightarrow \dot y^2=2gkx^2-\dot x^2, \text{which is also equal to } 4k^2x^2\dot x^2 \\
&\Rightarrow 0=\dot x^2-\frac{2gkx^2}{4k^2x^2+1}\\
&\Rightarrow \dot x = \sqrt{\frac{2gkx^2}{4k^2x^2+1}} \text{choosing positive sign, as speed increases}\\
&\Rightarrow\dot y^2 = 2gkx^2(1-\frac{1}{4k^2x^2+1})
\end{align*}
This now should be the velocities in each component. Is this coorect? However I don't get how this helps me with part 2 unfortunately.
 
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  • #7
Okay, so you have $$\dot x = \sqrt{\frac{2gkx^2}{4k^2x^2+1}}$$ which we can write as $$\dot x = \frac{\alpha x}{\sqrt{1 + \beta x^2}}$$ Now, the particle will stay in contact as long as ##\dot x## is increasing. You might have to think about this for a bit. In other words: if, in order to stay on the track, ##\dot x## must reduce, then the particle will lose contact. The one thing it cannot do is slow down in the x-direction.

Does that help?
 
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  • #8
PeroK said:
$$\dot x = \frac{\alpha x}{\sqrt{1 + \beta x^2}}$$ Now, the particle will stay in contact as long as ##\dot x## is increasing. You might have to think about this for a bit. In other words: if, in order to stay on the track, ##\dot x## must reduce, then the particle will lose contact. The one thing it cannot do is slow down in the x-direction.
Looking at ##\dot x## it seems to be monotonically increasing for any choice of k. So the particle would in fact never have to slow down. Does that then mean it never leaves the curve?
 
  • #9
randy said:
Looking at ##\dot x## it seems to be monotonically increasing for any choice of k. So the particle would in fact never have to slow down. Does that then mean it never leaves the curve?
Yes. You could beef up the argument I've given if you want. Originally, you were looking for a point where the normal force went to zero - you could build a further argument based on that.
 
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  • #10
PeroK said:
Yes. You could beef up the argument I've given if you want. Originally, you were looking for a point where the normal force went to zero - you could build a further argument based on that.
Thanks a lot! This was very helpful, I think I understood the concepts of that task now ways better :)
 

Related to Particle constrained on a curve

What is a particle constrained on a curve?

A particle constrained on a curve is a physical system in which a small object, known as a particle, is restricted to move only along a specific curve or path. This constraint can be imposed by various means, such as a physical barrier or a mathematical equation.

What are some examples of particles constrained on a curve?

Some common examples of particles constrained on a curve include a pendulum swinging back and forth, a rollercoaster car moving along a track, and a planet orbiting around a star. These systems all have a specific path that the particle must follow due to the constraints placed on it.

What factors affect the motion of a particle constrained on a curve?

The motion of a particle constrained on a curve is affected by several factors, including the shape and orientation of the curve, the initial conditions of the particle (such as its position and velocity), and any external forces acting on the particle.

How is the motion of a particle constrained on a curve described mathematically?

The motion of a particle constrained on a curve can be described using mathematical equations, such as the laws of motion and the equations of motion. These equations take into account the position, velocity, and acceleration of the particle as it moves along the curve.

What are the applications of studying particles constrained on a curve?

Studying particles constrained on a curve has many practical applications, such as in engineering design, robotics, and physics. Understanding the motion of particles in these constrained systems can help in the development of new technologies and improve our understanding of the natural world.

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