# Particle constrained on a curve

randy
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
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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• randy
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Hello Randy, !​
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.

$$L = \frac{1}{2} m (\dot x^2 + 4k^2x^2 \dot x^2) + mgkx^2 \Rightarrow \frac{\partial L}{\partial x} = 4mk^2x^2\dot x^2 + 2mgkx$$
isn't right.

randy
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?

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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##.

randy
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.

• PeroK
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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?

• randy
randy
$$\dot x = \frac{\alpha x}{\sqrt{1 + \beta x^2}}$$ 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?

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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.

• randy
randy
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 :)