 #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 = TU
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} (ykx^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.
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} (ykx^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.