# Equation of motion of a mass on a 2d curve

Homework Statement:
A mass ##m## slides on a curve described by the equation ##y=f(x)## under the action of gravity (neglect any other forces). Write the equation of motion in the general case and in the case of small angle between the tangent to the curve and the horizontal (i.e. when ##dy/dx## is small).
Relevant Equations:
One way is to use conservation of energy ##T=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2}),U=-mgy## so ##T+U=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})-mgy=constant##.
So ##T+U=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})-mgy=constant##. If I derive this with respect to ##t##

$$\dot{x}\ddot{x}+\dot{y}\ddot{y}-g\dot{y}=0$$

Then I use ##\dot{y}=\dot{x}\frac{dy}{dx},\ddot{y}=\ddot{x}\frac{dy}{dx}+\dot{x}^{2}\frac{d^{2}y}{dx^{2}}##
to get

$$\ddot{x}+\frac{dy}{dx}\left[\ddot{x}\frac{dy}{dx}+\dot{x}^{2}\frac{d^{2}y}{dx^{2}}\right]-g\frac{dy}{dx}=0$$

This is the general equation. From here not sure how to simplify in the small limit without making several assumptions.

Last edited:

mfb
Mentor
When dy/dx is very small compared to 1 then its square will be negligible in that equation.
While not directly given by the small dy/dx the problem statement might also assume that d2y/dx2 is small.

• Jenny Physics
Chestermiller
Mentor
What is wrong with $$\left(\frac{dx}{dt}\right)^2\left[1+\left(\frac{dy}{dx}\right)^2\right]-gy(x)=C$$ or $$\frac{dx}{dt}=\sqrt{\frac{C+gy}{\left[1+\left(\frac{dy}{dx}\right)^2\right]}}$$

• Jenny Physics
pasmith
Homework Helper
Problem Statement: A mass ##m## slides on a curve described by the equation ##y=f(x)## under the action of gravity (neglect any other forces). Write the equation of motion in the general case and in the case of small angle between the tangent to the curve and the horizontal (i.e. when ##dy/dx## is small).
Relevant Equations: One way is to use conservation of energy ##T=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2}),U=-mgy## so ##T+U=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})-mgy=constant##.

You have a sign error: if $y$ increases then the potential energy should also increase.

So ##T+U=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})-mgy=constant##. If I derive this with respect to ##t##

$$\dot{x}\ddot{x}+\dot{y}\ddot{y}-g\dot{y}=0$$

Easier to use $\dot y = f'(x) \dot x$ before differentiating with respect to $t$. Then $\ddot y$ doesn't appear.

Then I use ##\dot{y}=\dot{x}\frac{dy}{dx},\ddot{y}=\ddot{x}\frac{dy}{dx}+\dot{x}^{2}\frac{d^{2}y}{dx^{2}}##
to get

$$\ddot{x}+\frac{dy}{dx}\left[\ddot{x}\frac{dy}{dx}+\dot{x}^{2}\frac{d^{2}y}{dx^{2}}\right]-g\frac{dy}{dx}=0$$

This is the general equation. From here not sure how to simplify in the small limit without making several assumptions.

What is wrong with $$\left(\frac{dx}{dt}\right)^2\left[1+\left(\frac{dy}{dx}\right)^2\right]-gy(x)=C$$ or $$\frac{dx}{dt}=\sqrt{\frac{C+gy}{\left[1+\left(\frac{dy}{dx}\right)^2\right]}}$$

I think you are missing a factor of 1/2 from the KE, and again the PE term has the wrong sign. But this does show most clearly what happens when $|f'(x)| \ll 1$.

• Jenny Physics
Chestermiller
Mentor
I think you are missing a factor of 1/2 from the KE, and again the PE term has the wrong sign. But this does show most clearly what happens when $|f'(x)| \ll 1$.
Yes, you're right. I left out a factor of 2. It should read: $$\frac{dx}{dt}=\sqrt{\frac{C+2gy}{\left[1+\left(\frac{dy}{dx}\right)^2\right]}}$$The sign of the PE term is correct if y is measured downwards. And, of course, it is trivial to get the equation when f' is negligible; I didn't feel I needed to add that.

• Jenny Physics