- #1
PhysiSmo
[SOLVED] Motion of a particle along a curve under gravity
Assume that a curve [tex]F(x,y)=0[/tex] is given. Find the condition that [tex]F[/tex] satisfies, under which a point particle of mass [tex]m[/tex] is moving along the curve (and never falls of it) under gravity.
I suppose that Newton's second law plays a role somewhere in the problem.
[tex]m\ddot{r}=-mg\hat{y},[/tex]
where [tex]\vec{r}[/tex] is the vector position of the particle.
I have absolutely no clues on this one. The only idea I had was that since the particle is always on the curve, its velocity is tangential to the curve, thus perpendicular to a vector [tex]\vec{a}[/tex] that is perpendicular to the curve at a point (x,y).
Such a vector [tex]\vec{a}[/tex] is the gradient of the curve, that is [tex]\nabla F(x,y)[/tex]. So that one condition could be
[tex]\vec{a}\cdot \dot{r}=0[/tex], or
[tex]\dot{x}\frac{\partial F}{\partial x}+\dot{y}\frac{\partial F}{\partial y}=0,[/tex]
with [tex]\dot{x},\dot{y}[/tex] satisfying Newton's second law.
I don't really like my approach, any better ideas?
Homework Statement
Assume that a curve [tex]F(x,y)=0[/tex] is given. Find the condition that [tex]F[/tex] satisfies, under which a point particle of mass [tex]m[/tex] is moving along the curve (and never falls of it) under gravity.
Homework Equations
I suppose that Newton's second law plays a role somewhere in the problem.
[tex]m\ddot{r}=-mg\hat{y},[/tex]
where [tex]\vec{r}[/tex] is the vector position of the particle.
The Attempt at a Solution
I have absolutely no clues on this one. The only idea I had was that since the particle is always on the curve, its velocity is tangential to the curve, thus perpendicular to a vector [tex]\vec{a}[/tex] that is perpendicular to the curve at a point (x,y).
Such a vector [tex]\vec{a}[/tex] is the gradient of the curve, that is [tex]\nabla F(x,y)[/tex]. So that one condition could be
[tex]\vec{a}\cdot \dot{r}=0[/tex], or
[tex]\dot{x}\frac{\partial F}{\partial x}+\dot{y}\frac{\partial F}{\partial y}=0,[/tex]
with [tex]\dot{x},\dot{y}[/tex] satisfying Newton's second law.
I don't really like my approach, any better ideas?