# Mechanics, Tangential force and potential of a curvilinear path

1. May 16, 2013

### heycoa

1. The problem statement, all variables and given/known data

a) Prove that m (d^2s/dt^2) = Ftang, the tangential component of the net force on the bead. [hint] one way to do this is to take the time derivative of the equation v^2=v(dot)v. The left side should lead you to (d^2s/dt^2), and the right side should lead to Ftang.

b) One force on the bead is the normal force of the wire (which constrains the bead to stay on the wire). If we assume that all other forces (gravity, etc) are conservative, then their resultant can be derived from a potential energy U. Prove that Ftang= -(dU/ds). This shows that one-dimensional systems of this type can be treated just like linear systems, with x replaced by s and Fx by Ftang.

2. Relevant equations

3. The attempt at a solution

for problem a) I took v^2=v(dot)v and replaced v with (dx/dt)(x hat) + (dy/dt)(y hat), I then dotted the two together and got v^2= (d^2x/dt^2)+(d^2y/dt^2). Then I multiplied both sides of that equation by (d/dt). This lead to the equation equaling: (d^2s/dt^2)= sqrt((d^2x/dt^2)+(d^2y/dt^2)). Which makes sense. I then multiplied both sides by the mass and wound up with the correct term on the left side (m*(d^2s/dt^2)) and sqrt(m^2(d^2x/dt^2)+m^2(d^2y/dt^2)) on the right hand side. But I do not know if this is correct, I have no idea what the tangential force is supposed to look like.

For problem b) I am stuck and dont really know where to begin. I am calling the potential energy of this system m*g*y, where y is the height of the bead on the wire. I do not know where to go from here.

2. May 17, 2013

### tiny-tim

hi heycoa!

(try using the X2 button just above the Reply box )
no, they want you do it quickly using vector equations, not using coordinates

hint: what is the relation between F and dv/dt ?

what is the relation between F and Ftang ?

3. May 17, 2013

### heycoa

Tiny-tim, thank you very much for the response.

When you say they want me to do it quickly using vector equations, are you saying that m*(d2s/dt2) =m*sqrt(d2x/dt2)+(d2x/dt2)) is incorrect? I'm not sure how else I can show what s'' equals.

I am thinking that the tangential force is equal to the force: m*s'' in the direction of the curved wire.

Last edited by a moderator: May 17, 2013
4. May 17, 2013

### tiny-tim

what is the relation between F and dv/dt ?

what is the relation between F and Ftang ?