Mechanics, Tangential force and potential of a curvilinear path

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Homework Help Overview

The discussion revolves around a mechanics problem involving the tangential force on a bead constrained to move along a curvilinear path. The original poster seeks to prove relationships involving the tangential component of net force and potential energy in the context of conservative forces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the equation m (d^2s/dt^2) = Ftang using vector equations and questions the correctness of their approach. They express uncertainty about the form of the tangential force.
  • Participants suggest using vector equations instead of coordinate-based methods and inquire about the relationship between force and acceleration.
  • There is a discussion about the potential energy of the system and its relation to the tangential force, with the original poster expressing confusion about how to proceed.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. Some guidance has been offered regarding the use of vector equations, but there is no explicit consensus on the correct method or interpretation of the tangential force.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on proving relationships rather than deriving solutions.

heycoa
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Homework Statement



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.

Homework Equations





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 don't 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.

Please help me,
thank you for your time
 
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hi heycoa! :smile:

(try using the X2 button just above the Reply box :wink:)
heycoa said:
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.


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.

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

what is the relation between F and Ftang ?
 

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