Mechanics, Tangential force and potential of a curvilinear path

[heycoa]

Homework Statement

a) Prove that m (d^2s/dt^2) = F_tang, 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 F_tang.

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 F_tang= -(dU/ds). This shows that one-dimensional systems of this type can be treated just like linear systems, with x replaced by s and F_x by F_tang.

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

 

[tiny-tim]

hi heycoa!

(try using the X² button just above the Reply box )

a) Prove that m (d^2s/dt^2) = F_tang, 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 F_tang.

…
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 F_tang ? ​

 

[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.

 

[tiny-tim]

what is the relation between F and dv/dt ?

what is the relation between F and F_tang ?

 

[paranormal]

and effort.


Dear student,

Great job with problem a)! Your approach is correct, and you have correctly derived the equation (d^2s/dt^2) = Ftang. This means that the tangential component of the net force on the bead is equal to the second derivative of the position with respect to time, multiplied by the mass.

For problem b), you are on the right track by considering the potential energy of the system. However, instead of using m*g*y, let's use a more general form for the potential energy, U(s), where s is the position along the curvilinear path. This means that the potential energy is a function of the position along the path.

Now, the normal force of the wire can be represented as a conservative force, which means that it can be derived from a potential energy. This potential energy can be written as U(s). This means that the normal force can be written as -dU(s)/ds, since the derivative of the potential energy with respect to the position gives us the force.

We can also write the tangential force as Ftang = m*(d^2s/dt^2). Now, from problem a), we know that (d^2s/dt^2) = Ftang. Combining this with the expression for the normal force, we get Ftang = -dU(s)/ds. This shows that the tangential force can be treated just like the force in a linear system, with the position s replacing the variable x. This is a powerful result, as it allows us to use the same equations and principles for both linear and curvilinear systems.

I hope this helps clarify the problem for you. Keep up the good work!