Mechanics, Tangential force and potential of a curvilinear path

heycoa
Messages
73
Reaction score
0

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
 
Physics news on Phys.org
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 ?
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top