Calculating Workdone on a Helix using W=Integral F.Tds

  • Thread starter Thread starter cutemimi
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around calculating the work done on a particle moving along a helix defined by the parametric equations x=cos(t), y=sin(t), z=2t, from the point (1,0,0) to (1,0,4π) against a specified force vector. The force is given as F(x,y,z) = -y i + x j + z k, which is substituted into the work integral W = ∫ F · T ds.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the work integral and the evaluation of the parameter t at the endpoints of the helical path. There is an exploration of the implications of the parameterization on the limits of integration.

Discussion Status

Some participants have confirmed that the integral is set up correctly, while others are questioning the correct value of t at the endpoint of the path. There is an ongoing exploration of the implications of using t=2π and its effect on the coordinates derived from the parametric equations.

Contextual Notes

There is a noted confusion regarding the endpoint coordinates when applying the parameter t, specifically whether the endpoint should yield (1,0,4π) or (-1,0,4π). This highlights a potential misunderstanding of the parametric equations and their evaluation at the specified t values.

cutemimi
Messages
2
Reaction score
0

Homework Statement


Use W=Integral F.Tds
To find workdone when a particle is moved along the helix x=cost, y=sint, z=2t from (1,0,0) to (1,0,4pi) against a force F(x,y,z)=-yi+xj+zk = -sinti+costj+2tk ( substituting above)

Homework Equations





The Attempt at a Solution



Let rt = costi+sintj+2tk
r't= -sinti+costj+2k

Therefore W= Intergral(0tp4pi)[ <-sint,cost,2>.<-sint,cos t,2t>=Intergal[sin^2t+cos^2+2t] = [t+2t^2]from 0 to 4pi

Giving the answer 4pi+32pi^2
 
Last edited:
Physics news on Phys.org
cutemimi said:

Homework Statement


Use W=Integral F.Tds
To find workdone when a particle is moved along the helix x=cost, y=sint, z=2t from (1,0,0) to (1,0,4pi) against a force F(x,y,z)=-yi+xj+zk = -sinti+costj+2tk ( substituting above)

Homework Equations





The Attempt at a Solution



Let rt = costi+sintj+2tk
r't= -sinti+costj+2k
Therefore W= Intergral(0tp4pi)[ <-sint,cost,2>.<-sint,cos t,2t>=Intergal[sin^2t+cos^2+2t] = [t+2t^2]from 0 to 4pi

Giving the answer 4pi+32pi^2

Be careful here: you are integrating the work with respect to the value of the parameter t. What is the value of t at the endpoint of this helical path? (It's not 4·pi.)
 
Please assist...I am lost
 
You can't be all that lost: it looks like your work integral is set up correctly.

The path that the particle travels on is a helix (an ascending circular spiral) starting at (1, 0, 0) and ending at (1, 0, 4·pi). The parametric equation for this path is given as x = cos t, y = sin t, z = 2t . You said you're starting the path integral at t = 0, which corresponds to x = cos 0 = 1 , y = sin 0 = 0 , z = 2·0 = 0 , which checks against the starting coordinates. So what value must t have for the ending coordinates? That is the value you would end your integration of the dot product (1 + 4t) at.
 
I get the point, and I am now using 2pi as the ending coordinate. My next dump question is having (-1,0,4pi) not (1,0,4pi) when i apply 2pi to the parametric equations.
 
zimbob said:
I get the point, and I am now using 2pi as the ending coordinate. My next dump question is having (-1,0,4pi) not (1,0,4pi) when i apply 2pi to the parametric equations.

cos(2·pi) = ?
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K