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

  • Thread starter cutemimi
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In summary, the work done on a particle as it moves along a helix from (1, 0, 0) to (1, 0, 4·pi) is -sinti+costj+2tk.
  • #1
cutemimi
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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
 
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  • #2
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.)
 
  • #3
Please assist...I am lost
 
  • #4
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.
 
  • #5
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.
 
  • #6
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) = ?
 

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

What is the formula for calculating workdone on a helix?

The formula for calculating workdone on a helix is W = ∫ F · T · ds, where W is the workdone, F is the force applied, T is the tangent vector, and ds is the infinitesimal displacement along the helix.

What do the variables in the formula represent?

The variable W represents the workdone, which is the amount of energy transferred to an object when a force is applied to it. The variable F represents the force applied to the object. The variable T represents the tangent vector, which is the direction of motion along the helix. The variable ds represents the infinitesimal displacement along the helix.

Why is it important to calculate workdone on a helix?

Calculating workdone on a helix is important because it allows us to understand the amount of energy being transferred to an object as it moves along a helix. This information can be useful in various fields such as engineering, physics, and biomechanics.

How is workdone on a helix different from workdone on a straight path?

Workdone on a helix is different from workdone on a straight path because the direction of motion is constantly changing along a helix, while it remains constant along a straight path. This means that the force and displacement vectors are not always parallel to each other, leading to a different calculation for workdone.

Can the formula for calculating workdone on a helix be applied to other curved paths?

Yes, the same formula for calculating workdone on a helix can be applied to other curved paths, as long as the force and displacement vectors are known at every point along the path. This formula is a generalization of the work-energy theorem and can be applied to any path, curved or straight.

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