How Is Work Calculated Along a Semicircle in Vector Calculus?

In summary, the homework equation for work done along the semicircle (x-1)2+y2=1 from the point (0,0,0) to (2,0,0) is-4/35.
  • #1
HeisenbergW
11
0

Homework Statement


Find the work done along the semicircle (x-1)2+y2=1 from the point (0,0,0) to (2,0,0)
When F=r3cos2φsinφ[itex]\hat{r}[/itex] + r3*cosφ*cos(2φ) [itex]\hat{φ}[/itex]

Homework Equations


Work
W=∫F*dr
where dr=dr= dr[itex]\hat{r}[/itex] + rsinφdφ[itex]\hat{φ}[/itex]

The Attempt at a Solution



I convert the equation of the line (x-1)2+y2=1 to cylindrical coordinates.
Where x=rsinφ
and y=rcosφ
so (x-1)2+y2=1
eventually becomes r=2cosφ

now i plug in this r value into the force vector given, which gives me F=8cos5φsinφ[itex]\hat{r}[/itex] + 8*cos4φ*cos(2φ) [itex]\hat{φ}[/itex]

W=∫F*dr

So now i dot product this with the dr, which is now dr= dr[itex]\hat{r}[/itex] + 2cosφ
sinφdφ[itex]\hat{φ}[/itex] due to the r value I found.

The dot product comes out to be

∫8cos5φsinφdr + 16*cos5φ*cos(2φ)sinφdφ

and because r=2cosφ, dr=-2sinφ dφ

so this is once again plugged into the integral of the dot product, creating

∫-16cos5φsin2φdφ + 16*cos5φ*cos(2φ)sinφdφ

where I integrate from pi/2 to 0, since it ends at x=2 and y=0, but begins at x=0 and y=0

This gives me -4/35, which does not seem to be a correct answer
Any mistakes?
Any help is appreciated
Thank You in advance
 
Physics news on Phys.org
  • #2
HeisenbergW said:
where dr=dr= dr[itex]\hat{r}[/itex] + rsinφdφ[itex]\hat{φ}[/itex]

The equation is not correct.The second terms has to be rdφ[itex]\hat{φ}[/itex].
And make the vector [itex]\vec{dr}[/itex] different from dr, change of the coordinate r.


ehild
 
  • #3
For d[itex]\vec{r}[/itex] I actually left out the second term, which is the elevation angle, since it is constant. I guess I should have clarified that I am using this in spherical coordinates, but I did produce a typo It should be
dr[itex]\hat{r}[/itex]+rsinθd[itex]\varphi[/itex][itex]\hat{\varphi}[/itex].
Which of course just simplifies to
dr[itex]\hat{r}[/itex]+rd[itex]\varphi[/itex][itex]\hat{\varphi}[/itex]
because θ constantly equals pi/2
So you can ignore the lone sin at the end of the force vector, since it goes to one. Yet it still goes to a negative value unfortunately.
 
  • #4
The integration path is not clear: You can reach from (0.0,0) to (2,0,0) along the upper semicircle or the bottom one, yielding integrals of opposite signs.

ehild
 
  • #5
My apologies, we must go along the circle where y>0
 
  • #6
The integral is negative then.

ehild
 

FAQ: How Is Work Calculated Along a Semicircle in Vector Calculus?

What is work done along a semicircle?

Work done along a semicircle refers to the amount of energy required to move an object along a semicircular path. It takes into account the distance traveled, the force applied, and the angle of displacement.

How is work done calculated along a semicircle?

The formula for calculating work done along a semicircle is W = Fd cosθ, where W is the work done, F is the applied force, d is the distance traveled, and θ is the angle of displacement.

What is the relationship between work done and energy along a semicircle?

Work done and energy are directly proportional along a semicircular path. This means that as the work done increases, so does the energy required to move the object along the semicircle.

How does the direction of the force affect work done along a semicircle?

The direction of the force applied affects the work done along a semicircle. If the force is applied in the same direction as the displacement, the work done will be positive. If the force is applied in the opposite direction, the work done will be negative.

Can work done along a semicircle be negative?

Yes, work done along a semicircle can be negative. This occurs when the force and displacement are in opposite directions, resulting in the object losing energy instead of gaining it.

Similar threads

Back
Top