Work Done by Force Field (Part 4)

Classical Mechanics Level 5

A particle in the x y xy -plane is acted upon by a force field described by the equation below: F ⃗ = ( 3 2 x − 1 2 y ) ı ^ + ( 1 2 x + 3 2 y ) ȷ ^ . \large{\vec{F} = \left(\frac{\sqrt{3}}{2}x - \frac{1}{2}y\right) \hat{\imath} + \left(\frac{1}{2}x + \frac{\sqrt{3}}{2}y\right) \hat{\jmath}}. How much work does the force field do on the particle if the particle makes one complete revolution over a unit circle centered on the origin?

Notes: ı ^ \hat{\imath} and ȷ ^ \hat{\jmath} denote unit vectors in the x x and y y directions, respectively. Give your answer as a positive number to 2 decimal places.


The answer is 3.14.

Steven Chase by Steven Chase , 37, USA

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3 solutions

Abhishek Sinha
Apr 8, 2017

Using the Kelvin-Stoke's theorem, we have ∮ F ⃗ ⋅ d l ⃗ = ∫ ∇ × F ⃗ ⋅ d S ⃗ = ∫ 1 d S = π . \oint \vec{F} \cdot d\vec{l}= \int \nabla \times \vec{F}\cdot d\vec{S}= \int 1 dS=\pi.

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Where S ⃗ \vec{S} is the unit disk. Nice approach!

Steven Chase - 4 years, 2 months ago

On the unit circle we have x = cos ⁡ ( θ ) x = \cos(\theta) and y = sin ⁡ ( θ ) y = \sin(\theta) . So the components of the given force are

F i = 3 2 cos ⁡ ( θ ) − 1 2 sin ⁡ ( θ ) = cos ⁡ ( π 6 ) cos ⁡ ( θ ) − sin ⁡ ( π 6 ) sin ⁡ ( θ ) = cos ⁡ ( θ + π 6 ) F_{i} = \dfrac{\sqrt{3}}{2}\cos(\theta) - \dfrac{1}{2}\sin(\theta) = \cos\left(\dfrac{\pi}{6}\right)\cos(\theta) - \sin\left(\dfrac{\pi}{6}\right)\sin(\theta) = \cos\left(\theta + \dfrac{\pi}{6}\right) and

F j = 1 2 cos ⁡ ( θ ) + 3 2 sin ⁡ ( θ ) = sin ⁡ ( π 6 ) cos ⁡ ( θ ) + cos ⁡ ( π 6 ) sin ⁡ ( θ ) = sin ⁡ ( θ + π 6 ) F_{j} = \dfrac{1}{2}\cos(\theta) + \dfrac{\sqrt{3}}{2}\sin(\theta) = \sin\left(\dfrac{\pi}{6}\right)\cos(\theta) + \cos\left(\dfrac{\pi}{6}\right)\sin(\theta) = \sin\left(\theta + \dfrac{\pi}{6}\right) .

Now on the unit circular path C C with r ⃗ = < cos ⁡ ( θ ) , sin ⁡ ( θ ) > \vec{r} = <\cos(\theta), \sin(\theta)> the work done by F ⃗ \vec{F} is

W = ∫ C F ⃗ ⋅ d r ⃗ = ∫ 0 2 π F ⃗ ⋅ d r d θ ⃗ d θ = W = \displaystyle\int_{C} \vec{F} \cdot \vec{dr} = \int_{0}^{2\pi} \vec{F} \cdot \vec{\dfrac{dr}{d\theta}} d\theta =

∫ 0 2 π < cos ⁡ ( θ + π 6 ) , sin ⁡ ( θ + π 6 ) > ⋅ < − sin ⁡ ( θ ) , cos ⁡ ( θ ) > d θ = \displaystyle\int_{0}^{2\pi} <\cos\left(\theta + \dfrac{\pi}{6}\right), \sin\left(\theta + \dfrac{\pi}{6}\right)> \cdot <-\sin(\theta), \cos(\theta)> d\theta =

∫ 0 2 π ( − cos ⁡ ( θ + π 6 ) sin ⁡ ( θ ) + sin ⁡ ( θ + π 6 ) cos ⁡ ( θ ) ) d θ = \displaystyle\int_{0}^{2\pi} (-\cos\left(\theta + \dfrac{\pi}{6}\right)\sin(\theta) + \sin\left(\theta + \dfrac{\pi}{6}\right)\cos(\theta)) d\theta =

∫ 0 2 π sin ⁡ ( ( θ + π 6 ) − θ ) d θ = ∫ 0 2 π sin ⁡ ( π 6 ) d θ = 1 2 × 2 π = π = 3.14 \displaystyle\int_{0}^{2\pi} \sin\left(\left(\theta + \dfrac{\pi}{6}\right) - \theta\right) d\theta = \int_{0}^{2\pi} \sin\left(\dfrac{\pi}{6}\right) d\theta = \dfrac{1}{2} \times 2\pi = \pi = \boxed{3.14} to 2 decimal places.

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Nice use of trig identities. I wasn't quite sure how hard it would be to go from start to finish. In making the problem, I went from "finish" to "start". We've seen one where F ⃗ \vec{F} is at a right angle with the path, yielding zero net work. We've also seen one where F ⃗ \vec{F} is perfectly aligned with the path, yielding 2 π 2 \pi as a result. If they were 60 degrees apart, that would give us a factor of c o s ( 6 0 ∘ ) = 1 2 cos(60^\circ) = \frac{1}{2} , yielding a final value of π \pi . So to get that, just shift the force by 30 degrees, starting with the radial configuration. Using complex numbers ( j = − 1 j = \sqrt{-1} ) to make sense of the arithmetic of multiplying vectors yields:

F ⃗ = ( x + j y ) ( 3 2 + j 1 2 ) = 3 2 x − 1 2 y + j ( 1 2 x + 3 2 y ) \large{\vec{F} = (x + jy)(\frac{\sqrt{3}}{2} + j\frac{1}{2}) = \frac{\sqrt{3}}{2}x - \frac{1}{2}y + j(\frac{1}{2}x + \frac{\sqrt{3}}{2}y)}

Converting back into unit vector form gives:

F ⃗ = ( 3 2 x − 1 2 y ) ı ^ + ( 1 2 x + 3 2 y ) ȷ ^ . \large{\vec{F} = \Big(\frac{\sqrt{3}}{2}x - \frac{1}{2}y\Big) \hat{\imath} + \Big(\frac{1}{2}x + \frac{\sqrt{3}}{2}y\Big) \hat{\jmath}}.

Steven Chase - 4 years, 2 months ago

If we rotate the system by 60°,the force equation would yield F=-yî+xj which is a force of magnitude 1 for a unit circle path,hence the torque being T=FR=1Nm! The work should then be T*@=2π shouldn't it be??

Chirag Shyamsundar - 4 years, 2 months ago

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In my approach, I rotated the force by 30 degrees relative to the radial direction. So it's now at 60 degrees with respect to the motion, giving us a factor of c o s ( 6 0 ∘ ) = 1 2 cos(60^\circ) = \frac{1}{2} . If we rotated the force 60 degrees relative to the radial direction, the force would be at 30 degrees with respect to the motion, giving a factor of c o s ( 3 0 ∘ ) = 3 2 cos(30^\circ) = \frac{\sqrt{3}}{2} . So we would have 3 π \sqrt{3} \pi .

Steven Chase - 4 years, 2 months ago

With the force field you specify, we would indeed get 2 π 2\pi . But that would come from a 90 degree rotation of the force field with respect to the radial. I've actually posted that very problem. It was Part 1, I believe.

Steven Chase - 4 years, 2 months ago
Nitin Marwah
May 22, 2017

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