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I am wondering if anyone can help to let me know if I am on the right track or making mistakes. I am really not so confident in my work here so I am really looking for someone to look over it. It is the following problem: Yet again this is another problem I have tried significantly on my own and by trying to show my work on here, yet have received no full answers or even advice to incorporate the friction

A point is attracted to the origin by a force that is proportional to the cube of the distance from the origin, how much work is done moving the point from O to $(2,4)$ along the path $y=x^2$ assuming we have some coefficient of friction $\mu$

First question: Will I need a positive or negative sign infront of the force? I assume since it is attracted to the origin and we are moving away from the origin wouldn't we need a positive? But I am not sure.

I think my Force would be $$F=-C(x^2+y^2)^{3/2} \frac{OP}{||OP||}$$=$$-C(x^2+y^2)^{3/2}\frac{xi+yj}{(x^2+y^2)^{1/2}}$$ =$-C(x^2+y^2)(ti+tj)$ Does that seem okay? Should I have a minus sign in front of the C?

So for the force before considering friction,

parametrized with $r(t)=(t,t^2)$, $r'(t)=(1,2t)$ with $0 \le t \le 2$

$F(r(t))=-C(t^2+t^4)(ti)+(t^2+t^4)(tj)$

$F(r(t)) \bullet r'(t)= -C(2t^7+3t^5+t^3)$

integrated from $0$ to $2$, gives $$-100C$$ But I am having the more trouble with the friction part.

Moving on to the friction component, I know that work against friction is given by $-\mu | \int_{c} F_{N}dS|$

My unit tangent T is $$(\frac{1}{\sqrt{1+4t^2}},\frac{2t}{\sqrt{1+4t^2}})$$

and so would by unit Normal just be

$$ N= ( \frac{-2t}{(4t^2+1)^{3/2}(\sqrt{1/(4t^2+1)}},\frac{1}{(4t^2+1)^{3/2}(\sqrt{1/(4t^2+1)}} $$

and then If I take the dot product with F, and integrate that result, that will give the work by friction? And then I use Net Work= Work from central force+ W from friction? But since the form of the normal is so messy I am not even sure how to do the dot product in the correct way. and how to write F in such a way to do the dot product. I know it is done by component, but in my F I have a xi+yi, etc

I am still looking for help on incorporating the frictional component on this problem.

Thank you

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  • $\begingroup$ The work done by the force will be negative, the work required to overcome the force will be positive. $\endgroup$ – WW1 Feb 29 '16 at 22:05
  • $\begingroup$ If we're interested in work done by the attractive force, friction is irrelevant. But my guess is that there is another force that causes the particle to go to $(2,4)$ against the attractive force and friction, and we're supposed to find the work done by that force. The normal force it seems would depend on the angle of the path relative to the line to the origin as well as the distance to the origin. (I hope I'm not making the problem more complicated than it was meant to be.) $\endgroup$ – David K Feb 29 '16 at 22:12
  • $\begingroup$ OK, the unit normal appears to be off by a factor of $(1+4t^2)^{-1/2}$. You have two powers of $(1+4t^2)$ in the denominator of each component; combine them. Then compute the magnitude of the vector; it is not $1$. $\endgroup$ – David K Mar 4 '16 at 23:49
  • $\begingroup$ @DavidK Thank you for your time. My teacher said that the force we use here is the original F but dotted with the Normal $\endgroup$ – Quality Mar 4 '16 at 23:59
  • $\begingroup$ I may have made an unwarranted assumption in an earlier comment (now deleted, since it was wrong). I imagined the particle starts at rest at $(0,0)$. Under that assumption, the problem doesn't make sense. But if the particle has sufficient velocity at the beginning, it can reach $(2,4)$ before the force causes it to reverse direction. The speed of the particle at $(0,0)$ will affect the elapsed time but not the work done. So now it's just a question of how to integrate the frictional force without too much complication. I'll think about it. $\endgroup$ – David K Mar 5 '16 at 2:08
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Work done by the attractive force

The work done by the attractive force $\newcommand{F}{\mathbf F}\F$ along the path $C$ is $$ \newcommand{T}{\mathbf T} W_\F = \int_C \F \cdot \T \;ds $$ where $\T$ is the unit tangent vector and $s$ is distance along the curve. If we write $\newcommand{r}{\mathbf r}\r$ for the radial (position) vector of the particle, $r = \lvert\r\rvert$ for the distance from the center, and $\hat\r = \frac1r \r$ for the unit radial vector, then the force $\F$, which is a central force, can be written in the form $\F = f(r) \,\hat\r$. Using the fact that $\frac{dr}{ds} = \hat\r \cdot \T$, $$ W_\F = \int_C f(r)\, \hat\r \cdot \T \;ds = \int_C f(r)\, \frac{dr}{ds} \;ds = \int_C f(r)\, dr = \int_{r_0}^{r_1} f(r)\, dr \tag1 $$ where $r_0$ and $r_1$ are the distances from the center to the starting and ending points of $C$.

In other words, $\F$ is a conservative force whose potential is a function of distance from the center. It's useful to be able to recognize that a force is conservative, because it frees us from having to trace the work done along a particular path. In this problem, we simply set $f(r) = -kr^3$, $r_0 = 0$, and $r_1 = \lVert (2,4) \rVert = 2\sqrt5$, and integrate \begin{align} W_\F &= \int_0^{2\sqrt5} -kr^3\, dr \tag2\\ &= -k \left[ \frac14 r^4 \right]_0^{2\sqrt5} = -k\left(\frac14\left( 2\sqrt5 \right)^4\right)\\ & = -100k. \end{align} (I used $k$ rather than $C$ as the constant of proportionality because we already used $C$ to name a curve.)

Of course this is the same result you found (except for the choice of the name of the constant), and I knew that when I wrote this up. So why bother with all this? Several reasons:

  1. It's one more confirmation that you had that part of the solution already.

  2. Equation $(1)$ helps solve many problems, not just this one. It's a general property of force fields that (like this one) have spherical symmetry. If you are dealing with such a field you can pretty much just write down something like Equation $(2)$ and evaluate it.

  3. It demonstrates a way to turn a path integral into an ordinary definite integral via a change of variables without having to decide in advance how you want to parameterize your curve.

Work done by the frictional force

The frictional force is not conservative, so we still have to integrate it over the specific path $C$.

Note that I wrote $(2,4)$ for the vector from the point $(0,0)$ to the point $(2,4)$. An alternative notation is $\newcommand{ihat}{\hat \imath}\newcommand{jhat}{\hat \jmath} 2\ihat + 4\jhat$, where $\ihat = (1,0)$ and $\jhat = (0,1)$ are an orthonormal basis of this two-dimensional vector space. In other words, when you wrote (with slightly different symbols) $$ \F = -k(x^2 + y^2)(x\ihat + y\jhat), $$ it's the same vector as $\left(-k(x^2+y^2)x,-k(x^2+y^2)y\right)$, which can also be written as the product of the scalar $-k(x^2+y^2)$ with the vector $(x,y)$: $$ \F = -k(x^2 + y^2)(x, y). $$ On the path $C$ defined by $y = x^2$, this is $$ \F = -k(x^2 + x^4)(x, x^2). \tag3 $$

Conversely, your unit tangent vector, $$\T = \left( \frac{1}{\sqrt{1+4x^2}}, \frac{2x}{\sqrt{1+4x^2}} \right),$$ could instead be written $$\T = \frac{1}{\sqrt{1+4x^2}} \ihat + \frac{2x}{\sqrt{1+4x^2}} \jhat.$$

Admittedly, it's a little confusing to mix and match these two notations when writing an inner ("dot") product of two vectors, so just convert both vectors to the same notation and evaluate the inner product in that notation: either add up the componentwise products (if you use component notation), or use the facts that $\ihat\cdot\ihat = \jhat\cdot\jhat = 1$ and $\ihat\cdot\jhat = 0$.

For your normal vector, you simply need to rotate the tangent vector through a right angle. In general, a right-angled rotation of the vector $(u,v)$ will give you either the vector $(-v,u)$ or the vector $(v,-u)$; in this problem, the unit normal vector you want is the one that points to the "downward" side of the path (closer to the origin), so it is $$ \newcommand{N}{\mathbf N} \N = \left( -\frac{2x}{\sqrt{1+4x^2}}, \frac{1}{\sqrt{1+4x^2}} \right) $$ (although you could use either vector since we're going to take an absolute value later), but I think it's even easier to use the fact that $\frac{ds}{dx} = \sqrt{1+4x^2}$ to write the vector $\N$ as a scalar multiple of the vector $(-2x,1)$: $$ \N = \left(\frac{dx}{ds}\right) (-2x, 1). \tag4 $$

If you use Equations $(3)$ and $(4)$ to make substitutions for $\F$ and $\N$ in the formula for the work done on the particle by friction, $$ W_{\mathrm {friction}} = -\mu \int_C \lvert \F \cdot \N \rvert \;ds, $$ the scalar factor $\frac{dx}{ds}$ allows a convenient change of integration variable (as $\frac{dr}{ds}$ did before), and you're left with a scalar multiplied by the inner product $(x,x^2)\cdot(-2x,1)$ in the integrand: \begin{align} W_{\mathrm {friction}} & = -\mu \int_C \left\lvert -k(x^2 + x^4)(x, x^2) \cdot \left(\frac{dx}{ds}\right) (-2x, 1) \right\rvert \;ds \\ & = -\mu \int_C k(x^2 + x^4) \;\left\lvert(x, x^2) \cdot (-2x, 1)\right\rvert\; \frac{dx}{ds}\,ds \\ & = -\mu \int_0^2 k(x^2 + x^4) \;\left\lvert(x, x^2) \cdot (-2x, 1)\right\rvert\; dx \\ \end{align} The rest is just a matter of the usual simplifications and evaluations; here's the final result in Wolfram Alpha to check your work against.

By the way, using the scalar factor $\frac{dx}{ds}$ in $\N$ was not just an inspired guess or some weird stroke of genius. It was simply motivated by really not wanting to use $s$ as the variable of integration. So I looked around at what the derivatives of more convenient variables such as $r$, $x$, or $y$ would be and whether I could turn some part of the integrand (especially an othewise inconvenient part like the factor of $1/\sqrt{1+4x^2}$) into such a handy differential so that I could make a change to a more convenient variable of integration.

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  • $\begingroup$ Thank you. Could you please update with what the final numerical result is. I will go over all of it in detail, but I want to be able to check at the end that the final is correct. $\endgroup$ – Quality Mar 5 '16 at 5:48
  • $\begingroup$ because if you do take a look at what I had done on my own, I had gotten mostly all of this. My question is really about computing the actual value at the end ie the friction integral. I see how you got F to be written as -K(x^2+x^4)(x,x^2) but I dont understand next, what you are dotting it with, how, etc $\endgroup$ – Quality Mar 5 '16 at 19:07
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In general you get for this force field $$ F(x,y)·(\dot x,\dot y)=C(x^2+y^2)(x\dot x+y\dot y)=\frac C4\frac d{dt}(x^2+y^2)^2 $$ so that for $(x(t),y(t))=(t,t^2)$ for $t\in [0,2]$ the potential difference is $$ \left[\frac C4(t^2+t^4)^2\right]_{t=0}^{t=2}=100C $$


The sign should probably be negative, since the force field point from $(x,y)$ towards the origin, i.e., along the direction $(-x,-y)$.

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  • $\begingroup$ Thank you. Do you see where I went wrong in my post? $\endgroup$ – Quality Feb 29 '16 at 22:07
  • $\begingroup$ Yes, you somehow concluded that $(x,y)·(\dot x,\dot y)=(\dot x + \dot y)$ instead of $(x\dot x+y\dot y)$. $\endgroup$ – Dr. Lutz Lehmann Feb 29 '16 at 22:12
  • $\begingroup$ Sorry I dont understand this notation $\endgroup$ – Quality Feb 29 '16 at 22:13
  • $\begingroup$ Updated my post $\endgroup$ – Quality Mar 1 '16 at 17:03
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You were correct up to the line that should have read .. $$F(r(t)) \bullet r'(t)= C(t^3+t^5+2t^5+2t^7)$$

your mistake was that you didn't sub in $x=t$ and $y=t^2$ for the the $xi$ and $yj$ in the formula above that one.

$$F(r(t)) \bullet r'(t)= C(t^2+t^4)(t(1)+t^2(2t))$$

( you do end up with $100C$ when you do this )

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  • $\begingroup$ Thanks I will correct this and update after when I have done the part involving friction $\endgroup$ – Quality Mar 1 '16 at 1:49
  • $\begingroup$ I updated my post, I am stuck on the part with the friction $\endgroup$ – Quality Mar 1 '16 at 17:06
  • $\begingroup$ do you know how I could continue with the friction part? $\endgroup$ – Quality Mar 1 '16 at 20:58

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