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enter image description here

My question concerns (a).

I'm not sure how to plot this particle's distance from the origin as a function of time as it asks, since I don't know how to reconcile $\theta$ into the particle's motion. My guess would be that given $r=\frac{\alpha}{t}$, the object would follow a path similar to the image of $y = \frac{1}{x}$, with the domain restriction $x \ge 0$, however, at $t=0$, $r$ should also be $0$, and that idea falls flat as a result. Since the expressions are in polar, I'm not sure how to properly graph this $r(t)$. Any guidance on what I should do would be very helpful.

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  • $\begingroup$ Ask yourself this: in polar coordinates, how do you express the distance to the origin? $\endgroup$ – Javier Sep 27 '17 at 20:40
  • $\begingroup$ Why should $\theta(t)$ have any bearing on $r(t)$. They are independent, perpendicular coordinates in a polar system. The time $t$ is simply a thrid parameter which would allow you to plot a trajectory. That's not what (a) is asking for. And why do you say "at $t=0$, $r$ should also be $0$"? $\endgroup$ – Bill N Sep 27 '17 at 21:42
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Let's just think about what $r$ and $\theta$ tell us about he position of the particle. It is important that you do not forget $(r,\theta)$ are your polar coordinates. The key difference between is that compared to a cartesian coordinate system $(x,y)$, is that $\theta$ is just an angle, it is not a variable that tells you about a distance from some where like, $x$ is the distance from the $x$-axis, $y$ the distance from the $y$-axis and $r$ the distance from the origin.

You said, you want to know the particles distance $r$ from the origin at a specific time $t$. That is exactly $r(t)$.

$\theta(t)$ will will just give you the angle with respect to your coordinate system of the object at time $t$.

Suppose you have calculated $r(t_0)=r_0$, now you know the object must be on a circle with radius $r_0$ around the origin. If you now calculate the angle $\theta(t_0)=\theta_0$ then you know what point of that circle marks the position of the particle BUT this will not change the radius of the circle. This means the distance to the origin does not change. I hope this clears it.

Now let's think about the trajectory of the particle. $r$ for small $t$ is very large but for bigger $t$ it goes slower and slower to $0$. This means the particle comes from infinity rushing towards the origin and the closer it gets the slower it approaches the origin. However for $\beta\neq 0$it does not come in a straight line. It spirals around the origin. At first its path is fairly straight but as time goes on it rotates faster and faster around the origin.

Here is a plot for $\alpha=1,\beta=3,\gamma=0$. The center of it should also be blue, the progam just plots it for a small intervall for $t$.

EDIT: What you see as the input of the plot is a parametric representation for the path of the object. Sorry if it looks weird, I just don't know how to input a parametric plot with polar coordinates into wolfram alpha. - I just converted your polar coordinates $(r(t),\theta(t))$ into carthesian coordinates $(x(t),y(t))$. A little more precise: we made the transformation $$r\hat{r}+\theta\hat{\theta}\rightarrow x\hat{x}+y\hat{y},$$ where the hatted once are the corresponding unit vectors. Explicitly, I used the following relationships $$r\cos{\theta}=x\quad r\sin{\theta}=y.$$ Putting this into the transform we get

$$\begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\begin{pmatrix}r(t)\cos{\theta(t)}\\r(t)\sin{\theta(t)}\end{pmatrix}=\begin{pmatrix}\frac{\alpha}{t}\cos{\left(\frac{\beta t^3}{3}+\gamma\right)}\\\frac{\alpha}{t}\sin{\left(\frac{\beta t^3}{3}+\gamma\right)}\end{pmatrix}$$

EDIT2: I made a mistake, saying that the object has to spiral around the origin. That is only true if $\beta\neq 0$. If $\beta=0$ this means the angle $\theta$ is constant, i.e. no rotation. For a positive $\beta$ it spirals anti clockwise and for a negative one, it spirals clockwise.

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  • $\begingroup$ Looking at your graph, what you've stated should happen clearly is modelled well in the graph. However, I don't know what exactly you graphed to show this. Like, I don't know where the sine and cosine functions are in there for exactly, other than maybe you're converting those conditions you stated in Cartesian, and why their arguments are $t^3$. $\endgroup$ – sangstar Sep 29 '17 at 23:52
  • $\begingroup$ @sangstar I added the explanation to my answer, see the edit. $\endgroup$ – ty. Sep 30 '17 at 21:43
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Lets try to imagine how it moves. The function of the distance from the origin (r) is infinite at t=0 as you said and goes to 0 at long enough times. As t increases, θ increases also thus forming a counter-clockwise spiral. I don't know what they want you to do next, but nobody said at t=0, r=0. The correct answer is just a hyperbola y=1/x as you said.

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  • $\begingroup$ This doesn't seem to agree with the other answer.. hmmm.. $\endgroup$ – sangstar Sep 30 '17 at 0:02
  • $\begingroup$ It actually agrees perfectly. If you look at the end result of ty. and get the distance ( sqrt( x^2 + y^2)) , then you get the same function as the given one. $\endgroup$ – Luthelin Oct 2 '17 at 16:26

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