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I'm given a position vector $\overrightarrow r$ which depends on $\theta=\theta(t)$. I found the velocity vector by simply taking the derivative. But then I noticed that the problem also mentions that the particle is subjected to a gravitational acceleration in the $-z$ direction.

Does a position vector already account for this gravity? My guess is yes. If I were to simply put in a value of time, I would find out exactly where the particle is, regardless of anything else.

The problem:

A small bead of mass m is constrained to move on a helix: $\overrightarrow r(θ) = (R \cos(θ), R \sin(θ), q θ)$ where $R$ and $q$ are constants, and $θ=θ(t)$ describes the position of the bead along the helix at time $t$. The bead is also subjected to a gravitational acceleration $g$ downward ($-z$ direction). Find the following quantities in terms of $θ$ and $dθ/dt$ :

c) potential energy U

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  • $\begingroup$ $\mathbf{v} = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}$ is the defintion of velocity and so is always true. Having said that, what type of coordinates are you working in? Whether the time dependence of $\theta(t)$ includes any motion in the $z$ direction depends on what $\theta$ is. $\endgroup$ – By Symmetry Oct 16 '16 at 15:05
  • $\begingroup$ The problem doesn't specify what coordinate system to use, so I've got to choose the best one. $\theta$ is only defined as I have it above. There is no other definition for it. $\endgroup$ – whatwhatwhat Oct 16 '16 at 15:11
  • $\begingroup$ The problem as you quote it does explicitly state the path the particle is taking, so BySymmetry's definition can be applied straightforward. You only need to account for gravity if you look at the force to find the path of the particle. $\endgroup$ – Sanya Oct 16 '16 at 15:19
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Does a position vector already account for this gravity?

What a difference a small word makes, in this case "a"!

A position vector evolving in time is simply an expression of where the particle is w.r.t. its coordinate system. E.g. for a 2D Cartesian coordinate system:

$$\vec{r(t)}=x(t).\hat{i}+y(t).\hat{j}$$

Where $ \hat{i}$ and $\hat{j}$ are the unit vectors.

In your case you would use Cylindrical coordinates.

An sich a position vector tells you nothing about whether gravity and/or other forces acting on the particle have been properly 'accounted' for.

Now, if we refer to the position vector, as in specific to your problem, then it becomes a matter of having set up the correct equations of motion or not.

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  • $\begingroup$ a position vector tells you nothing about whether gravity and/or other forces acting on the particle have been properly 'accounted' for - but hang on, if the particle is at position A then it is at position B, we can't say that nothing happened to the particle during its travel. It was certainly affected by gravity and anything else that might be present. It only reached position B (that exact position) after being affected by external forces. For example, if gravity had been a little weaker, it might made it to a farther position C instead of position B. $\endgroup$ – whatwhatwhat Oct 16 '16 at 15:51
  • $\begingroup$ And therefore, the position vector accounts for this because it claims that the particle must be at position [blank] after traveling. What do you think? $\endgroup$ – whatwhatwhat Oct 16 '16 at 15:53
  • $\begingroup$ No one is saying 'nothing happened', You're confusing position vector with change in position. Yes, if $\vec{r(t)}$ is an expression accounting for all forces acting then of course it will account for gravity. When you computed $\vec{r(t)}$, did you take gravity into account? From your question, gravity seems to pop up almost as an 'afterthought'. $\endgroup$ – Gert Oct 16 '16 at 16:25

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