- #1
StudentOfScience
- 47
- 3
Homework Statement
Bead on spoke:
constant speed ##u## along spoke
it starts at center at ##t=0##
angular position is given by ##\theta=\omega t##, where ##\omega## is a constant
Homework Equations
## \frac{d\hat r}{dt} = \dot \theta \hat \theta ## (1)
## \frac{d\hat \theta}{dt} = -\dot \theta \hat r ## (2)
## \vec r = | \vec r| \hat r ## (3)
Note: velocity and acceleration in polar coordinates can be obtained by differentiating the above equation with resptect to time and using equations (1) and (2)
The Attempt at a Solution
I'm more or less confused as to why ## \vec r (t) ## doesn't take the form
## \vec r(t) = r_r \hat r + r_\theta \hat \theta ## (4).
They derived the velocity and acceleration equations by using a general vector in the plane given by
## \vec r = |\vec r|\hat r ## (5)
and differentiating it with respect to time. I realize that (5) is just a vector that is not time dependent; we would write ##\vec r(t)## if it were a function of time. Here is the approach that I took (I'm not exactly sure what is wrong with it, but here we go):
Define ## \vec r(t)=r_r \hat r + r_\theta \hat \theta ## (6). We are given radial velocity:
## v_r (t)=u ## (7)
We also know that
## \mathbf r_0 = \vec 0 ## (8)
since the bead starts at the pole. Hence,
## r_r =\int_0^t udt' = ut ##. (9)
In the problem statement, we are given angular position:
##r_\theta=\omega t ## (10)
Hence, the position of the bead is given by
## \vec r(t) = ut\hat r + \omega t \hat \theta ## (11).
I will go ahead and confirm that this is an Archimedean spiral:
Because
## \hat r = \cos\theta \hat i + \sin\theta \hat j, \hat \theta = -\sin\theta \hat i + \cos\theta \hat j ##, (12)
it follows that
## \vec r(t) = (ut\cos\theta - \omega t \sin \theta) \hat i + (ut\sin\theta + \omega t \cos\theta)\hat j ## (13)
And we defined ##\theta## earlier as
##\theta = \omega t## (14). Ok, here is a question I have: was it incorrect for me to assume that ##r_\theta## and ##\theta## in (13) were the same (##=\omega t##)? Perhaps I misinterpreted angular position as ## r_\theta##? We can check this though on a graphing calculator in parametric mode and, by inspection, see that it is an Archimedean spiral (ok, I was too lazy to graph it by hand)
For velocity, I differentiated (11) to get
$$\vec v(t)##= (u-\omega t \frac{d\theta}{dt})\hat r + (\omega+ut\frac{d\theta}{dt})\hat\theta $$ (14)
[and LaTeX apparently is not being so nice right now]
I didn't use
$$\dot \theta = \omega $$ (15) (maybe this is another mistake?); instead, since we have defined
$$ \vec r(t) = r_r \hat r + r_\theta \hat \theta $$, (16)
we can differentiate this to obtain
$$ \vec v(t) = (\dot r_r -r_\theta \dot \theta)\hat r + (r_r\dot \theta +\dot r_\theta)\hat \theta $$ (17)
By equality of components for (14) and (17), we have
$$ v_r= u-\omega t \dot \theta = \dot r_r - r_\theta \dot\theta $$
implies
$$ \dot \theta = \frac{u-\dot r_r}{\omega t-r_\theta}=\frac{0}{0}, $$
which is indeterminate. I tried equating the other component - ##v_\theta ## - but, to no avail, I also got an indeterminate form. And thus a road block.
The solution in Kleppner, though, is much cleaner:
They derived earlier that
$$ \vec a(t)= (\ddot r - r(\dot \theta)^2)\hat r + (r\ddot \theta+2\dot r \dot \theta)\hat \theta $$ (18)
and they essentially used ##\theta=\omega t## and ##r=ut## and differentiated when appropriate. If I take this approach, how can I find position? They started with the general vector ##\vec r=r\hat r ##, which, as I said earlier, is not a function of time. So would I have to convert to Cartesian coordinates, integrate, and then go back to polar coordinates, or is there a nicer way to find position (finding position wasn't part of the question, but I ask out of curiosity)? It seems as though (11) is incorrect for position.
Thank you all for your answers. Please let me know if you would like me to clarify something. I appreciate your time!