[Espy]

Nope, what's propelling the ball in a circle is inertia. And since the ball is swinging around at a constant speed, there's no acceleration in the "around" direction, and hence no force.

And, ick, that's some weird wording. How I understand it, the Cartesian coordinates (0, 1) [x=0, y=1] translates to (pi/2, 1) [theta=pi/2, r=1] in polar, and (0, -1) translates to (-pi/2, 1), so if x=0, theta is either -pi/2 or pi/2.

[Potironette]

Thanks for clearing that up!

Err, I thought that polar coordinates were in the form of (r, theta) though?

[Coda]

Regarding the orange force line: The tangential force line is 100% real -- it has a (net) magnitude of zero if it's not accelerating, but it's real. If you try to swing the object faster, then the magnitude will be nonzero until it reaches the new speed. That said, you and Espy are both right in that the force isn't necessary to be there for it to keep spinning, only for it to start spinning.

An interesting exercise is to think about how that force GETS there. When you're spinning the object around on the rope, you're not pushing it from the side in order to get it moving (unless you're playing tetherball) so how is it that you can accelerate it along that axis?

I personally prefer to notate polar coordinates either parametrically -- that is, r=1 θ=π/2 -- or using an angle symbol like 1∠π/2. That said, the standard notation is indeed (r, θ) in 2D and (r, θ, ϕ) in 3D.

[Potironette]

Ohh I see. So it's like the moon where the moon just needed to be set with a velocity around the Earth at some point long long ago and now it's just technically falling towards Earth?

Does it get there because..
first the rope is pulled up
the ball gets pulled in the same direction as the rope
then the rope is pulled perpendicular to the ball's motion
then the ball starts going in a circle pattern because of the ball's inertia?
And the direction of velocity of the ball keeps getting changed which is why it can't just get pulled into the center of the circle it's making?

I've not learned notating polar coordinates those ways :o

[Coda]

Pulling the rope perpendicular to the ball's motion will just make it fall in towards the center. Pulling the ball up in the first place is also not putting any force in the direction of the orbit.

I'll give you a hint: It's a little bit of a trick question. It's a question that does indeed have a real answer, but the process of trying to find it will reveal the shortcomings in your model.

The angle symbol notation is used by some graphing calculators. I've not ever seen it used in a textbook. The parametric notation is mostly used when it's not obvious what coordinate system is in use or when you're defining functions in the coordinate space instead of individual points. (In the latter case, you'll often see "r(t) = ..., θ(t) = ..." to specify the coordinate in terms of a parameter.)

[Potironette]

Then, is it just that the force comes from the person just pulling on the string to pull on the ball into the direction the person wants the ball to go to once in a while?

[Espy]

Originally Posted by Potironette

Err, I thought that polar coordinates were in the form of (r, theta) though?

Uh whoops, yeah. Been a while since I've done anything with actual polar coords.

[Coda]

Originally Posted by Potironette

Then, is it just that the force comes from the person just pulling on the string to pull on the ball into the direction the person wants the ball to go to once in a while?

More or less, yeah. You can't model it as an object on a fixed pivot with an infinitely thin, infinitely flexible, perfectly inelastic tether. You either have to give the tether some stiffness (at which point you're modeling an object on a rod instead of an object on a string), or you model it more realistically:

When you're spinning an object around on a string, you aren't spinning it around a fixed central pivot. You're pulling it at an angle, applying the force in a circle around the center of rotation. The string isn't parallel to the centripetal force vector or perpendicular to the tangential velocity vector; you're pulling on it to one side in addition to towards the middle.

[Potironette]

What about when you swing a thing then stop and let it swing on its own? The centripetal force keeps decreasing this way though.

Also, what's a "polynomial?" I have a review sheet asking me to identify if random expressions are polynomials or not, their degrees, and the "type" of polynomial they are, but I have no idea what a polynomial actually is. Why isn't cosine or sine something a polynomial? What does it mean that the "type" of polynomial is "based on the number of terms?"

[Coda]

If you stop, and you let it keep swinging on its own, then in an ideal system it would just keep moving forever like a planet in orbit. In reality, there's friction that will bleed off that momentum in the form of heat, and it'll spiral towards the center.

A polynomial is just ax^m + bxⁿ + ... + k. Nothing fancier about it than that. The degree of the polynomial is the highest exponent that appears in it. Cosine and sine aren't polynomials because they're not sums of multiples of powers of a single variable. (That said, there are polynomial approximations of those functions, such as Taylor series.) I'm not sure what the "type" of a polynomial is because that's a rather vague term and it could mean several things.

[Espy]

Coda, maybe "type" has to do with second order, third order, etc? Though that doesn't necessary dictate the number of terms, just the maximum number of terms.

Or perhaps, erm, even and odd polynomials?

[Potironette]

Thanks for the responses! I dunno what the "type" refers to and still need to ask about that.
So.. if something is ax^3/2 is it a polynomial? What about ax^-1?

Also, how do you find possible imaginary roots of a polynomial? I think I can find not-imaginary ones by having any factor of the constant over any factor of the coefficient of the largest degree, but what about imaginary roots?

Also, when I'm given a problem that tells me to find a polynomial with real coefficients that have a root 2 - i, I know that I have to do something like (x + 2 - i)(x - (2 - i)) but I'm not sure where to go from there. I forgot how to get rid of the i's.

[Coda]

Non-negative integer exponents. Those two examples are not polynomials. A term including x⁰ (that is, a term not including x at all) on the other hand can still be in a polynomial.

You get rid of the i's using i² = -1.

As for finding imaginary roots, we covered that already, you might need to go back a few pages.

[Potironette]

Is it reasonable for a car to accelerate at -18 mi/s^2 while it's braking?

(Context: I have a homework problem that asks me to find the acceleration of a car while it's braking: "According to the recent test data, an automobile travels 0.250 mi in 19.9 s starting from rest. The same car, when braking from 60.0 mi/h on dry pavement, stops in 146 ft. Assume constant acceleration in each part of the motion, but not necessarily the same acceleration when slowing down or speeding up. (a)Find the acceleration of the car when it is speeding up and when it is braking.
I found 1.51 mi/h^2 while speeding up so I figured maybe braking would be similar.
Tried to use x_f = x_i + v_i*t + .5at^2 and got -(60.0 mi/h)^2/(2*.0277 mi) = 6.50 * 10^4 mi/h^2. I tried to turn it to mi/s^2 because clearly the car isn't braking in a whole hour (still the number's huge?), but it was -18 mi/s^2. I'm not sure where I went wrong on this problem. The car couldn't possibly be accelerating that fast right..? Even earth rotates slower than that, right..?)

[Serra Britt]

Without doing the actual calculation to check your answer, I'm looking at it in reverse. At an acceleration of -18 mi/s^2, it would mean you would stop nearly instantly even going 60 mi/h. So your initial thoughts are correct that the acceleration is too fast. It does still need to travel the 146 ft at least.

Using formulas here http://www.dummies.com/education/sci...-and-velocity/

A speed of 60 mi/h is 88 ft/s.
146 ft = (88 ft/s)x(time) so time to stop is 1.659 seconds.
So then:
1.659 sec = (-88 ft/s)/(acceleration) so acceleration would be -53 ft/s^2, which is a lot less than -18 mi/s^2

[Potironette]

Thanks for replying!
(Also I actually used v_f^2 = v_i^2 + 2a(x_f-x_i) oops)

But I'm confused, why is distance = velocity x time ?
Isn't in this case distance = velocity x time + acceleration affecting the velocity during the time x time?

Edit: not confused anymore, thank-you!
Edit2: actually, I'm still confused why I can use the 60.0 m/s as the only speed when speed is changing.