I know this isn’t a physics website, but I’m working on some pretty simple physics atm, and this question came up, no idea how to start. So I come to the Straight Dope! and my second question is, should physics be this confusing?
The thought process is given in the question. You just need to do the math. You’ll need to integrate mv^2/2 along the spring to get the kinetic energy.
Suppose that instead of the classic situation of a spring anchored at one end with a concentrated mass at its free end, we have a spring anchored at one end with tiny weights glued onto it at regular intervals. Take it to the extreme of an infinite number of infinitesimal weights glued on at infinitesimal intervals. You know the weight of each “slice” of spring, and you also know the speed of each “slice,” based on its linear position within the spring. So you can calculate the kinetic energy of each “slice” of spring. Then you just integrate that expression.
If you’re strong on basic calculus, this is trivial. If you, like me, forgot all the details about 15 years ago but still recognize the terminology, then you’re hosed.
I agree with JWT Kottekoe, they completely told you exactly what to do, provided you can do simple integration word problems. These aren’t exercises in understanding, they’re mechanical plug *this *into *that *& integrate over the other thing problems.
I’m not belittling you; I’m as stuck as you are without breaking out my very dusty Calc 101 book from the Dark Ages. But that’s the thing it seems you need to get any traction on these.
McLuhan’s Thunder, since this appears to be some form of college homework, we understand you aren’t trying to get an answer per se, but want some help on how to understand what to do, so you can get about doing it right.
It would help us to do that if we knew just exactly where your own effort to solve the problem fails you? Is it lack of comprehension of what the words mean? Is it lack of understanding of what the physics is? Is it lack of the neccessary math to complete the requested calculations?
Walk us through what you DO know how to do, or at least what you are thinking about the problem, and then we can help you see where to refocus your efforts.
For what it is worth, and not saying it is applicable here, but increasingly it seems to me that students are losing the ability to read a problem and parse from it what they are expected to accomplish, as well as what information the problem has offered them. As a math teacher, this can be quite frustrating. Of course, for all I know, my math teachers when I was in high school/college felt the same way about my generation.
Kinetic energy of some mass m with speed v is (1/2) m v[sup]2[/sup].
In this case the trick is that each piece of the spring is moving at a different speed. They say the speed varies linear (that is, it’s given by the equation of a line), and one end (let’s say it’s at x = 0) is fixed, while the other end (let’s say it’s at x = L) is moving with speed v.
So I know that the speed of a point at position x is given by V(x) = v x / L.
Do you see why? I needed V to be of the form V(x) = ax + b, because it’s linear. I had to make b = 0 to satisfy V(0) = 0. And then I had to make a = v/L to satisfy V(L) = v.
Now if the spring were divided into discrete chunks of mass m at the various positions, we’d just find the kinetic energy of the chunk at position x to be K(x) = (1/2) m V(x)[sup]2[/sup] = (1/2) m (v x / L)[sup]2[/sup]. Then to find the total kinetic energy we just sum over all the discrete chunks.
However, the spring isn’t in discrete chunks, it’s a continuous object. So instead of a sum, we take an integral. Integrals are basically sums for continuous things. And where for the discrete case, say a spring split into 5 pieces, the mass of a piece would M/5 (where M is the total mass), for the continuous case we instead use M / L (the mass divided by the length of the spring). Make sense? (Note that if you were to just integrate M / L from x = 0 to x = L, you’d get back the total mass M. So that shows we’ve divided up the mass in the right way. Well, that and the fact that we were told the mass was distributed uniformly, meaning we couldn’t use some function of x for mass that just happened to integrate to the total mass M. M / L is the only constant that works.)
So the problem basically becomes Take the integral with respect to x of (1/2) (M/L) V(x)[sup]2[/sup] from x = 0 to x = L.
I disagree, I think there’s some understanding required here. You have to know what “the speed varies linearly” means. You have to know what to do with the fact that “the mass is distributed uniformly”. You have to know how to calculate the kinetic energy of a point mass. You have to know that you can treat the spring like a collection of point masses and then perform an integral along its length. And that’s all just for part (a).
None of these things were explicitly stated in the problem. Of course, these things are obvious if you have a good handle on how calculus is used to solve physics problems. But of course many students don’t, at least at first, and that’s probably just what the problem is supposed to test.
One of my physics professors was fond of saying that the hardest part of most problems was transforming the words into equations to be solved. After that, it’s just arithmetic. (Or just calculus, at any rate.)
That is almost the definition of physics.
First you discover a puzzle, then you write down the equations, then you work out the answer. The hard part, and the fun part, is the first step!
