Parabolic mirrors have interesting optical properties (eg. parallel rays all get
ID: 1950795 • Letter: P
Question
Parabolic mirrors have interesting optical properties (eg. parallel rays all get focused to a single point) that make them very useful in applications from car headlights to large telescope mirrors. One ingenious way of constructing a parabolic mirror is a method known as spin casting, where a pool of liquid (usually molten borosilicate glass) is spun and then cooled - forming a parabolic shape. We’ll see how this is accomplished in this problem.
I know how to do part (a) and how to get to the that equation. I only need help on part (b), which I don't understand how to do because the pressure is not the same at every point in y=0.
(a) We start by filling a cylindrical container with our molten mirror liquid of constant density , and rotate it about it’s axis of symmetry (the y axis) at an angular velocity . Show that the pressure of the liquid at a given y coordinate varies with radius as
P/r = 2r
Why is this a partial derivative as opposed to a total derivative (dP/dr)?
(b) Integrate this partial differential equation to find the pressure as a function of r at the bottom of
the container, y = 0.
Explanation / Answer
(b) Integrate this partial differential equation to find the pressure as a function of r at the bottom of
the container, y = 0.
"I know how to do part (a) and how to get to the that equation. I only need help on part (b), which I don't understand how to do because the pressure is not the same at every point in y=0."
Since you know how to integrate it in part (a), all you need to do is take the integrated value and plug in y = 0, aka solve for P when r = 0. It doesn't matter if the answer isn't the same for all y = 0 because, in this case, we're **keeping the constants** AND using r as our "y value." This is why you need to include the constant in your answer.
Why can we plug in 0 for r when the question asks for y? Consider this: If it's at the bottom of the container, the r value is automatically zero. Moreover, we know that we're solving for pressure - thus we must plug in the given value for our other variable, r, which will always equal 0 when we're at the "bottom" of the container.
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