conical glass

**pacman** · July 28th 2009, 12:45 AM

A conical glass of depth h and generating angle A, there is carefully dropped spherical iron ball of such size as to cause the greatest overflow.
Show that the radius of the sphere is equal to (h sin A)/(sin A + cos 2A).

**Opalg** · July 28th 2009, 12:26 PM

Originally Posted by pacman

A conical glass of depth h and generating angle A, there is carefully dropped spherical iron ball of such size as to cause the greatest overflow.
Show that the radius of the sphere is equal to (h sin A)/(sin A + cos 2A).

Let $x$ denote the height of the centre of the ball above the vertex of the cone, and let $r$ denote the radius of the ball, so that $r = x\sin A$ . The volume displaced is the volume of that part of the ball that lies below height h. I'll cheat by quoting the formula given here for the volume of a frustum of a sphere, which tells you that it is $V = \tfrac{\pi}3(r+h-x)^2(2r-h+x).$

To save writing, it's convenient to write $s$ for $\sin A$ . So replace $r$ by $xs$ in the above formula, and you get $V = \tfrac{\pi}3\bigl(h-x(1-s)\bigr)^2\bigl(x(2s+1)-h\bigr).$

To find the value of $x$ that maximises $V$ , differentiate (with respect to $x$ ) and put the derivative equal to 0. After some simplification you should find that the maximum occurs when $x = \frac h{s+1-2s^2}$ , from which
$r = xs = \frac {hs}{s+1-2s^2}$ . Finally, replace $s$ by $\sin A$ and remember that $\cos(2A) = 1-2\sin^2A$ .

**pacman** · July 30th 2009, 03:12 AM

Thank you very much opalg, it's great deal to have some members of this forum who dares to solve challenging problems.

. May your tribe increase!

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