[Math] Solving for radius of a combined shape of a cone and a cylinder where the cone is base is concentric with the cylinder

algebra-precalculustrigonometry

I have a solid that is a combined shape of a cylinder and a concentric cone

(a round sharpened pencil would be a good example)

Know values are:

Total Volume = 46,000

Height to Base Ratio = 2/1
(Height = cone height + cylinder height) (Base = Diameter)

Angle of cone slope = 30 degrees (between base and slope of cone)

How do you solve for the Radius?

2 let say the ratio is from the orignal length

Know value:

Total Volume = V = 46,000

Original Height to Base Ratio = t = 2/1
(Height = cone height + cylinder height + distance shortened) (Base = Diameter)

Angle of cone slope = θ = 30 degrees (between base and slope of cone)

distance shortened from original Height x = 3

(Let h be the height of the cylinder. Then h+rtanθ+x=2(2r)

How do you solve for the Radius?

Best Answer

I assume that by height to base ratio you mean the ratio of total height (cylinder plus cone) to the diameter of the base.

Let $\theta$ be the angle between the base of the cone and the sloping walls. Then the height of the cone is $r\tan\theta$. In our case $\tan\theta=\frac{1}{\sqrt{3}}$. Kind of a stubby cone. We keep on writing $\tan\theta$ instead of $\frac{1}{\sqrt{3}}$, it makes typing easier.

Let $h$ be the height of the cylinder. Then $h+r\tan\theta=2(2r)$, and therefore $h=r(4-\tan\theta)$.

The combined volume is $\pi r^2h+\frac{1}{3}\pi r^3\tan\theta$. Substituting for $h$, and simplifying a little, we find that the volume is $$\pi\left(4-\frac{2}{3}\tan\theta \right)r^3.$$ Now we can set this equal to $46000$ and solve for $r$. I get something close to $15.94$, but do check!