Math Help - [SOLVED] Difficult statics question?

1. [SOLVED] Difficult statics question?

If the greatest possible cube, is cut out of a solid hemisphere of uniform density, prove that the remainder can rest with its curved surface on a perfectly rough inclined plane with its base inclined to the horizon at an angle:

$\arcsin\left[\frac{8(3\pi - \sqrt{6})}{9\pi - 8}\sin \alpha\right]$

where $\alpha$ is the slope of the inclined plane.

2. Originally Posted by fardeen_gen
If the greatest possible cube, is cut out of a solid hemisphere of uniform density, prove that the remainder can rest with its curved surface on a perfectly rough inclined plane with its base inclined to the horizon at an angle:

$\arcsin\left[\frac{8(3\pi - \sqrt{6})}{9\pi - 8}\sin \alpha\right]$

where $\alpha$ is the slope of the inclined plane.
This really needs a diagram, but you must be able to position the body so that the centre of mass is directly above the point of contact.

CB

3. The text does not provide a diagram Which makes this confusing problem all the more confusing.

4. Solution

Here's how to do it:
Spoiler:

Introduction

I've included a diagram below.

You will note that, as CaptainBlack correctly said, for equilibrium the weight of the object must be directly over the point of contact.

In the diagram, you can see that $O$ is the point that would be the centre of the flat face of the hemisphere, $OS$ is the axis of rotational symmetry of the original hemisphere, $P$ is the point of contact with the inclined plane and $C$ is the location of the centre of mass of the hemispherical object.

Geometry

$CP$ is a vertical line and since $OP$ is at right angles to the inclined plane we must have that the angle $OPC$ is $\alpha$. Also the angle that the flat face of the hemisphere makes with the horizontal is the same as the angle between $OS$ and the vertical i.e. we need to find the angle $PCS$. Let this angle, $PCS$, be $\theta$.

Using the sine rule for the triangle $PCO$, we find that

$\frac{\sin (\pi - \theta)}{r} = \frac{\sin \alpha}{OC}$

since $OP = r$. Since we know $\theta$ is acute we can rearrange this so

$\color[rgb]{0,0,1} \boxed{\theta = \arcsin \left[ \frac{r}{OC} \sin \alpha \right]}$.

So all we need to do is determine $OC$.

Dimensions of the cube

The greatest cube that can be cut out of the hemisphere has its centre on $OS$ and so if we imagine expanding a small cube centred directly on top of $O$ we will no longer be able to expand it when its height is such that its four other corners touch the curved surface of the hemisphere. This cube then has one face in the same plane as the flat surface of the hemisphere and the opposite face's corners touching the curved suface. At any one of these corners that meet the curved surface the distance from $O$ must be $r$ so using Pythagoras Theorem we must have

$r^2 = a^2 +l^2$

where $a$ is the side length of the cube and $l$ is the distance from $O$ to the corresponding corner of the cube that lies in the plane of the flat surface of the hemisphere.

Again using Pythagoras we must have (for a cube) that

$l^2 = \left( \frac{a}{2} \right)^2 + \left(\frac{a}{2} \right)^2$,

hence

$r ^2 = a^2 + 2 \left(\frac{a}{2} \right)^2$,

$\color[rgb]{0,0,1} \boxed{\iff a = \sqrt{\frac{2}{3}} r}$.

Centre of mass calculation

I'm going to assume that you know the result that the centre of mass of a uniform hemispherical solid is on the rotational axis of symmetry at $3/8 r$ from $O$.

Let $\rho$ be the density of the solid, $m$ the mass and $\mu$ be the distance of the centre of mass from $O$ on $OS$ where the solid referred to is denoted by the subscripts $h$, $c$ and $hc$ for hemisphere (uncut), cube and hemisphere with cube cut out respectively.

Then we must have (in terms of moments about $O$) that

$m_{hc} \mu_{hc} + m_c \mu_{c} = m_h \mu_h$

and thus we can determine $\mu_{hc}$ since

$m_{h} = \frac{2}{3} \pi r^3 \rho$,
$\mu_{h} = \frac{3}{8} r$,

$m_{c} = \rho a^3 = \frac{2}{3} \sqrt{\frac{2}{3}} \rho r^3$,
$\mu_{c} = \frac{a}{2} = \frac{1}{\sqrt{6}} r$,

$m_{hc} = m_{h} - m_{c} = \frac{2}{3} \pi r^3 \rho -\frac{2}{3} \sqrt{\frac{2}{3}} \rho r^3 = \frac{2}{3} \left( \pi -\sqrt{\frac{2}{3}} \right) \rho r^3$.

So we have

$\mu_{hc} = \frac{m_{h} \mu_{h} - m_c \mu_c }{m_{hc}}
=\frac{\left( \frac{2}{3} \pi r^3 \rho \right) \cdot \left( \frac{3}{8} r \right)
- \left(\frac{2}{3} \sqrt{\frac{2}{3}} \rho r^3\right) \cdot \left( \frac{1}{\sqrt{6}} r\right)}{\frac{2}{3} \left( \pi -\sqrt{\frac{2}{3}} \right) \rho r^3}
= \frac{\frac{3}{8} \pi -\frac{1}{3}}{\pi -\sqrt{\frac{2}{3}}} r$
,
$= \frac{9 \pi - 8}{8 \left(3 \pi - \sqrt{6} \right)} r$

with $OP = \mu_{hc}$.

Putting it together

Substituting our value of $OP$ into our trigonometric result gives us that

$\color[rgb]{1,0,1} \boxed{\theta = \arcsin \left[ \frac{8 \left(3 \pi - \sqrt{6} \right)}{9 \pi - 8} \sin \alpha \right]}$.

5. Thanks again, the_doc! Your problem solving abilities are uncanny.

6. Shucks!

Your welcome, Fardeen - any time.