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Math Help - [SOLVED] Difficult statics question?

  1. #1
    Super Member fardeen_gen's Avatar
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    [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.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by fardeen_gen View Post
    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
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  3. #3
    Super Member fardeen_gen's Avatar
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    The text does not provide a diagram Which makes this confusing problem all the more confusing.
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    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}} <br />
=\frac{\left( \frac{2}{3} \pi r^3 \rho \right) \cdot  \left(  \frac{3}{8} r \right)<br />
- \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} <br />
= \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]}.

    Attached Thumbnails Attached Thumbnails [SOLVED] Difficult statics question?-statics_diagram_b.jpg  
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  5. #5
    Super Member fardeen_gen's Avatar
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    Thanks again, the_doc! Your problem solving abilities are uncanny.
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  6. #6
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    Shucks!

    Your welcome, Fardeen - any time.
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