2.6: Three-dimensional Solid Figures. Spheres, Cylinders, Cones.
- Page ID
- 6937
Sphere, mass \(m\), radius \(a\).
The volume of an elemental cylinder of radii \(x\), \( x + \delta x \), height \( 2y \) is \( 4 \pi yx \delta x = 4\pi(a^2-x^2 )^{1/2} x \delta x \). Its mass is \( m \times \frac{4\pi(a^2-x^2)^\frac{1}{2} x \delta x}{\frac{4}{3}\pi a^{3}} = \frac{3m}{a^{3}} \times (a^2-x^2)^\frac{1}{2} x \delta x. \) It' second moment of intertia is \( = \frac{3m}{a^{3}} \times (a^2-x^2)^\frac{1}{2} x^3 \delta x. \) The second moment of inertia of the entire sphere is
\( = \frac{3m}{a^{3}} \times \int_{0}^{a} (a^2-x^2)^\frac{1}{2} x^3 \delta x = \frac{2}{5} ma^2. \)
The moment of inertia of a uniform solid hemisphere of mass \( m\) and radius \( a\) about a diameter of its base is also , \( \frac{2}{5} ma^{2} \), because the distribution of mass around the axis is the same as for a complete sphere.
A hollow sphere is of mass \( M \), external radius \( a\) and internal radius \( xa \). Its rotational inertia is \( 0.5 Ma^2 \). Show that \(x\) is given by the solution of
\( 1 - 5x^3 + 4x^5 = 0 \)
and calculate \( x\) to four significant figures.
(Answer = 0.6836.)
Solid cylinder, mass \( m\), radius \( a\), length \( 2l\)
The mass of an elemental disc of thickness \( \delta x \) is \( \frac {m \delta x} {2l} \). Its moment of inertia about its diameter is \( \frac{1}{4} \frac{m \delta x }{2l} a^2 = \frac{m a^2 \delta x }{8l} \). Its moment of inertia about the dashed axis through the centre of the cylinder is \( \frac{m a^2 \delta x }{8l}+ \frac{m \delta x }{2l} x^2 = \frac{m(a^2+4x^2) \delta x}{8l}. \) The moment of inertia of the entire cylinder about the dashed axis is \( 2 \int_{0}^{1} \frac{m(a^2+4x^2) \delta x}{8l} = m(\frac{1}{4}a^2 + \frac{1}{3} l^2)\).
In a similar manner it can be shown that the moment of inertia of a uniform solid triangular prism of mass \( m \), length \( 2l \), cross section an equilateral triangle of side \(2a \)about an axis through its centre and perpendicular to its length is \(m(\frac{1}{6}a^2 + \frac{1}{3} l^2)\).
Solid cone, mass \( m\), height \( h\), base radius \( a\).
The mass of elemental disc of thickness \( \delta x \) is
\( m \times \frac{\pi y^2 \delta x}{\frac{1}{3} \pi a^2 h} = \frac{3my^2 \delta x } {a^2h}\).
Its second moment of inertia about the axis of the cone is
\( \frac{1}{2} \times \frac{3my^2 \delta x } {a^2h}\times y^2 = \frac{3my^4 \delta x } {2a^2h}\).
But \( y \) and \( x \) are related through \( y = \frac{ax}{h} \), so the moment of inertia of the elemental disk is
\( \frac{3ma^2x^4 \delta x } {2h^5}\).
The moment of inertia of the entire cone is
\(\frac{3ma^2} {2h^5} \int_{0}^{h}x^{4} dx = \frac{3ma^2} {10}\).
The following, for a solid cone of mass \(m\), height \(h\), base radius \(a\), are left as an exercise: