Question

explain moment of interia of a solid sphere and solid cylinder?

Answer 1

explain moment of interia of a solid sphere and solid cylinder?

Solid ball of radius R and mass M:

$$I = 2 M R ^ {2} / 5$$

The expression for the moment of inertia of a sphere can be developed by summing the moments of infinitesmally thin disks about the $z$ axis. The moment of inertia of a thin disk is

$$d I = \frac {1}{2} y ^ {2} d m = \frac {1}{2} y ^ {2} \rho d V = \frac {1}{2} y ^ {2} \rho \pi y ^ {2} d z$$

and the integral becomes

$$I = \frac {1}{2} \rho \pi \int_ {- R} ^ {R} y ^ {4} d z = \frac {1}{2} \rho \pi \int_ {- R} ^ {R} (R ^ {2} - z ^ {2}) ^ {2} d z = \frac {8}{1 5} \rho \pi R ^ {5}$$

Radius $= R$

Mass $= M$

Density $= \rho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi R^3}$

Substituting the density expression gives

$$I = \frac {8}{1 5} \left[ \frac {M}{\frac {4}{3} \pi R ^ {3}} \right] \pi R ^ {5} = \frac {2}{5} M R ^ {2}$$

Solid cylinder of radius R, height H and mass M:

$$I = M R ^ {2} / 2$$

The expression for the moment of inertia of a solid cylinder can be built up from the moment of inertia of thin cylindrical shells. Using the general definition for moment of inertia:

$$I = \int_ {0} ^ {M} r ^ {2} d m$$

The mass element can be expressed in terms of an infinitesimal radial thickness dr by

$$d m = \rho d V = \rho L 2 \pi r d r$$

Substituting gives a polynomial form integral:

$$I = 2 \pi \rho L \int_{0}^{R} r^{3} dr = 2 \pi \rho L \frac{R^{4}}{4}$$ $$I = 2 \pi \left[ \frac{M}{\pi R^{2} L} \right] L \frac{R^{4}}{4} = \frac{1}{2} M R^{2}$$