Explanation
The angular velocity is given as,
$$\omega = \dfrac{{\Delta {\rm{\theta }}}}{{\Delta t}}$$
$$\omega = \dfrac{{2\pi }}{{1\;{\rm{day}}}}$$
$$\omega = \dfrac{{2\pi }}{{24 \times 60 \times 60}}\;{\rm{rad/sec}}$$
Thus, the angular velocity of earth about its axis of rotation is $$\dfrac{{2\pi }}{{24 \times 60 \times 60}}\;{\rm{rad/sec}}$$.
The moment of inertia is given as,
$$I = \dfrac{2}{5}M{R^2}$$
$$I = \dfrac{2}{5} \times \dfrac{4}{3}\pi {R^3} \times \rho \times {R^2}$$
$$I \propto {R^5}$$
$$\dfrac{{{I_1}}}{{{I_2}}} = {\left( {\dfrac{{{R_1}}}{{{R_2}}}} \right)^5}$$
The ratio of moments of inertia about their diameters is given as,
$$\dfrac{{{I_1}}}{{{I_2}}} = {\left( {\dfrac{2}{1}} \right)^5}$$
$$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{32}}{1}$$
Moment of inertia of metre stick about its centre is given by:
$$I = \dfrac{{M{L^2}}}{{12}}$$
$$I = \dfrac{{0.4 \times {1^2}}}{{12}}$$
Now moment of inertia about axis perpendicular to length of stick at 20 cm mark
$$I = \dfrac{{0.1}}{3} + 0.4 \times {\left( {\dfrac{3}{{10}}} \right)^2}$$
$$I = \dfrac{{0.1}}{3} + \dfrac{{3.6}}{{100}}$$
$$I = \dfrac{{10 + 10.8}}{{300}}$$
$$I = \dfrac{{20.8}}{{300}}$$
$$I = 0.069\;{\rm{kg}}{{\rm{m}}^{\rm{2}}}$$
$$I = 6.9 \times {10^5}\;{\rm{gc}}{{\rm{m}}^{\rm{2}}}$$
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