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Moment of inertia of a sphere about its diameter is

Question:

The kinetic energy $\mathrm{K}$ of a rotating body depends on its moment of Inertia I and its angular speed $\omega$. Assuming the relation to be $\mathrm{K}=\mathrm{kl}^{\mathrm{a}} \omega^{\mathrm{b}}$, where $\mathrm{k}$ is a dimensionless constant, find $\mathrm{a}$ and $\mathrm{b}$. Moment of inertia of a sphere about its diameter is $\frac{2}{5} M r^{2}$

Solution:

$\mathrm{K}=\mathrm{kl}^{\mathrm{a}} \mathrm{\omega}^{\mathrm{b}}$, where $\mathrm{k}=$ constant and $\mathrm{K}=$ Kinetic energy

So, $\mathrm{K}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]$

$\mathrm{l}^{\mathrm{a}}=\left[\mathrm{ML}^{2}\right]^{\mathrm{a}}$ and $\omega^{\mathrm{b}}=\left[\mathrm{T}^{-1}\right]^{\mathrm{b}}$

So, $\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{2 \mathrm{a}} \mathrm{T}^{-\mathrm{b}}\right]$

So, $a=1$ and $b=2$

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