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Two rotating bodies $A$ and $B$ of masses $m$ and $2\,m$ with moments of inertia $I_A$ and $I_B (I_B> I_A)$ have equal kinetic energy of rotation. If $L_A$ and $L_B$ be their angular momenta respectively, then
$L_B>L_A$
$L_A>L_B$
$L_A=$$\frac{{{L_B}}}{2}$
$L_A=2L_B$
Solution
Here, ${m_A} = m,{m_B} = 2m$
Both bodies $A$ and $B$ have equal kinetic energy of rotation
${k_A} = {k_B} \Rightarrow \frac{1}{2}{I_A}\omega _A^2 = \frac{1}{2}{I_B}\omega _B^2$
$ \Rightarrow \frac{{\omega _A^2}}{{\omega _B^2}} = \frac{{{I_B}}}{{{I_A}}}\,…\left( i \right)$
Ratio of angular momenta,
$\frac{{{L_A}}}{{{L_B}}} = \frac{{{I_A}{\omega _A}}}{{{I_B}{\omega _B}}} = \frac{{{I_A}}}{{{I_B}}} \times \sqrt {\frac{{{I_B}}}{{{I_A}}}}$ $……….(i)$
$= \sqrt {\frac{{{I_A}}}{{{I_B}}}} < 1 ( {{I_B} > {I_A}}$
${L_B}>{L_A}$
