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Perpendicular Axis Theorem

The moments o...

Question

The moments of inertia of two freely rotating bodies $A$ and $B$ are $I_{A}$ and $I_{B}$ respectively( $I_{A} > I_{B}$ ) and their angular momenta are equal. If $K_{A}$ and $K_{B}$ are their kinetic energies, then

$K_{A} = K_{B}$

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$K_{A} \neq K_{B}$

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$K_{A} < K_{B}$

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$K_{A} = 2 K_{B}$

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Solution

The correct option is C
$K_{A} < K_{B}$

Step 1: Given data
The moments of inertia of two freely rotating bodies $A$ and $B$ are $I_{A}$ and $I_{B}$ respectively and $I_{A} > I_{B}$
Their angular momenta are equal
Step 2: Formula used and calculation
We know, that the relation between kinetic energy, angular momentum, and moment of inertia is
$K = \frac{L^{2}}{2 I}$
Since it is given that angular momentum is the same for both the bodies, we can write
$K \propto \frac{1}{I}$
i.e., kinetic energy is inversely proportional to the moment of inertia.
Since $I_{A} > I_{B}$ , we get $K_{A} < K_{B}$ is correct
Since $K_{A} < K_{B}$ we can say $K_{A} \neq K_{B}$ is correct
Options B and C is correct

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