JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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If '

I

' is the moment of inertia and '

L

' is angular momentum of a rotating body, then

\frac{L^2}{2 I}

is its

View answer and explanation

Correct option: B

rotational kinetic energy

Option B (rotational kinetic energy) is correct because the formula for rotational kinetic energy is

\frac{{1}}{{2}}I\omega^2

, and

L = I\omega

, so

\frac{{L^2}}{{2I}} = \frac{{1}}{{2}}I\omega^2

More Questions from SYSTEM OF PARTICLES

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\mathrm{AB}

\mathrm{L}

\mu(x)=a+\frac{b x}{L}

x

\mathrm{A}

\mathrm{CM}

\left(\frac{7}{12}\right) \mathrm{L}

\mathrm{A}

\mathrm{b}

1.0 \mathrm{~g}

8 \mathrm{~cm}

\omega = {t^4} + 2{t^3}

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)