Question

Angular momentum $\left( L \right) = mvr$
Moment of inertia $\left( I \right) = m{r^2}$
(m: mass, v: velocity, r: radius)
Then energy is equal to
A. $\dfrac{{{L^2}}}{{2{I^2}}}$
B. $2{L^2}I$
C. $\dfrac{{{L^2}}}{{2I}}$
D. $\dfrac{{2I}}{{{L^2}}}$

Answer 1

In the problem, formulae for angular momentum and moment of inertia are given. Angular momentum is the quantity of angular motion of the particle. Also, angular momentum can be written in terms of angular velocity. This means that the object is performing rotational motion. The energy thus obtained is the rotational kinetic energy.

Formula Used:
The rotational kinetic energy of any object is given as
${K_r} = \dfrac{1}{2}I{\omega ^2}$
where, ${K_r}$ is the rotational kinetic energy of the given object, $I$ is the moment of inertia and $\omega$ is the angular velocity.

Complete step by step answer:
For an object of mass $m$rotating with a constant speed $v$ in a circle of radius $r$, the angular momentum $L$ is given as $L = mvr$. This can also be written in terms of angular velocity as
$L = m{r^2}\omega$ $\to \left( 1 \right)$
where, $\omega$ is the angular velocity.
The rotational kinetic energy of any object is given as
${K_r} = \dfrac{1}{2}I{\omega ^2}$
In the problem, the moment of inertia is given as $I = m{r^2}$.Therefore, is the rotational kinetic energy will be,
${K_r} = \dfrac{1}{2}\left( {m{r^2}} \right){\omega ^2}$
This can be also written as,
${K_r} = \dfrac{1}{2}\left( {m{r^2}\omega } \right)\omega$
From, equation (1), $L = m{r^2}\omega$
Therefore, ${K_r} = \dfrac{1}{2}\left( L \right)\omega$ $\to \left( 2 \right)$
Multiplying both sides of equation (2) by $L$

\Rightarrow{K_r}= \dfrac{{{L^2}\omega }}{{2L}}$$ According, to equation (1) $$L$$can be written as $$m{r^2}\omega $$ (This substitution is only done in the denominator) $${K_r} = \dfrac{{{L^2}\omega }}{{2L}} \\\ \Rightarrow{K_r} = \dfrac{{{L^2}\omega }}{{2m{r^2}\omega }} \\\ \Rightarrow{K_r} = \dfrac{{{L^2}}}{{2m{r^2}}}$$ But, $$I = m{r^2}$$ Therefore, $${K_r} = \dfrac{{{L^2}}}{{2m{r^2}}} = \dfrac{{{L^2}}}{{2I}}$$ **Hence, option C is the correct answer.** **Note:** The angular momentum of an object performs rotational motion. If the object is in motion, the Kinetic energy is considered. And if the motion is rotary, the energy thus obtained is called the Rotational Kinetic Energy.According to the conservation of angular momentum, in the absence of external torque, the angular momentum of the body remains constant. Therefore, $$L = m{r^2}\omega = {\text{constant}}$$