Question

Two rotating bodies A and B of masses m and 2m 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 momentum respectively, then,

& A.\,{{L}_{A}}>{{L}_{B}} \\\ & B.\,{{L}_{A}}=\dfrac{{{L}_{B}}}{2} \\\ & C.\,{{L}_{A}}=2{{L}_{B}} \\\ & D.\,{{L}_{B}}>{{L}_{A}} \\\ \end{aligned}$$

Answer 1

Firstly we will consider the formulae of the kinetic energy of the rotating bodies and the angular momentum. As the kinetic energies are equal, thus, we will equate the same and represent the angular frequency in terms of the angular momentum and the moment of inertia.
Formula used:
$KE=\dfrac{1}{2}I{{\omega }^{2}}$
$L=I\omega$

Complete answer:
From the given information, we have the data as follows.
Two rotating bodies A and B of masses m and 2m with moments of inertia ${{I}_{A}}$and${{I}_{B}}({{I}_{A}}>{{I}_{B}})$ have the equal kinetic energy of rotation. ${{L}_{A}}$and ${{L}_{B}}$are their angular momentum respectively.
The kinetic energy of the rotating bodies is given as follows.
$KE=\dfrac{1}{2}I{{\omega }^{2}}$
Where I is the moment of inertia (rotational inertia) and w is the angular frequency.
Now, consider the kinetic energy of rotation for 2 bodies A and B.
$K{{E}_{A}}=\dfrac{1}{2}{{I}_{A}}{{\omega }_{A}}^{2}$and $K{{E}_{B}}=\dfrac{1}{2}{{I}_{B}}{{\omega }_{B}}^{2}$
The angular momentum of the rotating bodies is given as follows.
$L=I\omega$
Where I is the moment of inertia and w is the angular frequency.
Now, consider the angular momentum of rotation for 2 bodies A and B.
${{L}_{A}}={{I}_{A}}{{\omega }_{A}}$and ${{L}_{B}}={{I}_{B}}{{\omega }_{B}}$
Represent the above equations in terms of the angular frequency.
${{\omega }_{A}}=\dfrac{{{L}_{A}}}{{{I}_{A}}}$and ${{\omega }_{B}}=\dfrac{{{L}_{B}}}{{{I}_{B}}}$
The kinetic energy of rotation of these bodies are equal, thus, we have,

& \dfrac{1}{2}{{I}_{A}}{{\omega }_{A}}^{2}=\dfrac{1}{2}{{I}_{B}}{{\omega }_{B}}^{2} \\\ & \therefore {{I}_{A}}{{\omega }_{A}}^{2}={{I}_{B}}{{\omega }_{B}}^{2} \\\ \end{aligned}$$ Substitute the expressions of the angular frequencies in the above equations. $$\begin{aligned} & {{I}_{A}}{{\left( \dfrac{{{L}_{A}}}{{{I}_{A}}} \right)}^{2}}={{I}_{B}}{{\left( \dfrac{{{L}_{B}}}{{{I}_{B}}} \right)}^{2}} \\\ & \therefore \dfrac{{{L}_{A}}^{2}}{{{I}_{A}}}=\dfrac{{{L}_{B}}^{2}}{{{I}_{B}}} \\\ \end{aligned}$$ As $${{I}_{B}}>{{I}_{A}}$$ $$\begin{aligned} & L_{B}^{2}>L_{A}^{2} \\\ & \therefore {{L}_{B}}>{{L}_{A}} \\\ \end{aligned}$$ $$\therefore $$ The angular momentum of the rotating body B is greater than the angular momentum of the rotating body B. **Thus, option (D) is correct.** **Note:** Using the given condition, that is, the moment of inertia of the rotating body B is greater than the moment of inertia of the rotating body A, so, using the same, we have explained that, the angular momentum of the rotating body B is greater than that of rotating body A.