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# Two bodies with moments of inertia ${I_1}$ and ${I_2}$$\left( {{I_1} > {I_2}} \right)$ have equal angular momentum. If their kinetic energies of rotation areand, respectively, then:A) ${E_1} > {E_2}$B) ${E_1} < {E_2}$C) ${E_1} = {E_2}$D) ${E_1} = 2{E_2}$

Last updated date: 13th Jun 2024
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Hint: In question relation between moment of inertia of two bodies is given. Also given that their angular momentums are equal. Then, the relation between kinetic energy can be calculated.
We use angular momentum formula to find relation between angular speeds of two bodies. After that we use the kinetic energy formula from the relation between angular speeds.

Complete step by step solution:
Given: ${I_1}$ and ${I_2}$ are moments of inertia of two bodies.
In question given that both moments of inertia have following relation
${I_1} > {I_2}$
Let ${\omega _1}$ and ${\omega _2}$ are two angular velocities.
Angular momentum of first body, ${L_1} = {I_1}{\omega _1}$
Angular momentum of second body,${L_2} = {I_2}{\omega _2}$
According to the question, angular momentums are the same for two bodies.
Numerically, we can write as ${L_1} = {L_2}$
$\Rightarrow {I_1}{\omega _1} = {I_2}{\omega _2}$
$\therefore \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{\omega _2}}}{{{\omega _1}}}$
$\because {I_1} > {I_2}$
$\therefore \dfrac{{{I_1}}}{{{I_2}}} > 1$
Which gives $\dfrac{{{\omega _2}}}{{{\omega _1}}} > 1$
$\Rightarrow {\omega _2} > {\omega _1}$
Rotational kinetic energy is given by, $K{E_{Rotational}} = \dfrac{1}{2}I{\omega ^2} = E$
Rotational kinetic energy for first body is given by, ${E_1} = \dfrac{1}{2}{I_1}\omega _1^2$
Similarly rotational kinetic energy for first body is given by, ${E_2} = \dfrac{1}{2}{I_2}\omega _2^2$
Dividing ${E_1}$ by ${E_2}$, we get
$\Rightarrow$ $\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{\dfrac{1}{2}{I_1}\omega _1^2}}{{\dfrac{1}{2}{I_2}\omega _2^2}}$
$\Rightarrow$ $\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_1}\omega _1^2}}{{{I_2}\omega _2^2}}$
$\Rightarrow \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_1}}}{{{I_2}}} \times \dfrac{{\omega _1^2}}{{\omega _2^2}}$
$\Rightarrow$ $\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_1}}}{{{I_2}}} \times {\left( {\dfrac{{{\omega _1}}}{{{\omega _2}}}} \right)^2}$
$\Rightarrow \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_1}}}{{{I_2}}} \times {\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)^2}$ $\left[ {\because \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{\omega _2}}}{{{\omega _1}}}} \right]$
$\therefore \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_2}}}{{{I_1}}}$
After rearranging LHS, we get
$\Rightarrow$ $\dfrac{{{E_2}}}{{{E_1}}} = \dfrac{{{I_1}}}{{{I_2}}}$
$\Rightarrow \dfrac{{{E_2}}}{{{E_1}}} > 1$ $\left[ {\because we{\text{ }}have{\text{ }}calculated{\text{ }}as{{}}\dfrac{{{I_1}}}{{{I_2}}} > 1} \right]$
Hence, we get ${E_2} > {E_1}$ or we can write${E_1} < {E_2}$.

Hence, the correct option is B.

Additional information: Moment of inertia is defined as the quantitative measure of the rotational inertia. It can be expressed as after applying by a torque or turning force, it is known as an opposition that the body exhibits to having its speed of rotation about an axis. In a moment of inertia. the axis may be internal or external. Axis may or may not be fixed. Other names of moments of inertia are rotational inertia, the mass moment of inertia, angular mass.
The rotational kinetic energy of a rotating body is expressed in terms of the moment of inertia and angular velocity. The total kinetic energy of a rotating object can be given by the sum of the translational kinetic energy and the rotational kinetic energy.

Note: Students must be careful about linear kinetic energy and rotational kinetic energy. Both have different meanings. In linear kinetic energy, as the name shows the object is moving in a straight line (or in other way moving linearly). But in case of rotational kinetic energy, torque is the main reason for motion. In case of linear kinetic energy motion occurs due to force.