Question

Two bodies rotate with kinetic energies $E_1$ and $E_2$. Moment of inertia about their axis of rotation is $I_1$ and $I_2$. If $I_1 = \dfrac{I_2}{3}$ and $E_1 = 27E_2$, then the ratio of the angular momentum $L_1$ to $L_2$ is

• A. $3 : 1$
• B. $1 : 3$
• C. $1 : 9$
• D. $9 : 1$

Hint:

Use $E = \dfrac{L^2}{2I}$ to relate angular momentum with rotational energy.

Answer 1

The Correct Option is A

Step 1: Relation between kinetic energy and angular momentum.
Rotational kinetic energy is given by: \[ E = \dfrac{L^2}{2I} \] Step 2: Write ratio of energies.
\[ \dfrac{E_1}{E_2} = \dfrac{L_1^2 I_2}{L_2^2 I_1} \] Step 3: Substitute given values.
\[ 27 = \dfrac{L_1^2 I_2}{L_2^2 (I_2/3)} \] \[ 27 = \dfrac{3L_1^2}{L_2^2} \] Step 4: Solve for angular momentum ratio.
\[ \dfrac{L_1^2}{L_2^2} = 9 \] \[ \dfrac{L_1}{L_2} = 3 : 1 \]