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Two bodies A and B having same angular m...

Two bodies A and B having same angular momentum and `I_(A) gt I_(B)`, then the relation between `(K.E.)_(A) and (K.E.)_(B) ` will be :

A

`(K.E.)_(A) gt (K.E.)_(B)`

B

`(K.E.)_(A)=(K.E.)_(B)`

C

`(K.E.)_(A) lt (K.E.)_(B)`

D

`(K.E.)_(A)""^(1)(K.E.)_(B)`

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To solve the problem, we need to analyze the relationship between the kinetic energies of two bodies A and B, given that they have the same angular momentum and that the moment of inertia of A is greater than that of B. ### Step-by-Step Solution: 1. **Understanding Angular Momentum**: The angular momentum (L) of a rotating body is given by the formula: \[ L = I \omega \] where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. 2. **Given Conditions**: We are given that: - \(L_A = L_B\) (Angular momentum of A is equal to angular momentum of B) - \(I_A > I_B\) (Moment of inertia of A is greater than that of B) 3. **Expressing Angular Velocities**: Since the angular momenta are equal, we can express the angular velocities in terms of the moments of inertia: \[ I_A \omega_A = I_B \omega_B \] Rearranging gives us: \[ \omega_A = \frac{L_A}{I_A} \quad \text{and} \quad \omega_B = \frac{L_B}{I_B} \] Since \(L_A = L_B\), we can express \(\omega_A\) and \(\omega_B\) as: \[ \omega_A = \frac{L}{I_A} \quad \text{and} \quad \omega_B = \frac{L}{I_B} \] 4. **Kinetic Energy Formula**: The kinetic energy (K.E.) of a rotating body is given by: \[ K.E. = \frac{1}{2} I \omega^2 \] Thus, for bodies A and B, we have: \[ K.E._A = \frac{1}{2} I_A \omega_A^2 \quad \text{and} \quad K.E._B = \frac{1}{2} I_B \omega_B^2 \] 5. **Substituting Angular Velocities**: Substituting the expressions for \(\omega_A\) and \(\omega_B\) into the kinetic energy formulas: \[ K.E._A = \frac{1}{2} I_A \left(\frac{L}{I_A}\right)^2 = \frac{1}{2} \frac{L^2}{I_A} \] \[ K.E._B = \frac{1}{2} I_B \left(\frac{L}{I_B}\right)^2 = \frac{1}{2} \frac{L^2}{I_B} \] 6. **Comparing Kinetic Energies**: Now we can compare \(K.E._A\) and \(K.E._B\): \[ \frac{K.E._A}{K.E._B} = \frac{\frac{1}{2} \frac{L^2}{I_A}}{\frac{1}{2} \frac{L^2}{I_B}} = \frac{I_B}{I_A} \] Since \(I_A > I_B\), it follows that: \[ \frac{K.E._A}{K.E._B} < 1 \quad \Rightarrow \quad K.E._A < K.E._B \] ### Conclusion: Thus, the relation between the kinetic energies of the two bodies is: \[ K.E._A < K.E._B \]
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