The moments of inertia of two freely rotating bodies A and B are IA and IB respectively. IA IB and their angular momenta are equal. If Ka and Kb are their kinetic energies, then

Question

The moments of inertia of two freely rotating bodies A and B are IA and IB respectively. IA > IB and their angular momenta are equal. If Ka and Kb are their kinetic energies, then (A)  KA = KB  (B)  KA ≠ KB  (C)  KA < KB  (D)  KA > 2KB

Tardigrade Admin · Accepted Answer

Kinetic energy E=(L2/2 I) If angular momenta are equal, then E propto (1/I) Kinetic energy E=K (given in problem) If IA>IB then KA< KB

Q. The moments of inertia of two freely rotating bodies $A$ and $B$ are $I_{A}$ and $I_{B}$ respectively. $I_{A} > I_{B}$ and their angular momenta are equal. If $K_{a}$ and $K_{b}$ are their kinetic energies, then

$K_{A} = K_{B}$

$K_{A} \neq = K_{B}$

$K_{A} < K_{B}$

$K_{A} > 2 K_{B}$

Kinetic energy $E = \frac{L ^{2}}{2 I}$
If angular momenta are equal, then $E \propto \frac{1}{I}$
Kinetic energy $E = K$ (given in problem)
If $I_{A} > I_{B}$ then $K_{A} < K_{B}$

Q. The moments of inertia of two freely rotating bodies A and B are IA​ and IB​ respectively. IA​>IB​ and their angular momenta are equal. If Ka​ and Kb​ are their kinetic energies, then

KA​=KB​

KA​=KB​

KA​<KB​

KA​>2KB​