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 momenta respectively- thenL-A - L-BL-A - 2L-BL-B - L-AL-A - dfrac-L-B-2-

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Toppr Admin · Accepted Answer

The kinetic energy of rotation is given by 12I-x3C9-2-Since the kinetic energy of rotation is same-xA0-IA-x3C9-2A-IB-x3C9-2B-x27F9-x3C9-A-x3C9-B-x221A-IBIAThe ratio of the angular momentum becomes-LALB-IA-x3C9-AIB-x3C9-B-x221A-IAIB-lt-1-x27F9-LB-gt-LA

Two rotating bodies A and B of masses m and 2m with moments of inertia IA and IB(IB>IA) have equal kinetic energy of rotation. If LA and LB be their angular momenta respectively, thenLA>LBLA=2LBLB>LALA=LB2

The kinetic energy of rotation is given by 12Iω2.Since the kinetic energy of rotation is same, IAω2A=IBω2B⟹ωAωB=√IBIAThe ratio of the angular momentum becomes,LALB=IAωAIBωB=√IAIB<1⟹LB>LA

Two rotating bodies $A$ and $B$ of masses $m$ and $2 m$ 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 momenta respectively, then
$L_{A} > L_{B}$
$L_{A} = 2 L_{B}$
$L_{B} > L_{A}$
$L_{A} = \frac{L_{B}}{2}$