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System Of Particles And Rotational Motion

Question

Two rigid bodies A and B rotate with rotational kinetic energies E, and E, respectively. The moments of inertia of A and B about the axis of rotation are I_A and I_B respectively.
If $I_{A} = \frac{I_{B}}{2}$ and E_A = 100 = E_B, the ratio of angular momentum (L_A ) of A to the angular momentum ( L_B ) of B is

25

5/4

5

1/4

Solution

From the formula

$E_{rotational} = \frac{1}{2} I ω^{2}$

= $\frac{L^{2}}{2 I}$ $(∵ L = Iω)$

Therefore, L² = 2 EI

Hence $L_{A}^{2} = 2 E_{A} I_{A}$ .....(i)

and $L_{B}^{2} = 2 E_{B} I_{B}$ ......(ii)

From equatin (i) and (ii)

$\frac{L_{A}}{L_{B}} = \sqrt{\frac{2 E_{A} I_{A}}{2 E_{B} I_{B}}}$

= $\sqrt{\frac{100}{4}}$

$\frac{L_{A}}{L_{B}} = 5$

$(∵ \frac{E_{B}}{E_{A}} = 100 and \frac{I_{A}}{I_{B}} = 4)$