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JIPMER 2013
Exam
JIPMER
Year
2013
Subject
Physics
Chapter
Rotational Motion

JIPMER 2013 · Physics · Rotational Motion

Two rigid bodies A and B rotate with rotational kinetic energies...

Two rigid bodies AandBA and B rotate with rotational kinetic energies EAandEBE_A and E_B respectively. The moments of inertia of AandBA and B about the axis of rotation are IAandIBI_A and I_B respectively. If IA=IB4I_A=\frac{I_B}{4} and E_A=100=E_B, the ratio of angular momentum \left(L_A\right)ofAtotheangularmomentum(LB)of A to the angular momentum \left(L_B\right) of B is

Options

  1. A25
  2. B5/45 / 4
  3. C5
  4. D1/41 / 4

Correct answer

C. 5

Step-by-step solution

From, the formula, rotational KE Erotational =12lω2=L22∣(∵L=∣ω)E_{\text{rotational }}=\frac{1}{2} l \omega^2=\frac{L^2}{2 \mid}(\because L=\mid \omega) Therefore, L2=2El\quad L^2=2 E l Hence LA2=2EAIA…(i)L_A^2=2 E_A I_A\ldots(i) and LB2=2EBIB…(ii)L_B^2=2 E_B I_B\ldots(ii) From Eq. (i) and (ii) LALB=2EAIB2EBIA=1004=5(∵EBEA=100 and  IAIB=4)\begin{aligned} & \frac{L_A}{L_B}=\sqrt{\frac{2 E_A I_B}{2 E_B I_A}}=\sqrt{\frac{100}{4}}=5 \\ & \quad\left(\because \frac{E_B}{E_A}=100 \text{\text{ and } } \frac{I_A}{I_B}=4\right)\end{aligned}
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