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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

A ring of mass \(M\) and radius \(R\) is rotating about its axis with angular velocity \(\omega \). Two identical bodies each of mass \(m\) are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be

  1. A \(\frac{{m\left( {M + 2m} \right)}}{M}\,{\omega ^2}{R^2}\)
  2. B \(\frac{{Mm}}{{\left( {M + m} \right)}}\,{\omega ^2}{R^2}\)
  3. C \(\frac{{Mm}}{{\left( {M + 2m} \right)}}\,{\omega ^2}{R^2}\)
  4. D \(\frac{{\left( {M + m} \right)M}}{{\left( {M + 2m} \right)}}\,{\omega ^2}{R^2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{{Mm}}{{\left( {M + 2m} \right)}}\,{\omega ^2}{R^2}\)

Step-by-step Solution

Detailed explanation

\begin{array}{l} Kinetic\,energ{y_{\left( {rotational} \right)}}{K_R} = \frac{1}{2}I{\omega ^2}\\ kinetic\,energ{y_{(translational)}}{K_r} = \frac{1}{2}M{v^2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\…

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