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Rotational Mechanics — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsRotational Mechanics1 Mark
Question
A thin circular ring of mass M and radius r is rotating about its axis with an angular speed $\omega.$ Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become:

  1. $\frac{\omega\text{M}}{\text{M}+\text{m}}$

  2. $\frac{\omega\text{M}}{\text{M}+\text{2m}}$

  3. $\frac{\omega(\text{M}-\text{2m})}{\text{M}+\text{2m}}$

  4. $\frac{\omega(\text{M}+\text{2m})}{\text{M}}$

✓

Answer

  1. $\frac{\omega\text{M}}{\text{M}+\text{2m}}$

Explanation:

No external torque is applied on the ring; therefore, the angular momentum will be conserved.

$\text{I}\omega=\text{I}'\omega'$

$\Rightarrow\omega'=\frac{\text{I}\omega}{\Gamma}\ \dots(\text{i})$

$\text{I}=\text{Mr}^2$

$\text{I}'=\text{Mr}^2+\text{2mr}^2$

On putting these values in equation (i), we get:

$\omega'=\frac{\omega\text{M}}{\text{M}+\text{2m}}$

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