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  • A circular hoop of mass and radius a rolls without slipping with constant angular speed <img alt="\ma

A circular hoop of mass \mathrm{M} and radius a rolls without slipping with constant angular speed \mathrm{\omega} along the horizontal x-axes in the \mathrm{x-y} plane. when the centre of the hoop is at a distance \mathrm{d=\sqrt{2} a} from the origin, the magnitude of the total angular momentum of the hoop about the origion is -
 

Option: 1

\mathrm{m a^2 \omega}
 


Option: 2

\mathrm{\sqrt{2} M a^2 \omega}
 


Option: 3

\mathrm{2 m a^2 \omega}
 


Option: 4

\mathrm{3 M a^2 \omega}


Answers (1)

best_answer

use,\mathrm{\overrightarrow{L} =\overrightarrow{L}_{C M}+M \overrightarrow{\gamma}_0 \times \overrightarrow{v}_0 }

            \mathrm{=I \omega+M r_0 v_0 \sin \theta }

            \mathrm{ =M a^2 \omega+M \sqrt{2 a} \cdot \omega a \times \sin 45^{\circ} }

            \mathrm{ =M a^2 \omega+M \sqrt{2} a \times \omega a \times \frac{1}{\sqrt{2}} }

\mathrm{ \overrightarrow{L} =2 M a^2 \omega }

Hence option 3 is correct.







 

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