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A disc of mass \mathrm{m} and radius \mathrm{R} moves in the \mathrm{x-y} plane as shown in figure. The angular momentum of the disc about the origin\mathrm{ ' O '}at the instant shown is. -


 

Option: 1

\mathrm{\frac{-5}{2} m R^2 \omega \hat{k}}
 


Option: 2

\mathrm{\frac{7}{2} m R^2 m \hat{k}}
 


Option: 3

\mathrm{\frac{-9}{2} m R^2 \omega \hat{k}}
 


Option: 4

\mathrm{\quad \frac{5}{2} m R^2 \omega \hat{k}}


Answers (1)

best_answer

\mathrm{\overrightarrow{L} =m\left(\overrightarrow{r} \times \overrightarrow{v}_{c m}\right)+I_{c m} \overrightarrow{\omega} }
\mathrm{\overrightarrow{L} =-m vr \sin \theta(\hat{k})+\frac{m R^2}{2} \omega \hat{k} }

\mathrm{\overrightarrow{L}=-m vr\times \frac{3 R}{r} \hat{k}+\frac{m R^2}{2} \omega \hat{k} }

\mathrm{ L=-m(\omega R) 3 R \hat{k}+\frac{m R^2}{2} \omega \hat{k} }

\mathrm{\overrightarrow{L}=\frac{-5}{2} m R^2 \omega \hat{k}}

Hence option 1 is correct.


 

Posted by

Suraj Bhandari

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