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  • A spool of inner radius '  ' and outer radius ' <img alt="2 m" src="ht

A spool of inner radius ' 1 \mathrm{~m} ' and outer radius ' 2 m ' is lying on an inclined plane of inclination angle ' \theta ' as shown in the figure. A thread of negligible mass is wound on the spool and its other end is attached to an object of mass ' 1 \mathrm{~kg} '. If the mass of the spool is ' 2 \mathrm{~kg} ', then the angle for which the centre of gravity will remain at rest will be.

Option: 1

\sin ^{-1}\left(\frac{2}{3}\right)


Option: 2

\cos ^{-1}\left(\frac{2}{3}\right)


Option: 3

\sin ^{-1}\left(\frac{4}{9}\right)


Option: 4

\cos ^{-1}\left(\frac{4}{9}\right)


Answers (1)

best_answer

Applying the conservation of energy (Assuming, mass of the spool =M, mass of object =m, inner radius =r, outer radius =R

\begin{gathered} \frac{1}{2} m v^2+\frac{M v^2}{2} \frac{R^2}{r^2}=m g h \\ v=\sqrt{\frac{2 m g h}{m+M \frac{R^2}{r^2}}} \end{gathered}

[Moving object in the downward direction by 'h']

T=M g \sin \theta=m(g-a)

where, 

        a=\frac{m g}{m+\frac{M R^2}{r^2}}

      \begin{aligned} & \sin \theta=\frac{1}{\frac{M}{m}+\frac{r^2}{R^2}}=\frac{1}{\left(\frac{2}{1}\right)+\left(\frac{1}{4}\right)}=\frac{4}{9} \\ & \theta=\sin ^{-1}\left(\frac{4}{9}\right) \end{aligned}

Posted by

Nehul

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