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Two particles of masses \mathrm{m} and \mathrm{M(m< <<M)} are separated a distance \mathrm{1} . The system is to be rotated to get an angular velocity \mathrm{\omega}, about an axis perpendicular to the line joining the particles. The position of axes, if work done in the process is minimum.

Option: 1

\mathrm{x=\left[\frac{m r}{m+m}\right]}


Option: 2

\mathrm{x=\left[\frac{m r}{M}\right]}


Option: 3

\mathrm{x=\left[\frac{m r}{m-m}\right]}


Option: 4

\mathrm{x=0}


Answers (1)

best_answer

\mathrm{MI} of the system about axis shown - 
     
\mathrm{ I=m x^2+m(r-x)^2 }
\mathrm{ \omega=k=\frac{1}{2}\ I\ \omega^2 }
\mathrm{ =\frac{1}{2}\left[M x^2+m(r-x)^2\right] \omega^2 }
For minimum work, \mathrm{ =\frac{dw}{dx}=0 }   
\mathrm{ \frac{d}{d x}\left[\frac{1}{2}\left\{M x^2+m(r-x)^2\right\}\right]=0 }
or,
\mathrm{ \\ \begin{array}{r} M \times 2 n-2 m(r-x)=0 \\\\ x=\left[\frac{m r}{m+M}\right] \end{array} }
As \mathrm{m < <<m}, so \mathrm{x \rightarrow 0} so system is to be rotated about \mathrm{M}.

Posted by

Gautam harsolia

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