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Find the moment of inertia of thin and massless rod about an axis passing through its centre of mass of rod and pair of mass is suspended on both end of this rod.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :A system of a ligh rod of length l with a pair of small masses is shown as in figure. <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> mass of system is M. <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P1_C07_E01_046_S01.png" width="80%"/> <br/> System is rotating about an axis through the centre of mass of system and perpendicular to the rod. <br/> <a href="https://interviewquestions.tuteehub.com/tag/suppose-656311" style="font-weight:bold;" target="_blank" title="Click to know more about SUPPOSE">SUPPOSE</a> C is the centre of mass. So distance of small mass from its ends is `(l)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`. <br/> <a href="https://interviewquestions.tuteehub.com/tag/momentum-1100588" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENTUM">MOMENTUM</a> of inertia of masses about mention axis is `((M)/(2))((l)/(2))^(2)` : <br/> `therefore` Total moment of inertia of a system <br/> `I=((M)/(2))((l)/(2))^(2)+((M)/(2))((l)/(2))^(2)` <br/> `=2((M)/(2))((l)/(2))^(2)` <br/> `=(Ml^(2))/(4)`</body></html>


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