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A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 100 rpm. A bob of wax of mass 20 g falls on the disc and sticks to it a distance of 5 cm from the axis. If the moment of inertia of the disc about the given axis is 2xx10^(-4)kgm^(2), find new frequency of rotation of the disc.

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> :`I_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)` Momnet of inertia of the disc `=2xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)kgm^(2)` <br/> `I_(2)` = <a href="https://interviewquestions.tuteehub.com/tag/moment-25786" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENT">MOMENT</a> of inertia of the disc+Moment of inertia of the bob of wax on the dise <br/> `=2xx10^(-4)+mr^(2)=2xx10^(-4)+20xx10^(-3)(0.05)^(2)` <br/> `=2xx10^(-4)+0.5xx10^(-4)=2.5xx10^(-4)kgm^(2)` <br/> `(n//t)_(1)=100" rpm" ,(n//t_(2))=?` <br/> By the principle of conservation of angular momentum, <br/> `I_(1)omega_(1)=I_(2)omega_(2)""I_(1)2pi((n)/(t))_(1)=I_(2)2pi((n)/(t))_(2)` <br/> `2xx10^(-4)xx100=2.5xx10^(-4)((n)/(t))_(2)` <br/> `((n)/(t))_(2)=(100xx2)/(2.5)=80" rpm"`</body></html>


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