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A uniform disk of radius r= 0.6 m and mass M = 2.5 kg is freely suspended from a horizontal pivot located a radial distance d = 0.30 m from its centre . Find the angular frequency of small amplitude oscillations of the disk .

Answer» <html><body><p></p>Solution :The <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 <a href="https://interviewquestions.tuteehub.com/tag/disk-955870" style="font-weight:bold;" target="_blank" title="Click to know more about DISK">DISK</a> about a <a href="https://interviewquestions.tuteehub.com/tag/perpendicular-598789" style="font-weight:bold;" target="_blank" title="Click to know more about PERPENDICULAR">PERPENDICULAR</a> axis <a href="https://interviewquestions.tuteehub.com/tag/passing-1148520" style="font-weight:bold;" target="_blank" title="Click to know more about PASSING">PASSING</a> through its centre is `I=1/2mr^(2)`<br/> Form the <a href="https://interviewquestions.tuteehub.com/tag/parallel-1146369" style="font-weight:bold;" target="_blank" title="Click to know more about PARALLEL">PARALLEL</a> axis theorem , the moment of inertia of the disk about the pivot point is <br/> `I'=I+Md^(2)=(2.5xx0.6xx0.6)/(2)+2.5xx0.30xx0.30`<br/> `=(0.9)/(2)+0.225` <br/> `=0.45+0.225` <br/> `=0.675kgm^(2)`<br/> The angular frequency of small amplitude oscillations of a compound pendulum is given by <br/> `omega =sqrt((Mgd)/(I'))=sqrt((2.5xx9.8xx0.30)/(0.675))` <br/> `=sqrt((7.35)/(0.675))=sqrt((0.675)/(10.9))=3.3"rad"/s`</body></html>


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