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The angular momentum of a body changes by 60 kg m^(2)s^(-1) ,when its angular velocity changges from 10 rads^(-1) rad s^(-1) .Find the cange in its kinetic energy of rotation.

Answer» <html><body><p></p>Solution :`omega_(1)=10 rd s^(-1) ,omega_(2)=<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a> rad s^(-1),Delta L=60 kgm^(2)s^(-1)`=change in angular momentum <br/>change in angular <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> `=Deltaomega=omega_(2)omega_(1)=30-10=20 rads^(-1)` <br/>`I=(<a href="https://interviewquestions.tuteehub.com/tag/deltal-2053528" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAL">DELTAL</a>)/(Deltaomega)=3kgm^(2),L_(1)Iomega_(1)=30 kg^(2)s^(-1),L_(2)=Iomega_(2)=90 kg m^(2)s^(-1)` <br/>Chanhe in K.E.=`(1)/(2)(L_(1)+L_(2))Deltaomega=(1)/(2)(30+90)xx20=1200J`</body></html>


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