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.

The angular momentum of a body changes by 60 kg m^(2)s^(-1) ,when its

1.	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» Solution :`omega_(1)=10 rd s^(-1) ,omega_(2)=30 rad s^(-1),Delta L=60 kgm^(2)s^(-1)`=change in angular momentum change in angular VELOCITY `=Deltaomega=omega_(2)omega_(1)=30-10=20 rads^(-1)` `I=(DELTAL)/(Deltaomega)=3kgm^(2),L_(1)Iomega_(1)=30 kg^(2)s^(-1),L_(2)=Iomega_(2)=90 kg m^(2)s^(-1)` Chanhe in K.E.=`(1)/(2)(L_(1)+L_(2))Deltaomega=(1)/(2)(30+90)xx20=1200J`