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The angular momentum of a rotating body is increased by 20 %. What will be the increase in its rotational kinetic energy?

Answer»

Solution :KINETIC ENERGY K.E=`(L^(2))/(2L),E ALPHA L^(2),(E_(1))/(E_(2))=(L^(2))/(L_(2)^(2)),L_(1)=L,L(2)=L+(20L)/(100)=(120L)/(100)=(120L)/(100)=(12L)/(10)`
`therefore (E_(1))/(E_(2))=L^(2)/((12L)/(10))=(L^(2)xx100)/(144L^(2)),(E_(1))/(E_(2))=(100)/(144)therefore(E_(2))/(E_(1))=(144)/(100)`
`(E_(2)-E_(1))/(E-(1))=(144-100)/(100)=(44)/(100)therefore `% increase in rotaional kinetic energy =44 %


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