The rotational kinetic energy of a body is E. In the absence of external torque, the mass of the body is halved and its radius of gyration is doubled.
The rotational kinetic energy of a body is E. In the absence of external torque, the mass of the body is halved and its radius of gyration is doubled. Its rotational kinetic energy is
A. 2 E
B. E/2
C. E
D. E/4
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Correct Answer - B
`I_(1)omega_(1)=I_(2)omega_(2)` if no ecternal torqe acts on the system total angular momentum remains constant.
`I_(1)=m_(1)K_(1)^(2) " " I_(2)=m_(2)K_(2)^(2)`
`(I_(1))/(I_(2))=(m_(1))/(m_(2))(K_(1)^(2))/(K_(2)^(2))=2xx(1)/(4)=(1)/(2)`
`therefore I_(2)=2I_(1)`
`therefore omega_(2)=(I_(1)omega_(1))/(I_(2))=(I_(1)omega_(1))/(2I_(2))=(omega_(1))/(2)`
`(KE_(1))/(KE_(2))=(I_(1))/(I_(2))xx((omega_(2))/(omega_(1)))^(2)`
`=(1)/(2)xx4=2`
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