Angular acceleration of a body of mass 50 kg under the action of a torque of magnitude 500 Nm is 25 `rad//s^(2)`. Then radius of gyration of the body
Angular acceleration of a body of mass 50 kg under the action of a torque of magnitude 500 Nm is 25 `rad//s^(2)`. Then radius of gyration of the body about its axis of rotation is nearly equal to
A. 0.2 m
B. 0.6 m
C. 0.4 m
D. 1 m
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Correct Answer - B
`tau = I prop`
`I = (tau)/(prop)=(500)/(25)=20`
`K = sqrt((I)/(M))=sqrt((20)/(50))=sqrt(0.4) = 0.6 m`.
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