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Physics - Rotation Question with Solution | TestHub

PhysicsRotationAngular momentumEasy2 minQB

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A thin circular metal disc of the radius $500 mm$ is set rotating about a central axis normal to its plane. Upon raising in temperature gradually, the radius increases to $507.5 mm$ . The percentage change in the rotational kinetic energy will be

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Answer:

Solution:

Given $\frac{dR}{R} \times 100 = \frac{(507.5 - 500.0)}{500.0} \times 100 = 1.5%$

Now, $K = \frac{1}{2} I ω^{2} = \frac{1}{2 I} {(I ω)}^{2} = \frac{L^{2}}{2 I}$

or, $K = \frac{L^{2}}{2 (\frac{{MR}^{2}}{2})} = \frac{L^{2}}{M (R^{2})}$

Taking log and differentiating, $\frac{dK}{K} = - \frac{2 dR}{R}$

or, the percentage change in $K = \frac{dK \times 100}{K} = - \frac{2 dR}{R} \times 100 = - 3 %$

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Angular momentum

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