A thin disc rotates about an axis passing through its centre and perpendicular to its plane with a constant angular velocity ' ω' I is the moment of inertia of that disc and 'L' is its angular momentum about the given axis. Then rotational kinetic energy of the disc 'E' is

A) E $\large \propto$ L²B) E $\large \propto$ L^-2C) E $\large \propto$ l D)E $\large \propto$ l^-1

Question

A thin disc rotates about an axis passing through its centre and perpendicular to its plane with a constant angular velocity ' ω' I is the moment of inertia of that disc and 'L' is its angular momentum about the given axis. Then rotational kinetic energy of the disc 'E' is

A) E $\large \propto$ L²B) E $\large \propto$ L^-2C) E $\large \propto$ l D)E $\large \propto$ l^-1

Infinity Learn · Accepted Answer

$$($$$$KE$$$$)r$$ $$=$$ $$1/2$$ $$IW2$$ $$=$$ $$1/2$$ $$($$ (IW)2$$ $$/$$ I $$)⇒ $$($$$$KE$$$$)r$$ $$=$$ $$L2$$ $$/$$ $$2I$$.: E $$=$$ $$1/2$$ $$IW2$$ and E $$=$$ $$L2$$ $$/$$ $$2IAlso$$ , L $$=$$ I WSince W is constant ,E α i and L α I only (c) is correct.

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