A quarter disc of radius R and mass m is rotating about an axis OO′ which is perpendicular to the plane of the disc as shown in figure. Rotational kinetic energy of the disc is
A
12mR2ω2
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B
14mR2ω2
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C
18mR2ω2
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D
116mR2ω2
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Solution
The correct option is B14mR2ω2 Rotational kinetic energy KERot=12Iω2...(i) For the complete disc, the mass will be 4m. By symmetry of quarter disc, its moment of inertia about axis OO′ will be: I=Moment of inertia of complete disc4 I=[(4m)R22]4 ∴I=mR22...(i) From Eq (i)&(ii) KERot=12(mR22)ω2 ∴KERot=14mR2ω2