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Question

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 B 14mR2ω2
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

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