A hollow sphere of mass M and radius R is rolling without slipping on a rough horizontal surface. Then the percentage of it's total KE which is translational, will be
A
72%
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B
28%
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C
60%
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D
40%
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Solution
The correct option is C60% Moment of inertia of hollow sphere about the axis of rotation passing through it's centre of mass, I=23MR2 Condition for pure rolling, v=Rω v= Translational speed of centre of mass ⇒KETrans=12Mv2 ⇒KERot=12Iω2=12×(23MR2)×v2R2 ∴KERot=13Mv2 Total Kinetic Energy ⇒KEtotal=KETrans+KERot=12Mv2+13Mv2=56Mv2 % of Translational kinetic Energy =KETransKEtotal×100 =(12Mv2)(56Mv2)×100 ∴% of Translational kinetic energy=60%