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A small sphere S of radius ,r and mass m rolls without slipping, inside a large hemispherical bowl B of radius R as shown in figure. S starts from rest at the top point of the hemisphere |
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Answer» The fraction of translational KINETIC ENERGY at the sphere is 5/7. At this position (bottom of B), the translational kinetic energy of the sphere is GIVEN by `E_(1)=1/2mv^(2)`, where m is the mass of the sphere. As the sphere rolls down without slipping, we have, `v = romega`, r is the radius of the sphere. `thereforeE_(2)=1/5mr^(2)omega^(2)=1/5mv^(2)` Therefore, total kinetic energy of the sphere at the bottom of the bowl is `E=E_(1)+E_(2)=1/2mv^(2)+1/5mv^(2)=7/10mv^(2)` Fraction of translational kinetic energy of the sphere `=(E_(1))/E=(1/2mv^(2))/(7/10mv^(2))=5/7` Fraction of rotational kinetic energy of the sphere `=(E_(2))/E=(1/5mv^(2))/(7/10mv^(2))=2/7` |
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