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Little more for rolling body : |
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Answer» Solution :Velocity of body rolling without sliding from a slope of height h and reaching at the bottom is, `v^(2)=(2gh)/([1+k^(2)//R^(2)])""....(1)` LET d be the length of the inclined PLANE. Suppose the object starting from the position of rest, travel a DISTANCE d over inclined plane with linear acceleration a and reaches the bottom `therefore v^(2)=2ad [because v_(0)=0]` but `d=(h)/(sintheta)` `therefore v^(2)=(2ah)/(sintheta)"".....(2)` comparing equations (1) and (2) `a=(gsintheta)/([1+(k^(2))/(R^(2))])""......(3)` a is directly parallel to the slope. Its magnitude should be equal to `gsintheta`, which is the component of g parallel to the inclined plane `therefore` Decrease in the acceleration `a.=gsintheta-{(gsintheta)/([1+k^(2)//R^(2)])}` `a.=gsintheta-{1-(1)/(1+(k^(2))/(R^(2)))}` `=gsintheta{1-(R^(2))/(R^(2)+k^(2))}` `=gsintheta{(k^(2))/(k^(2)+R^(2))}""......(4)` Thus decrease in the acceleration is due to the frictional force acting on the rolling body. The work done against the force of FRICTION results into the kinetic energy of rotational motion. Therefore even in the presence of friction the law of conservation of mechanical energy can be used. `therefore` Friction force `F=ma.` `therefore` Friction force `F=mgsintheta{(k^(2))/(k^(2)+R^(2))}....(5)` |
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