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A solid cylinder rolls up an inclined plane of angle of inclination 30^(@). At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5ms^(-1) How long will it take to return to the bottom ? |
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Answer» Solution :But`a=(gsintheta)/(1+(K^2)/(R^2))` i.e `K=(R^2)/2,a=(2gsintheta)/3` HENCE `0=u-at_a` or `t_a=u/a=(5xx3)/(2xx10xxsin30^(@))=3/2` |
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