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A hollow cylinder and a solid cylinder are rolling without slipping down an inclined plane, then which of these reaches earlier ? |
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Answer» For a circulary symmetric rigid body, starting from rest and rolling down an inclined plane without slipping, its speed after a vertical displacement h is `v=sqrt((2gh)/(1+(I//MR^(2))))" "`….(1) where M, R and I are respectively the mass, radius and moment of inertia of the body. (i) For a ring (or a thin-walled hollow cylinder), `I=MR^(2)" "`...(2) `:. (I)/(MR^(2))=1` `:. v=sqrt((2gh)/(1+(I)/(MR^(2))))" "`....(3) (ii) For a solid cylinder (or a disc), `I=(1)/(2)MR^(2)" "`....(4) `:. (I)/(MR^(2))=(1)/(2)` `:. v=sqrt((2gh)/(1+(I)/(MR^(2))))=sqrt((2gh)/(1+(1)/(2)))=sqrt((3)/(4)gh)" "`......(5) (iii) For a solid sphere, `I=(2)/(5)MR^(2)" "`....(6) `:. (I)/(MR^(2))=(2)/(5)` `:. v=sqrt((2gh)/(1+(I)/(MR^(2))))=sqrt((2gh)/(1+(2)/(5)))=sqrt((10)/(7)gh)" "`......(7) |
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