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)