A solid cylinder rolls without slipping down an inclined plane of height h. The velocity of the cylinder when it reaches the bouom is

A solid cylinder rolls without slipping down an inclined plane of heig

1.	A solid cylinder rolls without slipping down an inclined plane of height h. The velocity of the cylinder when it reaches the bouom is
Answer» `sqrt((2gh)/(3))` `sqrt((4gh)/(3))` `sqrt((3gh)/(2))` `sqrt(gh)` Solution :Let M and R be the mass and radius of the SOLID cylinder RESPECTIVELY . According to conservation of mechanical energy , `Mgh = (1)/(2) MV^(2) + (1)/(2) I OMEGA^(2) = (1)/(2) Mv^(2) + (1)/(2) ((1)/(2) MR^(2)) omega^(2)` = `(1)/(2) Mv^(2) + (1)/(4) MR^(2) ((v)/(R))^(2) = (1)/(2) Mv^(2) + (1)/(4) Mv^(2) = (3)/(4) Mv^(2)` or `v^(2) = (4)/(3) gh = v = sqrt((4)/(3) gh)`