A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom?

A solid cylinder of mass M and radius R rolls without slipping down an

1.	A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom?
Answer» `sqrt(4gh)` `sqrt(2gh)` `sqrt((3)/(4)gh)` `sqrt((4)/(3)gh)` SOLUTION :According to conservation of mechanical energy, `(("Total potential energy"),("of CYLINDER at the top"),("of inclined plane"))=(("Rotational KINETIC"),("energy at the bottom"),("of inclined plane"))+(("The linear kinetic energy"),("of centre of mass of cylinder"),("at the bottom of inclined plane"))` `therefore mgh=(1)/(2)Iomega^(2)+(1)/(2)mv^(2)` putting `omega=(v)/(R )` in this FORMULA, `mgh=(1)/(2)I(v^(2))/(R^(2))+(1)/(2)mv^(2)` Now Moment of inertia of SOLID cylinder `I=(mR^(2))/(2)` `therefore mgh=(1)/(2).(mR^(2))/(2).(v^(2))/(R^(2))+(1)/(2)mv^(2)` `therefore gh=(1)/(4)v^(2)+(1)/(2)v^(2)` `gh=(3)/(4)v^(2)` `therefore v=sqrt((4)/(3)gh)`