A solid cylinder when dropped from a height of 2 m acquires a velocity while reaching the ground. If the same cylinder is rolled down from the top of an inclined plane to reach the ground with same velocity, what must be the height of the inclined plane? Also compute the velocity

Explanation

In the first case, ,
potential energy - kinetic energy
`mgh=(1)/(2) mv^(2)`
`mgxx2=(1)/(2) mv^(2)" "...(1)`

In second case,
potential energy - translational kinetic energy + rotational kinetic energy
`mgh.=(1)/(2) mv^(2)+(1)/(2)1 omega^(2)`
`mgh=(1)/(2)+(1)/(2) ((mr^(2))/(2))((v^(2))/(r^(2)))`
`:. mgh.=(3)/(4) mv^(2) " "....(2)`
Dividing. (2) by (1),
`(mgh.)/(mg xx2)=((3)/(4)mv^(2))/((1)/(2)mv^(2))=(3)/(4)xx(2)/(1)=(3)/(2)`
`k=3m`
From equation (1) `2mg=(1)/(2) mv^(2)`
`v= sqrt(4)g=2sqrt(g)`
`v=2xx sqrt(9.81)`
`v=6.3 ms^(-1)`