Question

A solid cylinder of mass 2kg rolls down (pure rolling) an inclined plane from a height of 4m. Its rotational kinetic energy, when its reaches the foot of the plane is (Take g=10m`s^(-2))`
A. 20 J
B. 40 J
C. `80/3`J
D. 80 J

Answer 1

c) At foot of the plane,
For rotational motion about CM
`2Tr = 1/2Mr^(2)alpha`
Where, r = radius of cylinder
`rArr 2T = 1/2Ma`…………….(ii)
`From Eqs. (i) and (ii), we get
`a = 2/3g` and `T=(mg)/6`
Now, from equation of motion ,`v^(2) = u^(2) + 2gh`, we get
`v^(2) = 2(2/3g) h rArr v= sqrt((4gh)/3)`
`mgh = 1/2 komega^(2) + 1/2mu^(2)`
`80 = 3/4mu^(2)`
`therefore` Rotational KE = 1/2lomega^(2)`
`=1/4 mu^(2) = 1/4 xx (80 xx4)/3 = 80/3` J