A soild cylinder rolls down an inclined plane. Its mass is `2 kg and radius 0.1 m`. It the height of the inclined plane is `4m`, what is its rotational `K.E.` when it reaches foot of the plane ? Assume that the surfaces are smooth. Take `M.I.` of soild cylinder about its axis = `mr^(2)//2`.

Explanation

Here, `m = 2kg, r = 0.1 m`
Height of inclined plane, `h = 4 m`.
Rotational `K.E. = ?`
At the top of inclined plane, the soild cylinder has `P.E. = mgh`
At the bottom of inclined plane, the cylinder
has `K.E.` of translation `((1)/(2) m upsilon^(2))`
and `K.E.` of rotation `((1)/(2) I omega^(2))`
`:. (1)/(2)m upsilon^(2) + (1)/(2)I omega^(2) = m g h`
Now, `upsilon = r omega, I = (1)/(2) m r^(2)`
`:. (1)/(2)m (r omega)^(2) + (1)/(2) xx ((1)/(2) m r^(2)) omega^(2) = mgh`
`(3)/(4) m r^(2) omega^(2) = mgh`
`omega^(2) = (4mgh)/(3m r^(2)) = (4g h)/(3 r^(2))`
`K.E.` of rotation `= (1)/(2)I omega^(2) = (1)/(2) ((1)/(2) m r^(2)) xx (4 g h)/(3 r^(2))`
`(mgh)/(3) = (2xx9.8 xx4)/(3) = 26.13J`