A disc of mass M and radius R is rolling without slipping down an inclined plane. Show that the acceleration of the centre of mass of the disc is `(2)/(3)`g sin `theta`. Given angle of inclination of the plane is `theta` and moment of inertia of the disc is `(MR^(2))/(2)` .

Explanation

Let at A the velocity of the disc = 0 , i.e., Kinetic energy = 0. Now the disc sliding along the inclined plane reaches at point B . Suppose, The velocity of the disc be v, so the angular velocity is `omega = (v)/(R )`.
Hence kinetic energy,
`k = (1)/(2) Mv^(2) + (1)/(2) I omega^(2) = (1)/(2) Mv^(2) + (1)/(2)cdot (1)/(2)MR^(2) (v^(2))/(R^(2))`
= `(3)/(4)Mv^(2)`
So, increase in kinetic energy from A to B
= `(3)/(4)Mv^(2) - 0 = (3)/(4)Mv^(2)`
Decrease in potential energy from A to B
= MgH = Mg sin `theta`
`therefore (3)/(4)Mv^(2) = "Mgs sin"theta " " or, v^(2) = (4)/(3) "gs sin " theta`
When acceleration of cetre of mass of the disc is a, `v^(2) = ` 2 as.
`therefore" " 2 as = (4)/(3) " gs sin" theta`
or, a = `(2)/(3)`gsin`theta`