Ans.

When a circular disc of mass M and radius r rolls on a table, then its total kinetic energy E will be,

\( E \) = (Energy due to linear motion) + (energy due to rotational motion)

The moment of inertia of a circular disc about an axis passing through the centre of the disc and perpendicular to the plane of the disc is
\( I=\frac{1}{2}Mr^2 \)
[ To know the derivation (CLICK HERE) ]

Energy due to the linear motion is \( E_l=\frac{1}{2}Mv^2\\=\frac{1}{2}M{(r\omega)}^2\\=\frac{1}{2}Mr^2{\omega}^2 \)

Energy due to the rotational motion is \( E_r=\frac{1}{2}I{\omega}^2 \), [ To know the derivation of rotational kinetic energy, (CLICK HERE) ]

Putting the value of \( I \) we get, \( E_r=\frac{1}{4}Mr^2{\omega}^2 \).

So the total kinetic energy of the disc is

\( E \)=\( E_l+E_r\\=\frac{1}{2}Mr^2{\omega}^2+\frac{1}{4}Mr^2{\omega}^2\\=\frac{3}{4}Mr^2{\omega}^2 \)