From the conservation of energy, we know
Total initial energy = Total final energy
Initial potential energy +initial kinetic energy= final potential energy + final kinetic energy.
\({U_i} + K{E_i} = {U_f} + K{E_f}\)
Since we know the height of the object is zero and initially, the disk is at rest
Therefore,
The initial potential energy,\({U_i} = 0\)
The initial kinetic energy\(K{E_i} = 0\)
Since the finally the block is below the axis, and so distance is R.
Therefore,
Final potential energy\({U_f} = - mgR\)
The final kinetic energy will be the sum of translational kinetic energy and rotational kinetic energy
Final kinetic energy
\(\begin{aligned}{}K{E_f} = K{E_t} + K{E_r}\\ = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}\end{aligned}\)
Where m is mass, v is linear velocity, I is a moment of inertia and\(\omega \)is angular velocity.
The moment of inertial of the disk is
\(I = \frac{1}{2}m{R^2}\)
Where m is mass and R is the radius
Substituting all the values into the equation of energy conservation
\(0 + 0 = - mgR + \frac{1}{2}m{v^2} + \frac{1}{2}\left( {\frac{1}{2}m{R^2}} \right){\omega ^2}\)
We know,
\(v = r\omega \)
\(\begin{aligned}{}0 = - mgR + \frac{1}{2}m{\left( {R\omega } \right)^2} + \frac{1}{2}\left( {\frac{1}{2}m{R^2}} \right){\omega ^2}\\mgR = \frac{3}{4}m{\omega ^2}{R^2}\\\omega = \sqrt {\frac{{4g}}{{3R}}} \end{aligned}\)
The angular speed when the block is below the axis is \(\omega = \sqrt {\frac{{4g}}{{3R}}} \)
