91Ó°ÊÓ

Given in the question,

Mass of the uniform solid disk \(m\)

The radius of the solid disk\(R\)

Mass of the object\(m\)

The disk was initially at the rest

Law of conservation of energy

According to the conservation of energy, the initial total energy is always equal to the final total energy of the system.

\({E_F} = {E_I}\)

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 \(\omega = \sqrt {\frac{{4g}}{{3R}}} \)