Step 1

We are given the mass of the uniform disk is m, the angular velocity of the disk is ${\omega _0}$, and the coefficient of kinetic friction between the disk and floor is ${\mu _k}$.

We are asked to determine the time before it roll without slipping and the angular velocity of the disk.

 
Step 2

The diagram of the system is shown as:

 
Step 3

The formula to calculate the mass moment of inertia of the disk is,

\[{I_G} = \frac{1}{2}m{r^2}\]
 
Step 4

The formula to calculate the frictional force on the disk is,

\[{F_f} = {\mu _k}N\]
 
Step 5

The formula to calculate the force relation in vertical direction is,

\[\begin{array}{c} N - mg = 0\\ N = mg \end{array}\]
 
Step 6

The formula to calculate the force relation in vertical direction is,

\[\begin{array}{c} {F_f} = m{a_G}\\ {\mu _k}N = m{a_G}\\ {\mu _k}\left( {mg} \right) = m{a_G}\\ {a_G} = {\mu _k}g \end{array}\]

Here, g is the gravitational acceleration.

 
Step 7

The formula to calculate the moment about point G is,

\[\begin{array}{c} \Sigma {M_G} = {I_G}\alpha \\ - \left( {{F_f} \times r} \right) = \left( {\frac{1}{2}m{r^2}} \right)\alpha \end{array}\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{c} - \left( {{\mu _k}N \times r} \right) = \left( {\frac{1}{2}m{r^2}} \right)\alpha \\ - \left( {{\mu _k} \times mg} \right) = \left( {\frac{1}{2}mr} \right)\alpha \\ \alpha = - \frac{{2{\mu _k}g}}{r} \end{array}\]
 
Step 9

The formula to calculate the time before it roll without slipping by using kinematics is,

\[{v_G} = {\left( {{v_G}} \right)_0} + {a_G}t\]

Here, ${\left( {{v_G}} \right)_0}$ is the initial velocity and its value is zero.

 
Step 10

Substitute the values in the above expression.

\[\begin{array}{c} r\omega = 0 + \left( {{\mu _k}g} \right)t\\ t = \frac{{r\omega }}{{{\mu _k}g}} \end{array}\]
 
Step 11

The formula to calculate the angular velocity of the disk is,

\[\omega = {\omega _0} + \alpha t\]

Here, ${\omega _0}$ is the initial angular velocity.

 
Step 12

Substitute the values in the above expression.

\[\begin{array}{c} \omega = {\omega _0} + \left( { - \frac{{2{\mu _k}g}}{r}} \right)\left( {\frac{{r\omega }}{{{\mu _k}g}}} \right)\\ \omega = {\omega _0} + 2\omega \\ \omega = \frac{{{\omega _0}}}{3} \end{array}\]
 
Step 13

Substitute the value of ${\omega _0}$ in expression of t.

\[\begin{array}{l} t = \frac{r}{{{\mu _k}g}}\left( {\frac{{{\omega _0}}}{3}} \right)\\ t = \frac{{r{\omega _0}}}{{3{\mu _k}g}} \end{array}\]