Question 17.SP.7: A uniform sphere of mass m and radius r is projected along a......

While the sphere is sliding relative to the surface, it is acted upon by the normal force N, the friction force F, and its weight W of magnitude W = mg.
Principle of Impulse and Momentum. We apply the principle of impulse and momentum to the sphere from the time t_1=0 when it is placed on the surface until the time t_2=t when it starts rolling without sliding.

\mathbf{\text { Syst Momenta }} _1+\mathbf{\text { Syst Ext Imp}} _{1 \rightarrow 2}=\mathbf{\text { Syst Momenta }} _2
\begin{aligned}&+\uparrow y \text { components: }\\&\stackrel{+}{\rightarrow} x \text { components: }\\&+\downarrow \text { moments about } G:\end{aligned}\quad\quad\quad\begin{aligned}N t-W t&=0 &(1)\\m \bar{v}_1-F t&=m \bar{v}_2&(2) \\F t r&=\bar{I} \omega_2&(3)\end{aligned}

From (1) we obtain N = W = mg. During the entire time interval considered, sliding occurs at point C and we have F=\mu_k N=\mu_k m g. Substituting CS for F into (2), we write

m \bar{v}_1-\mu_k m \text{g} t=m \bar{v}_2 \quad\quad \bar{v}_2=\bar{v}_1-\mu_k \text{g} t              (4)

Substituting F=\mu_k m g \text { and } \bar{I}=\frac{2}{5} m r^2 into (3),

\mu_k m \text{g} t r=\frac{2}{5} m r^2 \omega_2 \quad\quad \omega_2=\frac{5}{2} \frac{\mu_k \text{g}}{r} t                (5)

The sphere will start rolling without sliding when the velocity \mathbf{v} _C of the point of contact is zero. At that time, point C becomes the instantaneous center of rotation, and we have \bar{v}_2=r \omega_2. Substituting from (4) and (5), we write

\bar{v}_2=r \omega_2 \quad\quad \bar{v}_1-\mu_k \text{g} t=r\left(\frac{5}{2} \frac{\mu_k \text{g}}{r} t\right)\quad \quad t=\frac{2}{7} \frac{\bar{v}_1}{\mu_k \text{g}}

Substituting this expression for t into (5),

\omega_2=\frac{5}{2} \frac{\mu_k \text{g}}{r}\left(\frac{2}{7} \frac{\bar{v}_1}{\mu_k \text{g}}\right) \quad \omega_2=\frac{5}{7} \frac{\bar{v}_1}{r} \quad \omega_2=\frac{5}{7}\frac{\bar{v}_1}{r} \downarrow
\bar{v}_2=r \omega_2 \quad \bar{v}_2=r\left(\frac{5}{7} \frac{v_1}{r}\right) \quad \overline{ \mathbf{v} }_2=\frac{5}{7} \bar{v}_1 \rightarrow