Question 16.SP.5: A uniform sphere of mass m and radius r is projected along a......
A uniform sphere of mass m and radius r is projected along a rough horizontal surface with a linear velocity \overline{ \mathbf{v} }_0 and no angular velocity. Denoting by \mu_k the coefficient of kinetic friction between the sphere and the floor, determine (a) the time t_1 at which the sphere will start rolling without sliding,(b) the linear velocity and angular velocity of the sphere at time t_1.
Equations of Motion. The positive sense is chosen to the right for \overline{ a } and clockwise for α. The external forces acting on the sphere consist of the weight W, the normal reaction N, and the friction force F. Since the point of the sphere in contact with the surface is sliding to the right, the friction force F is directed to the left. While the sphere is sliding, the magnitude of the friction force is F=\mu_k N. The effective forces consist of the vector m \overline{ a } attached at G and the couple \bar{I} \alpha. Expressing that the system of the external forces is equivalent to the system of the effective forces, we write
N=W=\text{mg} \quad F=\mu_k N=\mu_k\text{mg}
\begin{array}{llll}\stackrel{+}{\rightarrow} \Sigma F_x=\Sigma\left(F_x\right)_{ \text {eff } }: & -F=m \bar{a} & \mu_k \text{mg}=m \bar{a} & \bar{a}=-\mu_k \text{g} \\+\downarrow \Sigma M_G=\Sigma\left(M_G\right)_{ \text {eff } }: & F r=\bar{I} \alpha &\end{array}Noting that \bar{I}=\frac{2}{5} m r^2 and substituting the value obtained for F, we write
\left(\mu_k\text{mg}\right) r=\frac{2}{5} m r^2 \alpha \quad \alpha=\frac{5}{2} \frac{\mu_k \text{g}}{r}Kinematics of Motion. As long as the sphere both rotates and slides, its linear and angular motions are uniformly accelerated.
\begin{aligned}& t=0, \bar{v}=\bar{v}_0 \quad \bar{v}=\bar{v}_0+\bar{a} t=\bar{v}_0-\mu_k \text{g} t&(1)\\& t=0, \omega_0=0 \quad \omega=\omega_0+\alpha t=0+\left(\frac{5}{2} \frac{\mu_k\text{g}}{r}\right) t &(2)\end{aligned}
The sphere will start rolling without sliding when the velocity \mathbf{v}_C of the point of contact C is zero. At that time,t=t_1, point C becomes the instantaneous center of rotation, and we have
\bar{v}_1=r \omega_1 (3)
Substituting in (3) the values obtained for \bar{v}_1 \text { and } \omega_1 by making t=t_1 in (1) and (2), respectively, we write
\bar{v}_0-\mu_k \text{g} t_1=r\left(\frac{5}{2} \frac{\mu_k \text{g}}{r} t_1\right) \quad t_1=\frac{2}{7} \frac{\bar{v}_0}{\mu_k \text{g}}
Substituting for t_1 into (2), we have
\omega_1=\frac{5}{2} \frac{\mu_k \text{g}}{r} t_1=\frac{5}{2} \frac{\mu_k \text{g}}{r}\left(\frac{2}{7} \frac{\bar{v}_0}{\mu_k \text{g}}\right)\quad\quad\quad\quad\omega_1=\frac{5}{7} \frac{\bar{v}_0}{r} \quad \omega_1=\frac{5}{7} \frac{\bar{v}_0}{r}\downarrow
\bar{v}_1=r \omega_1=r\left(\frac{5}{7} \frac{\bar{v}_0}{r}\right) \quad \bar{v}_1=\frac{5}{7} \bar{v}_0 \quad \mathbf{v} _1=\frac{5}{7} \bar{v}_0 \rightarrow
