Question 16.SP.6: A uniform sphere with mass m and radius r is projected along...
A uniform sphere with 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 the coefficient of kinetic friction between the sphere and the floor by μ_k, determine (a) the time t_1 at which the sphere starts rolling without sliding, (b) the linear velocity and angular velocity of the sphere at time t_1.
STRATEGY: Since you have forces acting on the sphere, use Newton’s second law. To relate the acceleration to the velocity, you need to use the basic kinematic relationships. The sphere starts out rotating and sliding; it stops sliding when the instantaneous point of contact with the ground has a velocity of zero.
MODELING: Choose the sphere as your system and model it as a rigid body. The assumed positive directions for the acceleration of the mass center and the angular acceleration are shown in Fig. 1. Free-body and kinetic diagrams for this system are shown in Fig. 2. 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.
ANALYSIS:
Equations of Motion. Applying Newton’s second law in the x and y directions gives
\begin{array}{rlc}+\uparrow \Sigma F_y=m \bar{a}_y: & N-W=0 \\& N=W=m g \quad F=\mu_k N=\mu_k m g & \\\stackrel{+}{\rightarrow} \Sigma F_x=m \bar{a}_x: & -F=m \bar{a} \quad-\mu_k m g=m \bar{a} \quad \bar{a}=-\mu_k g\end{array}
Now taking moments about the center of gravity, you get
+\circlearrowright \Sigma M_G=\bar{I} \alpha: \quad F r=\bar{I} \alpha
Noting that \bar{I}=\frac{2}{5} m r^2 and substituting the given value for F, you have
\left(\mu_k m g\right) r=\frac{2}{5} m r^2 \alpha \quad \alpha=\frac{5}{2} \frac{\mu_k g}{r}
Kinematics of Motion. As long as the sphere both rotates and slides, its linear and angular accelerations are constant. Therefore, you can use the constant-acceleration equations to relate these accelerations to the linear velocity and angular velocity.
t=0, \bar{v}=\bar{v}_0 \quad \bar{v}=\bar{v}_0 + \bar{a} t=\bar{v}_0 – \mu_k g t (1)
t=0, \omega_0=0 \quad \omega=\omega_0+\alpha t=0 + \left(\frac{5}{2} \frac{\mu_k g}{r}\right) t (2)
The sphere starts rolling without sliding when the velocity v_C of the point of contact C is zero (Fig. 3). At that time, t = t_1, point C becomes the instantaneous center of rotation, and you have
\bar{v}_1=r \omega_1 (3)
Substituting in Eq. (3) the values obtained for \bar{v}_1 and \omega_1 by making t = t_1 in Eqs. (1) and (2), respectively, you obtain
\bar{v}_0-\mu_k g t_1=r\left(\frac{5}{2} \frac{\mu_k g}{r} t_1\right) \quad t_1=\frac{2}{7} \frac{\bar{v}_0}{\mu_k g}
Substituting for t_1 into Eq. (2), you have
\begin{aligned}&\omega_1=\frac{5}{2} \frac{\mu_k g}{r} t_1=\frac{5}{2} \frac{\mu_k g}{r}\left(\frac{2}{7} \frac{\bar{v}_0}{\mu_k g}\right) \quad \omega_1=\frac{5}{7} \frac{\bar{v}_0}{r} \quad \omega_1=\frac{5}{7} \frac{\bar{v}_0}{r} \circlearrowright \\&\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 \mathrm{v}_1=\frac{5}{7} \bar{v}_0 \rightarrow\end{aligned}
REFLECT and THINK: Notice we chose a different coordinate system then we usually do, with the positive rotation going clockwise. This means that you will not be able to use vector algebra solutions since it is not a right-handed coordinate system.
You could use this type of analysis to determine how long it takes a bowling ball to begin to roll without slip or to see how the coefficient of friction affects this motion. Instead of taking moments about the center of gravity, you could have chosen to take moments about point C, in which case your third equation would have been \Sigma M_C=\dot{H}_C \longrightarrow 0=m \bar{a} r + \bar{I} \alpha.
