Question 16.SP.6: A uniform sphere with mass m and radius r is projected along...

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   $\Sigma M_C=\dot{H}_C \longrightarrow 0=m \bar{a} r  +  \bar{I} \alpha$.