Question 16.SP.8: A sphere of radius r and weight W is released with no initia......
A sphere of radius r and weight W is released with no initial velocity on the incline and rolls without slipping. Determine (a) the minimum value of the coefficient of static friction compatible with the rolling motion, (b) the velocity of the center G of the sphere after the sphere has rolled 10 ft, (c) the velocity of G if the sphere were to move 10 ft down a frictionless 30° incline.
a. Minimum \mu_s for Rolling Motion. The external forces W, N, and F form a system equivalent to the system of effective forces represented by the vector m \overline{\mathbf{a}} and the couple \bar{I} \alpha. Since the sphere rolls without sliding, we have \bar{a}=r \alpha. \begin{aligned}+\downarrow \Sigma M_C=\Sigma\left(M_C\right)_{\text{eff}}: \quad\quad\quad (W \sin\theta) r&=(m \bar{a}) r+\bar{I} \alpha \\(W \sin \theta) r&=(m r \alpha) r+\bar{I} \alpha\end{aligned}
Noting that m = W/g and \bar{I}=\frac{2}{5} m r^2, we write
\begin{gathered}(W \sin \theta) r=\left(\frac{W}{\text{g}} r \alpha\right) r+\frac{2}{5} \frac{W}{\text{g}} r^2\alpha \quad \alpha=+\frac{5 \text{g} \sin \theta}{7 r} \\\bar{a}=r \alpha=\frac{5 \text{g} \sin \theta}{7}=\frac{5\left(32.2\text{ ft}/ s ^2\right) \sin 30^{\circ}}{7}=11.50 \text{ ft} / s ^2\end{gathered}
\begin{aligned}+\searrow \Sigma F_x=\Sigma\left(F_x\right)_{ \text{eff} }: \quad W \sin \theta-F&=m \bar{a} \\W \sin \theta-F&=\frac{W}{\text{g}} \frac{5 \text{g} \sin \theta}{7}\end{aligned}F=+\frac{2}{7} W \sin \theta=\frac{2}{7} W \sin 30^{\circ} \quad F =0.143 W ⦩ 30^{\circ}
\begin{aligned}& +\nearrow \Sigma F_y=\Sigma\left(F_y\right)_{\text{eff}}: \quad\quad N-W \cos \theta=0 \\& N=W \cos \theta=0.866 W \quad N =0.866 W ⦨ 60^{\circ}\end{aligned}\mu_s=\frac{F}{N}=\frac{0.143 W}{0.866 W} \quad \mu_s=0.165
b. Velocity of Rolling Sphere. We have uniformly accelerated motion:
\begin{aligned}\bar{v}_0&=0 \quad \bar{a}=11.50\text{ ft}/ s ^2 \quad \bar{x}=10 \text{ ft} \quad \bar{x}_0=0 \\\bar{v}^2&=\bar{v}_0^2+2 \bar{a}\left(\bar{x}-\bar{x}_0\right)\quad\bar{v}^2=0+2\left(11.50 \text{ ft} / s ^2\right)(10 \text{ ft} )\end{aligned}\bar{v}=15.17 \text{ ft} / s \quad \overline{ \text{v} }=15.17 \text{ ft} / s ⦪ {30}^{\circ}
c. Velocity of Sliding Sphere. Assuming now no friction, we have F = 0 and obtain
\bar{a}=+16.1\text{ ft}/ s ^2 \quad \overline{ \mathbf{a} }=16.1 \text{ ft} / s ^2 ⦪ 30^{\circ}
Substituting \bar{a}=16.1\text{ ft}/ s ^2 into the equations for uniformly accelerated motion, we obtain
\begin{aligned}& \bar{v}^2=\bar{v}_0^2+2 \bar{a}\left(\bar{x}-\bar{x}_0\right) \quad\quad\quad \bar{v}^2=0+2\left(16.1 \text{ ft} / s ^2\right)(10 \text{ ft} ) \\& \bar{v}=17.94 \text{ ft} / s& \overline{ \text {v} }=17.94 \text{ ft} / s ⦪ 30^{\circ}\end{aligned}
