Question 16.SP.9: A sphere with a radius r and a weight W is released with no ...
A sphere with a radius r and a weight W is released with no initial velocity on an 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.
STRATEGY: Use Newton’s second law to determine the acceleration of the center of gravity. Then determine the velocity from kinematics.
MODELING: Choose the sphere to be your system and model it as a rigid body. Recall that for rolling motion, the instantaneous point of contact has a velocity of zero, which leads to \bar{a}=r \alpha (Fig. 1). A free-body diagram and kinetic diagram for this system are shown in Fig. 2. The external forces W, N, and F form a system equivalent to the inertial terms represented by the vector m \overline{\mathbf{a}} and the couple \bar{I} \alpha.
ANALYSIS:
a. Minimum μ_s for Rolling Motion. Since the sphere rolls without sliding, you have \bar{a}=r \alpha and can sum moments about C:
\begin{aligned}+\circlearrowright \Sigma M_C=\bar{I} \alpha + m \bar{a} d_{\perp}: &(W \sin \theta) r=\bar{I} \alpha+(m \bar{a}) r \\&(W \sin \theta) r=\bar{I} \alpha + (m r \alpha) r\end{aligned}
Noting that m = W/g and \bar{I}=\frac{2}{5} m r^2, you have
\begin{aligned}&(W \sin \theta) r=\frac{2}{5} \frac{W}{g} r^2 \alpha + \left(\frac{W}{g} r \alpha\right) r \quad \alpha=+\frac{5 g \sin \theta}{7 r} \\&\bar{a}=r \alpha=\frac{5 g \sin \theta}{7}=\frac{5\left(32.2 \mathrm{ft} / \mathrm{s}^2\right) \sin 30^{\circ}}{7}=11.50 \mathrm{ft} / \mathrm{s}^2\end{aligned}
Applying Newton’s second law in the x and y directions gives
\begin{gathered}+\searrow \Sigma F_x=m \bar{a}_x: \quad W \sin \theta – F=m \bar{a} \\W \sin \theta – F=\frac{W}{g} \frac{5 g \sin \theta}{7} \\F=+\frac{2}{7} W \sin \theta=\frac{2}{7} W \sin 30^{\circ} \quad \mathbf{F}=0.143 W \text{⦩} 30^{\circ}\end{gathered}
+\nearrow \Sigma F_y=m \bar{a}_y: \quad N – W \cos \theta=0
N=W \cos \theta=0.866 W \quad \mathbf{N}=0.866 W \text{⦨} 60^{\circ}
\mu_s=\frac{F}{N}=\frac{0.143 W}{0.866 W} \quad \quad \mu_s=0.165
b. Velocity of Rolling Sphere. This is a case of uniformly accelerated motion, so
\begin{aligned}&v_0=0 \quad \bar{a}=11.50 \mathrm{ft} / \mathrm{s}^2 \quad \bar{x}=10 \mathrm{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 \mathrm{ft} / \mathrm{s}^2\right)(10 \mathrm{ft})\end{aligned}
\bar{v}=15.17 \mathrm{ft} / \mathrm{s} \quad \overline{\mathbf{v}}=15.17 \mathrm{ft} / \mathrm{s} \text {⦪ }30^{\circ}
c. Velocity of Sliding Sphere. Now assuming no friction, you have F = 0 and obtain
\begin{aligned}&+\circlearrowright \Sigma M_G=\bar{I} \alpha: \quad 0=\bar{I} \alpha \quad \alpha=0 \\&+\searrow \Sigma F_x=m \bar{a}_x: \quad W \sin 30^{\circ}=m \bar{a} \quad 0.50 W=\frac{W}{g} \bar{a} \\&\quad \bar{a}=+16.1 \mathrm{ft} / \mathrm{s}^2 \quad \overline{\mathrm{a}}=16.1 \mathrm{ft} / \mathrm{s}^2 \text { ⦪ } 30^{\circ}\end{aligned}
Substituting \bar{a}=16.1 \mathrm{ft} / \mathrm{s}^2 into the equations for uniformly accelerated motion, you obtain
\bar{v}^2=\bar{v}_0^2 + 2 \bar{a}\left(\bar{x} – \bar{x}_0\right) \quad \bar{v}^2=0 + 2\left(16.1 \mathrm{ft} / \mathrm{s}^2\right)(10 \mathrm{ft})
\bar{v}=17.94 \mathrm{ft} / \mathrm{s} \quad \overline{\mathbf{v}}=17.94 \mathrm{ft} / \mathrm{s} \text { ⦪ } 30^{\circ}
REFLECT and THINK: Note that the sphere moving down a frictionless surface has a higher velocity than the rolling sphere, as you would expect. It is also interesting to note that the expression you obtained for the acceleration of the center of mass, that is, \bar{a}=5 g \sin \theta / 7, is independent of the radius of the sphere and the mass of the sphere. This means that any two solid spheres, as long they are rolling without sliding, have the same linear acceleration.