Sphere rolling down incline, friction and velocity

Solid sphere of weight $W = 80$ kN, rolling without slipping on a $30^\circ$ incline: $\bar I = \tfrac25 mr^2$ and $\bar a = \alpha r$. Distances are in feet, so $g = 32.2$ ft/s².

(a) Minimum coefficient of friction

Take axes along and perpendicular to the incline. Let $F$ be friction up the incline.

$$\begin{aligned} W\sin30^\circ - F &= m\bar a\\ N - W\cos30^\circ &= 0\\ Fr &= \bar I\alpha = \tfrac25 mr^2\frac{\bar a}{r} \;\Rightarrow\; F = \tfrac25 m\bar a \end{aligned}$$

Substituting $F$ in the first equation: $W\sin30^\circ = \tfrac75 m\bar a$, so

$$\bar a = \tfrac57 g\sin30^\circ = \tfrac57(32.2)(0.5) = 11.50\ \text{ft/s}^2$$ $$F = \tfrac25 m\bar a = \tfrac27W\sin30^\circ = \tfrac27(80)(0.5) = 11.43\ \text{kN}, \qquad N = W\cos30^\circ = 80\cos30^\circ = 69.28\ \text{kN}$$

For rolling without slipping, $F \le \mu_s N$:

$$\mu_{min} = \frac{F}{N} = \tfrac27\tan30^\circ = 0.1650$$

(b) Velocity of G after rolling 15 ft

The acceleration is constant, starting from rest:

$$\bar v^2 = 2\bar a\,s = 2(11.50)(15) \;\Rightarrow\; \bar v = 18.57\ \text{ft/s} \;(= 5.66\ \text{m/s})$$

(Check by energy: $Ws\sin30^\circ = \tfrac12m\bar v^2 + \tfrac12\bar I\omega^2 = \tfrac12\cdot\tfrac75 m\bar v^2$, giving the same $\bar v$.)

(c) Frictionless incline, 20 ft

The sphere does not rotate; it slides with $a = g\sin30^\circ = 16.1$ ft/s²:

$$\bar v = \sqrt{2(16.1)(20)} = 25.38\ \text{ft/s} \;(= 7.73\ \text{m/s})$$

Answer: (a) $\mu_{min} = 0.1650$; (b) $\bar v = 18.57$ ft/s; (c) $\bar v = 25.38$ ft/s.