Cylinder rolling down an incline

Given: $m = 20$ kg, $r = 0.3$ m, $\theta = 30^\circ$, rolling without slipping. $I_G = \tfrac12 mr^2 = 0.90$ kg·m².

         \
          \  O (G)
           (o)--> a
            \ F (friction up slope)
             \
              \ 30 deg

Equations of motion

Let $F$ be the friction force up the slope at the contact point, $N$ the normal reaction, and $a$ the acceleration of G down the slope.

Along the slope: $mg\sin\theta - F = ma$ (1)

Perpendicular to the slope: $N = mg\cos\theta$ (2)

Moments about G: $F r = I_G\alpha$ (3)

Rolling without slipping: $a = \alpha r$ (4)

From (3) and (4): $F = \dfrac{I_G a}{r^2} = \tfrac12 ma$. Put in (1):

$$mg\sin\theta = ma + \tfrac12 ma = \tfrac32 ma \;\Rightarrow\; a = \tfrac23 g\sin\theta$$

(a) Linear acceleration

$$a = \tfrac23(9.81)(0.5) = 3.27\ \text{m/s}^2$$

(b) Angular acceleration

$$\alpha = \frac{a}{r} = \frac{3.27}{0.3} = 10.90\ \text{rad/s}^2$$

(c) Friction and normal reaction

$$F = \tfrac12 ma = \tfrac12(20)(3.27) = 32.7\ \text{N} \qquad N = 20(9.81)\cos30^\circ = 169.9\ \text{N}$$

(d) Minimum coefficient of friction

For no slipping $F \le \mu_s N$:

$$\mu_{min} = \frac{F}{N} = \frac{32.7}{169.9} = 0.1925 \;\left(= \tfrac13\tan30^\circ\right)$$

Answer: a = 3.27 m/s²; α = 10.90 rad/s²; F = 32.7 N; N = 169.9 N; μ_min = 0.192.