$A$ cylinder of mass $10 \; kg$ and radius $15 \; cm$ is rolling perfectly on a plane of inclination $30^{\circ}$. The coefficient of static friction $\mu_{S} = 0.25$.
$(a)$ How much is the force of friction acting on the cylinder?
$(b)$ What is the work done against friction during rolling?
$(c)$ If the inclination $\theta$ of the plane is increased,at what value of $\theta$ does the cylinder begin to skid and not roll perfectly?

(C) Given:
Mass of the cylinder,$m = 10 \; kg$
Radius of the cylinder,$r = 15 \; cm = 0.15 \; m$
Coefficient of static friction,$\mu_{S} = 0.25$
Angle of inclination,$\theta = 30^{\circ}$
Moment of inertia of a solid cylinder about its geometric axis,$I = \frac{1}{2} m r^{2}$
$(a)$ The acceleration of the cylinder rolling down an incline is given by:
$a = \frac{g \sin \theta}{1 + \frac{I}{mr^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{2}{3} g \sin 30^{\circ} = \frac{2}{3} \times 9.8 \times 0.5 = 3.27 \; m/s^{2}$
Using Newton's second law for translational motion: $mg \sin \theta - f = ma$
$f = m(g \sin \theta - a) = 10 \times (9.8 \times 0.5 - 3.27) = 10 \times (4.9 - 3.27) = 16.3 \; N$
$(b)$ During pure rolling,the point of contact is instantaneously at rest. Therefore,the work done against friction is $0 \; J$.
$(c)$ For rolling without skidding,the condition is $f \leq \mu_{S} N$,where $N = mg \cos \theta$.
$mg \sin \theta - ma \leq \mu_{S} mg \cos \theta$
Substituting $a = \frac{2}{3} g \sin \theta$:
$mg \sin \theta - m(\frac{2}{3} g \sin \theta) \leq \mu_{S} mg \cos \theta$
$\frac{1}{3} \sin \theta \leq \mu_{S} \cos \theta \implies \tan \theta \leq 3 \mu_{S}$
$\tan \theta \leq 3 \times 0.25 = 0.75$
$\theta \leq \tan^{-1}(0.75) \approx 36.87^{\circ}$
The cylinder begins to skid at $\theta \approx 36.87^{\circ}$.