A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°.

Solution:

Mass of the cylinder, m = 10 kg

Radius of the cylinder, r = 15 cm = 0.15 m

Co-efficient of kinetic friction, $\mu_{\mathrm{k}}=0.25$

Angle of inclination, $\theta=30^{\circ}$

Moment of inertia of a solid cylinder about its geometric axis, $I=\frac{1}{2} m r^{2}$

The various forces acting on the cylinder are shown in the following figure:

The acceleration of the cylinder is given as:

$a=\frac{m g \sin \theta}{m+\frac{I}{r^{2}}}$

$=\frac{m g \sin \theta}{m+\frac{1 m r^{2}}{2 r^{2}}}=\frac{2}{3} g \sin 30^{\circ}$

$=\frac{2}{3} \times 9.8 \times 0.5=3.27 \mathrm{~m} / \mathrm{s}^{2}$

(a) Using Newton's second law of motion, we can write net force as:

$f_{\text {net }}=m a$

$m g \sin 30^{\circ}-f=m a$

$=10 \times 9.8 \times 0.5-10 \times 3.27$

$=49-32.7=16.3 \mathrm{~N}$

(b) During rolling, the instantaneous point of contact with the plane comes to rest. Hence, the work done against frictional force is zero.

      (c) For rolling without skid, we have the relation:

$\mu=\frac{1}{3} \tan \theta$

$\tan \theta=3 \mu=3 \times 0.25$

$\therefore \theta=\tan ^{-1}(0.75)=36.87^{\circ}$