Question 9.12: Cylinder rolling down a plane A uniform hollow circular cyli...

Suppose that, at time t, the cylinder has displacement x down the plane (from some reference configuration) and that the centre of mass G of the cylinder has velocity  v(=\dot{x})  down the plane. The angular velocity ω of the cylinder is then determined by the rolling condition to be ω = v/b. The kinetic energy of the cylinder is therefore

T=\frac{1}{2} M v^{2}+\frac{1}{2} I \omega^{2}=\frac{1}{2} M v^{2}+\frac{1}{2} I\left(\frac{v}{b}\right)^{2}

where M is the mass of the cylinder, and I is its moment of inertia about its axis of symmetry. From the Appendix, we find that I = Mb² so that the kinetic energy of the cylinder is given by T = Mv².

The gravitational potential energy of the cylinder is given by V = −Mgx sin α.
We must now dispose of the constraint forces. The reaction forces that the inclined plane exerts on the cylinder act on particles of the cylinder which, because of  he rolling condition, have zero velocity. These reaction forces therefore do no work.
Also the internal forces that keep the cylinder rigid do no work in total. Hence the constraint forces do no work in total.
Conservation of energy therefore applies in the form

M v^{2}-M g x \sin \alpha=E,

where E is the total energy. If we now differentiate this equation with respect to t (and cancel by Mv), we obtain

\frac{d v}{d t}=\frac{1}{2} g \sin \alpha,

which is the equation of motion of the cylinder. Thus the acceleration of the cylinder down the plane is \frac{1}{2} g \sin \alpha . (A block sliding down a smooth plane would have acceleration g sin α.)