A Cylinder's Life

This can easily be solved by considering the conservation of mechanical energy.

Let the mass of the solid cylinder be M and its radius be r. Let the linear speed be v and the angular velocity be $\omega$ .

At rest, only potential energy is relevant. Resolving the component forces, we get:

$E_{1}$ = Mgh = Mg(5 $sin 30^{o}$ ) = $\frac{5M}{2g}$ ... (1)

When the cylinder reaches the bottom, we have to deal with kinetic energy in both rotational and translational mechanics.

$E_{2}$ = $\frac{1}{2}$ $I\omega^{2}$ + $\frac{1}{2}$ $Mv^{2}$

Moment of Inertia, I for a solid cylinder = $\frac{1}{2}$ $Mr^{2}$

In rolling, v = $r\omega$ ... (a)

Therefore, $E_{2}$ = $\frac{1}{2}$ . $\frac{1}{2}$ $Mr^{2}$ $\omega^{2}$ + $\frac{1}{2}$ $Mv^{2}$

= $\frac{1}{4}$ $Mv^{2}$ + $\frac{1}{2}$ $Mv^{2}$ ... Using (a)

= $\frac{3}{4}$ $Mv^{2}$ ... (2)

As $E_{1}$ = $E_{2}$ ,

Using (1) and (2): $\frac{5M}{2g}$ = $\frac{3}{4}$ $Mv^{2}$

$\implies$ v = $\sqrt(\frac{10g}{3}$ ) = 5.715 m/s