A uniform solid cylinder of radius $R = 15 cm$ rolls over a horizontal plane passing into an inclined plane forming an angle $a = 30°$ with the horizontal shown in the Figure. Find the maximum value of the $v_0$, which still permits the cylinder to roll onto the inclined plane section without a jump. The sliding is assumed to be absent.

Initial rolling kinetic energy of the cylinder

$$\begin{gathered} k_1=\frac{1}{2}m{v_0}^2+\frac{1}{2}I{\omega }^2 \\ =\frac{1}{2}m{v_0}^2+\frac{1}{2}\left(\frac{m{R}^2}{2}\right){\left(\frac{v_0}{R}\right)}^2 \\ =\frac{3}{4m{v}_0^2} \dots \left(1\right) \end{gathered}$$

When the cylinder passes onto the inclined plane its centre of mass descends through a distance $h=R(1-\cos \alpha )$

If $v$ is the velocity of its centre of mass now,

then rolling kinetic energy $=\frac{3}{4}m{v}^2$

From energy conservation, we have

$=\frac{3}{4}m{v_0}^2+mgR(1-\cos \alpha )=\frac{3}{4}m{v}^2$

At point $P$. $mg \cos \alpha =N+\frac{m{v}^2}{R}$

Cylinder passes the point P without a jump, if $N\ge 0$

For the maximum value of $v_0$. $N$ should have minimum $+ve$value i.e., $N=0$

$\therefore mg\cos \alpha =\frac{m{v}^2}{R}$

Solving equations $\left(1\right)$ and $\left(2\right)$ substituting $\alpha =30°$,$R=0.15m$ and $g=9.81 m/s^2$,we get

$v_0=\sqrt{\frac{gR}{3}(7\cos \alpha -4)}=1.0 m/s$