Question

A uniform solid cylinder of radius $R=15cm$ rolls over a horizontal plane passing into an inclined plane forming an angle $\alpha ={30}^{\circ }$ with the horizontal (figure shown above). Find the maximum value of the velocity $v_0$ in (m/s) which still permits the cylinder to roll onto the inclined plane section without a jump. The sliding is assumed to be absent.

Answer 1

Since the cylinder moves without sliding, the centre of the cylinder rotates about the point $O$ becomes the foot of the instantaneous axis of rotation of the cylinder.
If at any instant during this motion the velocity of the C.M. is $v_1$ when the angle (shown in figure below) is $\beta$, we have
$\frac{m{v}_1^2}{R}=mg\cos \beta -N$,
where $N$ is the normal reaction of the edge
or, $v_1^2=gR\cos \beta -\frac{NR}{m}$ (1)
From the energy conservation law,
$\frac{1}{2}{I}_0\frac{v_1^2}{R^2}-\frac{1}{2}{I}_0\frac{v_0^2}{R^2}=mgR(1-\cos \beta )$
But $I_0=\frac{m{R}^2}{2}+m{R}^2=\frac{3}{2}m{R}^2$,
(from the parallel axis theorem)
Thus, $v_1^2=v_0^2+\frac{4}{3}gR(1-\cos \beta )$ (2)
From (1) and (2)
$v_0^2=\frac{gR}{3}(7\cos \beta -4)-\frac{NR}{m}$
The angle $\beta$ in this equation is clearly smaller than or equal to $\alpha$ so putting $\beta =\alpha$ we get
$v_0^2=\frac{gR}{3}(7\cos \alpha -4)-\frac{N_{0}R}{m}$
where, $N_0$ is the corresponding reaction. Note that $N\ge N_0$. No
jumping occurs during this turning if $N_0>0$. Hence, v_0 must be less than
$v_{max}=\sqrt{\frac{gR}{3}(7\cos \alpha -4)}=1m/s$