Question

A uniform solid cylinder of radius $R=15\text{cm}$ rolls over a horizontal plane passing into an inclined plane forming an angle $\alpha ={30}^{\circ }$ with the horizontal. Find the maximum value of the velocity $v_{\circ }$ which still permits the cylinder to roll onto the inclined plane section without a jump. Assume there is no sliding between the cylinder and surfaces.

• A. $1\text{m/s}$
• B. $2\text{m/s}$
• C. $4\text{m/s}$
• D.
  $0.5\text{m/s}$

Answer 1

The correct option is A $1\text{m/s}$

At the instant shown in figure, the cylinder rotates about the point $O$.


Let the velocity of the center of mass of the cylinder be $v_1$ at this instant. As there is no sliding, we can say
$N=mg\cos \beta -\frac{m{v}_1^2}{R}$
where $N$ is normal reaction at the edge $O$.
or $v_1^2=gR\cos \beta -\frac{NR}{m}....\left(i\right)$

Change in vertical height of center of mass when the cylinder is at $O$ is $R(1-\cos \beta )$.

From energy conservation, we get
Increase in rotational KE = Decrease in gravitational PE
$\frac{1}{2}{I}_o\frac{v_1^2}{R^2}-\frac{1}{2}{I}_o\frac{v_0^2}{R^2}=mgR(1-\cos \beta )....\left(ii\right)$
Also, $I_o=\frac{1}{2}m{R}^2+m{R}^2=\frac{3}{2}m{R}^2$ (from the parallel axis theorem, about IAOR)
$\therefore$ Putting values of $I_o$ in eq. (ii)
$\frac{3}{4}\frac{m{R}^2}{R^2}[v_1^2-v_0^2]=mgR(1-\cos \beta )$
or $v_1^2=v_0^2+\frac{4}{3}gR(1-\cos \beta )..\left(iii\right)$

Comparing equation (i) $\&$ (iii),
$gR\cos \beta -\frac{NR}{m}=v_o^2+\frac{4}{3}gR(1-\cos \beta )$
or $v_0^2=\frac{gR}{3}(7\cos \beta -4)-\frac{NR}{m}$

The limitiing value for normal reaction $N$ is $N=0$, which will occur when $\beta =\alpha$.
For all other $\beta$ values, $N>0$
$\therefore$ we put $\beta =\alpha \&N=0$
{for these conditions $v_0=v_{max}$}

$v_{max}^2=\frac{gR}{3}(7\cos \alpha -4)$
$\therefore v_{max}=\sqrt{\frac{gR}{3}(7\cos \alpha -4)}$
Putting $\alpha ={30}^{\circ }\&R=0.15\text{m}$
$v_{max}\approx \sqrt{1}=1\text{m/s}$