Question

There is a rough track, a portion of which is in the form of a cylinder of radius $R$ as shown in the figure. Find the minimum linear speed of a uniform ring of radius $r$ with which it should be set rolling without sliding on the horizontal part so that it can complete round the circle without sliding on the cylindrical part.

Answer 1

Let the ring be projected with velocity $v$

For the Rolling: $v=w.r$

$\to$ The ring has the max chance of falling at the highest point as slow

As at thuse point $mg$ is distributed towards the centre so it can provide all centripetal force and $Ncm$ be zero

Limiting Case: $N=O\left(a\right)$ highest point let speed $=v$

$\Rightarrow mg=\frac{m.\left(v\right)}{(R-\pi )}\Rightarrow (v{)}^2=g(R-\pi )....\left(ii\right)$

As this is Pure rolling Mechanical Energy will be conserved in (Earth + Ray) system:

By Conservation of Energy:

$k_i+u_i=k_f+u_f$

$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}{I}_{ring}{w}^2+O=\frac{1}{2}m{v}^2+\frac{1}{2}{I}_{ring}{w}^2+mg2(R-r)$

Using $\left(i\right)$ & $\left(ii\right)$ for limiting case:

$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}m{r}^2.\frac{v^2}{r^2}$

$=\frac{1}{2}mg(R+i)+\frac{1}{2}.m{r}^2{\left(\frac{v}{r}\right)}^2+mg2(R-r)$

$\Rightarrow m{v}^2=\frac{1}{2}mg(R-r)+\frac{1}{2}mg(R-r)+2mg(R-r)$

$\Rightarrow v=\sqrt{3g(R-r)}$ This is the min speed of projection required.