Question

A sphere of mass $m$ and radius $r$ is pushed onto the fixed horizontal surface such that it rolls without slipping from the beginning. The minimum speed v of its mass centre at the bottom so that it rolls completely around the loop of radius $(R+r)$ without leaving the track in between is given as $v=\sqrt{\frac{x}{7}gR}$. Find $x$

Answer 1

Using Energy conservation ,let the final velocity at the top be $v_1$

$m{v}^2/2+\frac{1}{2}\frac{2m{r}^2{\omega }^2}{5}=m{v}_1^2/2+\frac{1}{2}\frac{2m{r}^2{\omega }_1^2}{5}+2mgR$ ,here H=2R ,center to center distance.

for rolling $v=\omega r$

so above becomes

$\frac{7m{v}^2}{10}=\frac{7m{v}_1^2}{10}+2mgR$...(1)

so satisfy circulation motion ,

$m{v}_1^2/R=mg\to v_1^2=Rg$

putting the value of $v_1$ we get

$v=\sqrt{\frac{27gR}{7}}$