Question

A solid sphere is projected up long an inclined plane of inclination $\theta ={30}^o$ with a velocity $v=2m/sec$. If it rolls without slipping from starting, find the maximum distance traversed by it. $(g=10m/se{c}^2)$.

Answer 1

Let the solid of mass $m$ and radius $r$ traverse a distance $\ell$ along the inclined plane. Since it rolls without sliding its initial $K.E.$ is given as
${KE}_I=\frac{1}{2}{mv}^2(1+\frac{k^2}{r^2})$, where $k=\sqrt{\frac{2}{5}}r$
$=\frac{1}{2}{mv}^2(1+\frac{2}{5})=\left(7/10\right){mv}^2$
As it comes to rest attaining a height $h=\ell \sin \theta$, its final $KE=0$
$\Delta KE=\frac{7}{10}m{v}^2$ ...$\left(a\right)$
Change in gravitational, $P.E.=mgh$ ...$\left(b\right)$
Conservation of energy yields,
$\left(\Delta PE\right)+\left(\Delta KE\right)=0$
$\Rightarrow mgh-\left(7/10\right)m{v}^2=0$
$\Rightarrow h=\frac{7v^2}{10g}=\frac{7\times (2{)}^2}{10\times 10}=0.28m$
$\therefore \ell =h\csc \theta =0.28\csc {30}^o=0.56m$