Question

A ball of mass m and radius r rolls along a circular path of radius R. Its speed at the bottom $(\theta =0^{\circ })$ of the path is $v_0$.Find the force of the path on the ball as a function of $\theta$.

Answer 1

$h=(R-r)(1-\cos \theta ).....\left(i\right)$
Kinetic energy at angle $\theta$ is,
$K=\frac{7}{5}\left(\frac{1}{2}m{v}_0^2\right)-mgh$
$\because$ In case orpure rolling
$K_T=\frac{5}{7}k$
$\therefore \frac{1}{2}m{v}^2=\frac{1}{2}m{v}_0^2-\frac{5}{7}mgh$
$\therefore v^2=v_0^2-\frac{10}{7}gh.....\left(ii\right)$
Equation of motion at angle $\theta$ is,
$n-mg\cos \theta =\frac{m{v}^2}{(R-r)}$
$\therefore N=mg\cos \theta +\frac{m}{(R-r)}(v_0^2-\frac{10}{7}gh)$
Substitnting value of h from Eq. (i)
$\therefore N=mg\cos \theta +\left(\frac{m}{R-r}\right)$
$\{v_0^2-\frac{10}{7}g(R-r\left)\right(1-\cos \theta \left)\right\}$
$=\frac{mg}{7}(17\cos \theta -10)+\frac{m{v}_0^2}{(R-r)}$
Force of friction,
$f=\frac{mg\sin \theta }{1+\frac{m{r}^2}{I}}$ (forpure rOlhng to take place)
$=\frac{mg\sin \theta }{1+\frac{5}{2}}$ $(I=\frac{2}{5}m{r}^2)$
$=\frac{2}{7}mg\sin \theta$