A drum of radius $R$ and mass $M$, rolls down without slipping along an inclined plane of angle $\theta$. The frictional force

The correct option is A Converts transnational energy to rotational energy

According to question

$mg\ast sinA-f=ma$--------------(1)

in pure rolling $\alpha =\frac{a}{r}$ where $\alpha$ = angular acceleration

Now

net, $\tau =I\alpha$ where I = moment of inertia

$f\ast r=I\alpha$

$f\ast r=Ia/r$

therefore

$f=Ia/r^2$. -------------------(2)

putting value of f in (1)

$mgsinA-Ia/r^2=ma$

here I of solid cylinder = $m{r}^2/2$

$mgsinA-ma/2=ma$

$a=\left(2gsinA\right)/3$

Now

$f=Ia/r^2$ from (2)

$f=ma/2$ which should be less than or equal to $\mu N$ otherwise body will slip (where u is coefficient of friction and N is normal acting on cylinder which is equal to mg cos A)

$ma/2<=\mu mgcosA$

putting value of a

$(m\ast g\ast sinA)/3<=\mu mgcosA$

$tanA<=3\mu$