Let's Roll

Let the acceleration of the sphere be $a$ , so along the incline, $mg\sin\theta-f=ma$ .

About the instantaneous axis of rotation,

$\tau=mg\sin\theta R=\left(\frac{2}{5}mR^2+mR^2 \right)\alpha$

$a=\alpha R$

So, $a=\frac{5}{7}g\sin\theta$ and $f=\frac{2}{7}mg\sin\theta$ .

For pure rolling to happen,

$f=\frac{2}{7}mg\sin\theta<\mu mg \cos\theta \\ \mu>\frac{2}{7}\tan\theta$

Moment of Inertia, I for a solid sphere = $\frac{2}{5}$ $mr^{2}$ , where r is the radius of the sphere.

Let f be the force of friction and let a be the linear acceleration. Let N be the normal contact force of the sphere.

Resolving the component of forces along the incline, we get:

mg $sin\theta$ - f = ma ... (1)

f = $\mu$ N = $\mu$ mg $cos\theta$ ... (2)

Solving for torque, T w.r.t. f:

T = I. $\frac{a}{r}$ = f.r

$\implies$ a = $\frac{f.r^{2}}{I}$

$\implies$ a = $\frac{f.r^{2}}{(2/5).m.r^{2}}$

$\implies$ a = $\frac{5f}{2m}$

$\implies$ f = $\frac{2}{5}$ ma ... (3)

Putting (3) in (1):

mg $sin\theta$ - $\frac{2}{5}$ ma = ma

$\implies$ g $sin\theta$ = $\frac{7}{5}$ a

$\implies$ a = $\frac{5}{7}$ g $sin\theta$ ... (4)

Now for pure rolling: (2) > (3):

$\mu$ mg $cos\theta$ > $\frac{2}{5}$ ma

$\implies$ $\mu$ g $cos\theta$ > $\frac{2}{5}$ a

$\implies$ $\mu$ g $cos\theta$ > $\frac{2}{5}$ $\frac{5}{7}$ g $sin\theta$ ... Using (4)

$\implies$ $\mu$ > $\frac{2}{7}$ $tan\theta$