Answer:
Part a)
[tex]f = \frac{2}{7}mgsin\theta[/tex]
Part b)
[tex]\mu = \frac{2}{7} gtan\theta[/tex]
Explanation:
Part a)
Force equation on the inclined plane is given as
[tex]mgsin\theta - f = ma[/tex]
now for torque equation of the ball
[tex]fR = \frac{2}{5}mR^2 ( \frac{a}{R})[/tex]
[tex]f = \frac{2}{5}ma[/tex]
now from above two equations
[tex]mg sin\theta - \frac{2}{5}ma = ma[/tex]
[tex]mg sin\theta = \frac{7}{5} ma[/tex]
[tex]a = \frac{5}{7} gsin\theta[/tex]
so frictional force is given as
[tex]f = \frac{2}{5}ma[/tex]
[tex]f = \frac{2}{5}m(\frac{5}{7}gsin\theta)[/tex]
[tex]f = \frac{2}{7}mgsin\theta[/tex]
Part b)
Also we know that in the normal direction of the motion we have
[tex]F_n = mgcos\theta[/tex]
so we have
[tex]f = \mu F_n[/tex]
[tex]\frac{2}{7} mg sin\theta = \mu (mg cos\theta)[/tex]
now we have
[tex]\mu = \frac{2}{7} gtan\theta[/tex]