Rotational mechanics

Hey !
Please help me out with this problem.

A hollow sphere of mass m is released from top on an inclined plane of inclination [tex]\theta[/tex].
(a) What should be the minimum coefficient of friction between the sphere and the plane to prevent skidding?
I did this:

mgsin[tex]\theta[/tex] - f = ma ________ 1
(f is frictional force)

torque = I[tex]\alpha[/tex] = r X f
(symbols stand for their usual meanings)

r*f = [tex]\frac{2mr^2}{\frac{a}{r}}[/tex]

f = (2/3)ma _______2

Subst. in 1 ,
a= (3/5)gsin[tex]\theta[/tex] ______ 3

subst. both 2 and 3 in 1,

f = (2/5) mgsin[tex]\theta[/tex]
but,
[tex]\mu[/tex]mgcos[tex]\theta[/tex] = f = (2/5)mgsin[tex]\theta[/tex]

[tex]\mu[/tex] = (2/5) tan [tex]\theta[/tex]

this matches with the text book answer.

(b) Find the Kinetic Energy of the ball as it moves down a length L on the incline if the friction coefficient is half the value calculated in part(a).
I did this :

[tex]\mu[/tex] = (1/5) tan [tex]\theta[/tex]

f = [tex]\mu[/tex]mgcos[tex]\theta[/tex]

putting this in 1(of part a)

a = (4/5) g sin[tex]\theta[/tex]

torque = I[tex]\alpha[/tex] = [tex]\frac{2mr^2}{3}[/tex]
[tex]\alpha[/tex] = [tex]\frac{3gsin\theta}{10r}[/tex]

KE = (1/2)mv^2 + (1/2)I[tex]\omega^2[/tex]

v^2 = 2aL = (8/5) gLsin[tex]\theta[/tex]

[tex]\omega^2[/tex] = 2[tex]\alpha\theta[/tex]

i found [tex]\omega^2[/tex] = (6gLsin[tex]\theta[/tex]) / (20[tex]\pi\r^2[/tex]

I get

KE = (4/5) mgLsin[tex]\theta[/tex] + mmgLsin[tex]\theta[/tex] / 2[tex]\pi[/tex]

KE = 0.831 * mgLsin[tex]\theta[/tex]

But the answer in the book is (7/8) mgLsin[tex]\theta[/tex]
which is 0.875 mgLsin[tex]\theta[/tex]


Where did I go wrong???

Might help a bit ;)