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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 is0.875mgLsin[tex]\theta[\tex]

Where did I go wrong???

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# Rotational mechanics

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