- #1
prasanna
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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 textbook 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?
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 textbook 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?