Rotational mechanics

[prasanna]

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???

[ponjavic]

Hey !
Please help me out with this problem.

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

mgsin\theta - f = ma ________ 1
(f is frictional force)

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

r*f = \frac{2mr^2}{\frac{a}{r}}

f = (2/3)ma _______2

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

subst. both 2 and 3 in 1,

f = (2/5) mgsin\theta
but,
\mumgcos\theta = f = (2/5)mgsin\theta

\mu = (2/5) tan \theta

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 :

\mu = (1/5) tan \theta

f = \mumgcos\theta

putting this in 1(of part a)

a = (4/5) g sin\theta

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

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

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

\omega^2 = 2\alpha\theta

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

I get

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

KE = 0.831 * mgLsin\theta

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


Where did I go wrong???

Might help a bit ;)

[prasanna]

Might help a bit ;)

Thank a lot !!
How did you do that?

What happened that my original thread did not show LaTEX ??

[spacetime]

Hey Prasanna, your name is like that of a cricketer.
Anyway, use [/tex] instead of [\tex] for ending LaTeX code.



spacetime
www.geocities.com/physics_all

[Justin Lazear]

You used backslashes in the [/tex] instead of the forward slashes.

Backslashes for TeX characters, forward slashes for tags.

--J