Solving Rotational Mechanics: Minimum Friction & Kinetic Energy

[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 \theta[\tex].<br /> <b>(a) What should be the minimum coefficient of friction between the sphere and the plane to prevent skidding?</b><br /> I did this:<br /> <br /> mgsin\theta[\tex] - f = ma ________ 1&lt;br /&gt; (f is frictional force)&lt;br /&gt; &lt;br /&gt; torque = I\alpha[\tex] = r X f &amp;lt;br /&amp;gt; (symbols stand for their usual meanings)&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; r*f = \frac{2mr^2}{\frac{a}{r}}[\tex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; f = (2/3)ma _______2&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; Subst. in 1 ,&amp;amp;lt;br /&amp;amp;gt; a= (3/5)gsin\theta[\tex] ______ 3&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; subst. both 2 and 3 in 1,&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; f = (2/5) mgsin\theta[\tex]&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; but,&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; \mu[\tex]mgcos\theta[\tex] = f = (2/5)mgsin\theta[\tex]&amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;gt; \mu[\tex] = (2/5) tan \theta[\tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; this matches with the textbook answer.&amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; (&amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;b&amp;amp;amp;amp;amp;amp;amp;amp;amp;gt;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).&amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;/b&amp;amp;amp;amp;amp;amp;amp;amp;amp;gt;&amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; I did this :&amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; \mu[\tex] = (1/5) tan \theta[\tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; f = \mu[\tex]mgcos\theta[\tex] &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; putting this in 1(of part a)&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; a = (4/5) g sin\theta[\tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; torque = I\alpha[\tex] = \frac{2mr^2}{3}[\tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; \alpha[\tex] = \frac{3gsin\theta}{10r}[\tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; KE = (1/2)mv^2 + (1/2)I\omega^2[\tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; v^2 = 2aL = (8/5) gLsin\theta[\tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; \omega^2[\tex] = 2\alpha\theta[\tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; i found \omega^2[\tex] = (6gLsin\theta[\tex]) / (20\pi\r^2[\tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; I get &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; KE = (4/5) mgLsin\theta[\tex] + mmgLsin\theta[\tex] / 2\pi[\tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; KE = &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;b&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt;0.831 &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;/b&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; * mgLsin\theta[\tex] &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; But the answer in the book is (7/8) mgLsin\theta[\tex] &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; which is &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;b&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt;0.875 &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;/b&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; mgLsin\theta[\tex] &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; 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 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 :

\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