Solving Rotational Mechanics: Minimum Friction & Kinetic Energy

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Homework Help Overview

The problem involves a hollow sphere released on an inclined plane, focusing on the minimum coefficient of friction required to prevent skidding and the calculation of kinetic energy as it moves down the incline. The subject area includes rotational mechanics and dynamics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of forces and torques to derive the minimum coefficient of friction. They also explore the relationship between friction, acceleration, and kinetic energy as the sphere descends the incline.

Discussion Status

Some participants have successfully matched their calculations with textbook answers for the coefficient of friction. However, there is ongoing discussion regarding discrepancies in the kinetic energy calculations, with participants questioning where errors may have occurred in their reasoning.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There are also issues with formatting LaTeX in the thread that have been noted.

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

Thank a lot !
How did you do that?

What happened that my original thread did not show LaTEX ??
 
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
 
You used backslashes in the [/tex] instead of the forward slashes.

Backslashes for TeX characters, forward slashes for tags.

--J
 

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