# Rotational mechanics

by prasanna
Tags: mechanics, rotational
 P: 45 Hey ! Please help me out with this problem. A hollow sphere of mass m is released from top on an inclined plane of inclination $$\theta[\tex]. (a) What should be the minimum coefficient of friction between the sphere and the plane to prevent skidding? I did this: mgsin[tex]\theta[\tex] - f = ma ________ 1 (f is frictional force) torque = I[tex]\alpha[\tex] = r X f (symbols stand for their usual meanings) r*f = [tex]\frac{2mr^2}{\frac{a}{r}}[\tex] f = (2/3)ma _______2 Subst. in 1 , a= (3/5)gsin[tex]\theta[\tex] ______ 3 subst. both 2 and 3 in 1, f = (2/5) mgsin[tex]\theta[\tex] but, [tex]\mu[\tex]mgcos[tex]\theta[\tex] = f = (2/5)mgsin[tex]\theta[\tex] [tex]\mu[\tex] = (2/5) tan [tex]\theta[\tex] this matches with the text book answer. (b) Find the Kinetic Energy of the ball as it moves down a length L on the incline if the friction coefficient is half the value calculated in part(a). I did this : [tex]\mu[\tex] = (1/5) tan [tex]\theta[\tex] f = [tex]\mu[\tex]mgcos[tex]\theta[\tex] putting this in 1(of part a) a = (4/5) g sin[tex]\theta[\tex] torque = I[tex]\alpha[\tex] = [tex]\frac{2mr^2}{3}[\tex] [tex]\alpha[\tex] = [tex]\frac{3gsin\theta}{10r}[\tex] KE = (1/2)mv^2 + (1/2)I[tex]\omega^2[\tex] v^2 = 2aL = (8/5) gLsin[tex]\theta[\tex] [tex]\omega^2[\tex] = 2[tex]\alpha\theta[\tex] i found [tex]\omega^2[\tex] = (6gLsin[tex]\theta[\tex]) / (20[tex]\pi\r^2[\tex] I get KE = (4/5) mgLsin[tex]\theta[\tex] + mmgLsin[tex]\theta[\tex] / 2[tex]\pi[\tex] KE = 0.831 * mgLsin[tex]\theta[\tex] But the answer in the book is (7/8) mgLsin[tex]\theta[\tex] which is 0.875 mgLsin[tex]\theta[\tex] Where did I go wrong??? P: 226  Hey ! Please help me out with this problem. A hollow sphere of mass m is released from top on an inclined plane of inclination [tex]\theta$$. (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, $$\mu$$mgcos$$\theta$$ = 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 = $$\mu$$mgcos$$\theta$$ 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 ;)
P: 45
 Quote by ponjavic Might help a bit ;)
Thank a lot !!
How did you do that?

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

P: 119

## Rotational mechanics

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
 P: 291 You used backslashes in the [/tex] instead of the forward slashes. Backslashes for TeX characters, forward slashes for tags. --J

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