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Find the conditions for the static coef. for nonslip conditions 
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#1
Jun2812, 07:06 PM

P: 263

1. The problem statement, all variables and given/known data
It's a cilinder of mass M and radius R rolling without slipping and I'm asked to find the maximum value of the static friction coefficient for the cilinder to roll without slipping. 2. Relevant equations Nonslip conditions: [tex] \displaystyle v=\omega R[/tex] [tex] \displaystyle a=\alpha R[/tex] Torque: [tex] \displaystyle \tau =I\alpha [/tex] 3. The attempt at a solution Net force in axis X (my axis is along with the inclined plane): [tex] \displaystyle mg\sin \theta {{\mu }_{s}}N=ma[/tex] Equation in axis y will give me the normal force: [tex] \displaystyle N=mg\cos \theta [/tex] So: [tex] \displaystyle mg\sin \theta {{\mu }_{s}}mg\cos \theta =ma[/tex] Solving for u: [tex] \displaystyle {{\mu }_{s}}=\frac{g\sin \theta a}{g\cos \theta }=\frac{g\sin \theta \alpha R}{g\cos \theta }[/tex] So I need to find out the angular aceleration: [tex] \displaystyle \tau =I\alpha [/tex] [tex] \displaystyle {{\mu }_{s}}mg\cos \theta \cdot R=\frac{1}{2}m{{R}^{2}}\cdot \alpha [/tex] [tex] \displaystyle \alpha =\frac{2g{{\mu }_{s}}\cos \theta }{R}[/tex] Replacing alpha in the equation I get that: [tex] \displaystyle {{\mu }_{s}}\le \tan \theta [/tex] But the option marked as correct is: [tex] \displaystyle {{\mu }_{s}}\ge \frac{\tan \theta }{3}[/tex] any help? Thanks!! 


#2
Jun2812, 08:09 PM

P: 963

Check your I



#3
Jun2812, 08:12 PM

P: 263

What's wrong out the inertia of the cilinder? Isn't it 1/2*MR^2? Thank you!



#4
Jun2812, 08:24 PM

P: 73

Find the conditions for the static coef. for nonslip conditions
Hello Hernaner28,
You have implicitly assumed the limiting friction case in your rolling situation (fixing f_{s}=μN ) First of all treat friction as an unknown independent of the normal reaction of the cylinder on the inclined plane to get an expression for friction in terms of Mass of object,the inclination of the plane ,The radius and the Moment of inertia.Now the maximum friction that your incline can offer at the given N is ____________?(answer) So your calculated friction must not exceed _________? Try to answer them and using the expression that comes as an interpretation calculate the condition again.I pretty much think this gives the actual answer as marked.Does this help? regards Yukoel 


#5
Jun2812, 08:27 PM

P: 263

But what was wrong about what I did? I worked with the limiting friction case because that would be the maximum coefficient of friction... that's why I finally get it has to be "less or equal".
Thanks! 


#6
Jun2812, 08:58 PM

P: 73

Less than or equal to what?The maximum value isnt it?There should be a separate equation for the general expression of friction isnt it?Here is an advised algorithm .Find the general expression for friction for the non limiting case (i.e. not involving the usage of N ;just treat f as an unknown fully and just apply the pure rolling rolling condition with Gravity and friction ;get an expression and compare it with the maximum value of friction sought).Now let us see What if the maximum value of friction is less than this required value of friction? There will be inevitable slipping right? By the way the expression for friction that comes from the general case will give you the general expression for friction for a pure rolling body on an inclined plane. Does this help? regards Yukoel 


#7
Jun2812, 09:04 PM

P: 263

Less or equal to tan(theta) as I finnaly worked out. I understand your thinking but I don't understand what is the difference between mine and yours, and what's the mistake about mine..
Thanks 


#8
Jun2812, 09:17 PM

P: 73

I think you have by mistake yielded your derivations to a≥Rα.Contradictory to pure rolling isn't it?(The last step quoted) regards Yukoel 


#9
Jun2812, 09:21 PM

P: 263

Ah yes, you're right. So what is what I got? Because it is not incorrect but maybe it is not what I was asked



#10
Jun2812, 09:24 PM

P: 963




#11
Jun2812, 09:27 PM

P: 263

No, both forces point the same direction. The cilinder rolls downward so friction force does an opposite torque (downward).
But Yukoel, what was what I got then? 


#12
Jun2812, 09:31 PM

P: 963

Where the torque comes from? The mgsinθ acts through the center of the cylinder. 


#13
Jun2812, 09:34 PM

P: 263

Ah yes you're right :D ! THANKS! Anyway, let's assume the last result is fine, then, what's the meaning of what I got? Is it the maximum coefficient of friction for the cilinder not to slip?



#14
Jun2812, 09:40 PM

P: 263

I don't know why but now I get the correct answer... even with my first thinking (nothing was wrong about it). I think it was because of the direction of the force which azizlwl corrected.
THANKS!!! 


#15
Jun2812, 09:42 PM

P: 263

I get:
[tex] \displaystyle F=\frac{mg\sin \theta }{3}[/tex] The friction F force is: [tex] \displaystyle {{\mu }_{s}}mg\cos \theta =\frac{mg\sin \theta }{3}[/tex] [tex] \displaystyle {{\mu }_{s}}=\frac{\tan \theta }{3}[/tex] For the limit case... so that would be the minimum value, or maximum? 


#16
Jun2812, 09:43 PM

P: 963

The exact answer is μ≥ (2/7)tanθ
The maximum ratio friction/N is μ_{s}. 


#17
Jun2812, 09:46 PM

P: 73

Let me get to point on signs here You can take friction arbitrarily let us suppose.For pure rolling the point of contact has to be at rest. Suppose the gravity and friction are in the same direction and look at the bottom .It will be accelerated.This is why friction has to be on the opposite of gravity. This however has nothing to do with limiting cases.As a matter fact it is quite comfortable to take things down the incline as positive. regards Yukoel 


#18
Jun2812, 09:48 PM

P: 263

Yukoel, I got the correct answer now:
[tex] \displaystyle {{\mu }_{s}}=\frac{\tan \theta }{3}[/tex] But that is a equality, not inequality, now I have to decide the sign of the inequality. That value is the minimum or maximum value for the cilinder not to slip? 


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