Slip condition for a pulled cylinder

AI Thread Summary
The discussion centers on determining the maximum force that can be applied to a string wrapped around a cylinder, allowing it to roll without slipping. The calculated answer is 24 N, and there is a query about whether the maximum force is generally three times the kinetic friction before slipping occurs. Participants suggest using symbolic representation to derive a general result, confirming the relationship between applied tension and friction. The analysis reveals that while the cylinder accelerates, it can still roll without slipping, clarifying the conditions under which this occurs. Overall, the conversation emphasizes the importance of understanding the dynamics of rolling motion in relation to applied forces and friction.
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Homework Statement



Massless and inextensible string is wrapped around the periphery of a homogeneous cylinder of radius R = 0.5 m and mass m = 2 kg. The string is pulled straight away from the upper part of the periphery of the cylinder, without relative slipping. The cylinder moves on a horizontal floor, for which the friction coefficient (μ) is 0.4. What is most nearly the maximum force that can be exerted on the free end of the string so that the cylinder rolls without sliding?

(A) 24 N
(B) 12 N
(C) 8 N
(D) 6 N
(E) 8/3 N

Homework Equations



Your general rotational equations.

The Attempt at a Solution



I know the answer is A. My question is, is it a completely general case that the maximum force you can apply to a cylinder is 3 times the kinetic friction before it starts to slip?
 
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Hi, Helmholtz.

Did you try applying the same method that you used to find the answer (A), but using symbols rather than numbers? That way, you can get a general result and see if your conjecture is true.
 
Thanks for the response. I have done that, but I didn't like what I did and I was wonder if this was something someone could confirm.

Two forces directed in the same direction, one on top of the wheel, the other on the bottom. Forces add as such: F_1+F_2=ma (F_1 is the tension, F_2 is the friction), torques subtract such as F_1-F_2 = ma/2 (skipped a few steps hope it's followable). Then add them together (2*F_1 = 3/2 * ma), solve for a (a=4/3 * F_1/m), plug back into the force equation and find that 3*F_2 = F_1.

What I don't like is that the slip condition isn't obvious to me. Why is 'a' non-zero?

(Edit: I guess the slip condition was in the fact when I subtracted the torques and allowed \alpha = a/r)
(Edit: I suppose 'a' can be positive and there still is no slipping, not a condition I needed to impose. But when I thought about this in my mind, I assumed a cylinder of constant velocity, which is incorrect)
 
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Right. The center of the cylinder is accelerating even though the cylinder is not slipping on the surface. Your analysis and comments look correct to me.
 
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