Maximum Applied Force for No Slippage in Angular Motion

AI Thread Summary
The discussion revolves around calculating the maximum force that can be applied to a cylinder without causing it to slip. Participants clarify that static friction is crucial in preventing slippage, and once the applied force exceeds the static friction, the cylinder will begin to roll. The conversation highlights the importance of analyzing forces and torque, particularly around the point of contact, to understand the conditions under which slippage occurs. There is some confusion regarding the moment of inertia of the cylinder, with participants debating its correct formulation. The thread emphasizes the relationship between applied force, static friction, and the resulting motion of the cylinder.
nocloud
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okay, I've thought about this for a while and still can't figure it out and its absolutely bugging me to death

here's the picture:
http://img307.imageshack.us/img307/3193/problemcr4.gif

Assume that the static friction coefficient between the cylinder and the ground is "u", the gravitational acceleration is "g", the mass of the cylinder is "M", and the radius is "r". What is the maximum force that can be applied without slippage occurring
 
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Well, what forces are preventing the cylinder from rolling?
 
nocloud said:
okay, I've thought about this for a while and still can't figure it out and its absolutely bugging me to death

here's the picture:
http://img307.imageshack.us/img307/3193/problemcr4.gif

Assume that the static friction coefficient between the cylinder and the ground is "u", the gravitational acceleration is "g", the mass of the cylinder is "M", and the radius is "r". What is the maximum value of F without causing the cylinder to roll?
I don't get this question. It would make more sense if it asked what maximum force can be applied without slippage occurring.

Any force F will cause the cylinder to roll.

AM
 
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AH, oops, my bad, yea, the question is supposed to be what is the maximum force that can be applied without slippage occurring
 
Under what conditions (force-wise) does an object slides on a surface?
 
when the torque caused by the force exceeds the torque generated by friction?
 
ahh, I'm so confused, can somebody help? I'm totally lost now
 
Draw a free body diagram. The forces are:

F on the top

Force of static friction on the bottom

However, the static friction can only apply so much force, right? So what happens after F is greater than the static friction on the bottom?
 
it begins to roll?
 
  • #10
nocloud, which direction is the static friction pointing in? (Hint: It's the opposite direction of motion).

You should be able to see that the static friction is applied in the correct direction for the cylinder to roll. However, what happens after the static friction is overcome? Then it's kinetic friction... now let's compare the two:

Static friction: Object doesn't slide (hence, it rolls instead)
Kinetic friction: Object slides
 
  • #11
nocloud said:
AH, oops, my bad, yea, the question is supposed to be what is the maximum force that can be applied without slippage occurring
Try analysing the forces and torque about the point of instantaneous point of contact. You can see that the force and torque does not change as the cylinder rolls - it is just the point of contact (the fulcrum) moves.

\tau_{top} = I_{rim}\alpha = (\frac{1}{2}MR^2 + MR^2)\alpha

where \alpha is the angular acceleration of the cylinder about the point of contact, which is just a/R where a is the acceleration of the centre of mass. So:

(1) \tau_{top} = \frac{3}{2}MRa

Since the net torque produces the angular acceleration of the cylinder about the cylinder's center of mass (which if it rolls is just a/R where a is the acceleration of the centre of mass), we have:

(2) \tau_{top} - \tau_{bottom} = I_{cm}\alpha = \frac{1}{2}MR^2\alpha = \frac{1}{2}MRa

Subtracting (2) from (1) we have:

(3) \tau_{bottom} = MRa \implies F_{bottom} = Ma

This makes sense, since the road provides the force which accelerates the centre of mass.

AM
 
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  • #12
I don't understand why I(rim)=1/2MR^2+MR^2. shouldn't that be just 1/2MR^2?
 
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