Maximum Applied Force for No Slippage in Angular Motion

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Homework Help Overview

The discussion revolves around a problem in angular motion involving a cylinder, static friction, and the maximum applied force that can be exerted without causing slippage. Participants are exploring the dynamics of forces acting on the cylinder, including the effects of static and kinetic friction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to clarify the conditions under which the cylinder begins to roll versus when it slips. There are questions about the roles of static and kinetic friction, as well as the torque generated by the applied force.

Discussion Status

Some participants have offered insights into analyzing forces and torques acting on the cylinder, while others express confusion about the problem setup and the implications of static friction. Multiple interpretations of the problem are being explored, particularly regarding the maximum force that can be applied without slippage.

Contextual Notes

There is a mention of a diagram that participants are referring to, which may be crucial for visualizing the forces involved. Additionally, some participants are questioning the definitions and assumptions related to the problem, such as the moment of inertia 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
 
Last edited by a moderator:
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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
 
Last edited by a moderator:
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.

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

where [itex]\alpha[/itex] 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) [tex]\tau_{top} = \frac{3}{2}MRa[/tex]

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) [tex]\tau_{top} - \tau_{bottom} = I_{cm}\alpha = \frac{1}{2}MR^2\alpha = \frac{1}{2}MRa[/tex]

Subtracting (2) from (1) we have:

(3) [tex]\tau_{bottom} = MRa \implies F_{bottom} = Ma[/tex]

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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