Tough angular motion problem

  • Thread starter 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 [Broken]

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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  • #2
Office_Shredder
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Well, what forces are preventing the cylinder from rolling?
 
  • #3
Andrew Mason
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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 [Broken]

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
 
  • #5
quasar987
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Under what conditions (force-wise) does an object slides on a surface?
 
  • #6
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when the torque caused by the force exceeds the torque generated by friction?
 
  • #7
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ahh, i'm so confused, can somebody help? i'm totally lost now
 
  • #8
Office_Shredder
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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?
 
  • #9
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it begins to roll?
 
  • #10
Office_Shredder
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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
Andrew Mason
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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
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I don't understand why I(rim)=1/2MR^2+MR^2. shouldn't that be just 1/2MR^2?
 

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