Rotation with Static Frictionby petitericeball Tags: disk, rolling, rotation, spool, static friction 

#1
Nov1109, 11:51 PM

P: 24

1. The problem statement, all variables and given/known data
A spool rests on a horizontal surface on which it rolls without slipping. The middle section of the spool has a radius r and is very light compared with the ends of the cylinder which have radius R and together have mass M. A string is wrapped around the middle section so you can pull horizontally (from the middle section's top side) with a force T. 2.1 Determine the total frictional force on the spool in terms of T, r and R. 2.2 What is the condition for the linear acceleration of the spool to exceed T/M? 2.3 Which direction does the friction force point when the acceleration is less than T/M? Which direction does the friction force point when the acceleration is greater than T/M? 2. Relevant equations [tex] \Sigma \vec{F} = m \vec{a} [/tex] [tex] \Sigma T = I \alpha = \vec{F} R [/tex] 3. The attempt at a solution So, [tex] T\vec {F}_{friction} = m \vec{a} [/tex] (This T is force) [tex] \Sigma T = Tr + F_{friction}R [/tex] (Net Torque = Force*radius  Friction*Radius) Since [tex] \Sigma T = I \alpha = \vec{F} R [/tex] [tex] \frac{1}{2}MR^2(a/R)=Tr + F_{friction}R [/tex] [tex] ma \Rightarrow TF_{friction} [/tex] [tex]\frac{1}{2}(TF_{friction})R=Tr+F_{friction}R [/tex] Rearrange [tex]TR2Tr=2F_{friction}R+F_{friction}R[/tex] I get this: [tex]T(R2r)/3R = F_{friction} [/tex] Is this right? At first I thought that the negative sign would have cancelled out somewhere and I had made a mistake, but it seems that it should be there, and shows that the [tex] \vec {F}_{friction} [/tex] opposes [tex] \vec {T} [/tex] 2.2 What is the condition for the linear acceleration of the spool to exceed T/M? So, this says that [tex] A>T/M \Rightarrow MA>T \Rightarrow TF_{friction}>T [/tex] This says that in order for this to be true, friction must be doing positive work on the object, or there must be another force acting on the spool. I was also reading around, and there something about if you were walking, you push downward and backward, and the ground must oppose and push forward and upward. In this case, would that mean the force friction is in the same direction as the velocity? 2.3 Which direction does the friction force point when the acceleration is less than T/M? Which direction does the friction force point when the acceleration is greater than T/M? Similar to the previous question, if [tex]MA<T [/tex], then friction is pointing against motion. If [tex] MA>T [/tex], then friction is pointing along with the motion. 



#2
Nov1209, 03:56 AM

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ehild 



#3
Nov1209, 07:39 PM

P: 24

Thanks for the reply. Yes I did write the equations with respect to the center of mass, but didn't realize my mistake.
[tex] \Sigma T = Tr + F_{friction}R [/tex] So I'm going to edit the original post to reflect these changes. Also, I emailed one of the graders about the interpretation of the moment of inertia, and he said "Yes there is static friction. I believe that the way to interpret the next part is to imagine that a 2D disk with the center cut out where mass exists on all parts except at a radius less then r," so I must change my moment of inertia. 



#4
Nov1209, 08:20 PM

P: 24

Rotation with Static Friction
So, according to a site I found, the moment of inertia of a disk with the center cut out will be [tex] \frac{1}{2}M(r^2+R^2) [/tex] so I'll post the revised results here along with the corrections from earlier.
So, [tex] T\vec {F}_{friction} = m \vec{a} [/tex] (This T is force) [tex] \Sigma T = Tr + F_{friction}R [/tex] (Net Torque = Force*radius  Friction*Radius) Since [tex] \Sigma T = I \alpha = \vec{F} R [/tex] [tex] \frac{1}{2}M(r^2 + R^2)(a/R)=Tr + F_{friction}R [/tex] [tex] ma \Rightarrow TF_{friction} [/tex] [tex]\frac{1}{2}(TF_{friction})(r^2+R^2)(\frac{1}{R})=Tr+F_{friction}R [/tex] Rearrange [tex]TR^2 +Tr^2F_{friction}R^2F_{friction}r^2=2TrR+2F_{friction}r^2[/tex] [tex]T(R^22TrR+r^2)=3F_{friction}R^2+F_{friction}r^2[/tex] And I end with this: [tex] \frac{T(Rr)^2}{(3R^2+r^2)}=F_{friction} [/tex] 



#5
Nov1209, 11:22 PM

HW Helper
Thanks
P: 9,818

It seems all right,
ehild 



#6
Nov1209, 11:26 PM

P: 24

Okay thanks a bunch. :)



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