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 T-F_{friction} [/tex] [tex]\frac{1}{2}(T-F_{friction})R=Tr+F_{friction}R [/tex] Rearrange [tex]TR-2Tr=2F_{friction}R+F_{friction}R[/tex] I get this: [tex]T(R-2r)/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 T-F_{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.
You assumed in your first equation that the the direction of friction is opposite to the direction of the pull. You have written the torque with respect to the center of mass, didn't you? than the torque of both forces point in the same clock-wise direction. ehild
Thanks for the reply. Yes I did write the equations with respect to the center of mass, but didn't realize my mistake. So if I change the - to a + sign, then the rest is correct? [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.
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 T-F_{friction} [/tex] [tex]\frac{1}{2}(T-F_{friction})(r^2+R^2)(\frac{1}{R})=Tr+F_{friction}R [/tex] Rearrange [tex]TR^2 +Tr^2-F_{friction}R^2-F_{friction}r^2=2TrR+2F_{friction}r^2[/tex] [tex]T(R^2-2TrR+r^2)=3F_{friction}R^2+F_{friction}r^2[/tex] And I end with this: [tex] \frac{T(R-r)^2}{(3R^2+r^2)}=F_{friction} [/tex]