(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Rotation with Static Friction

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