# Torque on rolling cylinder

Consider a rod with two equal forces acting parallel to the direction of velocity of a rolling rod, one at each end. When analysing problems involving torque, the choice of axis should not matter. However, when we consider the axis as one end of the rod, there seems to be an unbalanced torque at the other end which will cause angular acceleration. However, in reality there is no such acceleration. What causes this discrepancy?

Note: consider 2nd example in image with F1 = F2 and cylinder rolling along long end

• Is this question clear to anyone else? – M. Enns Aug 18 '17 at 11:32
• I think the confusion is coming from very unclear images... – bendl Aug 18 '17 at 14:30
• I think OP may be asking why in the second figure, considering the torque about one end of the rod leads to a net torque on the system, and why this doesn't cause an angular acceleration. – EigenFunction Aug 18 '17 at 16:47
• You are looking at a non-inertial frame, but not accounting for the fictitious forces that appear. physics.stackexchange.com/questions/288712/… – BowlOfRed May 21 at 7:01

Let us take the point through which $F_2(=F_1=F)$ acts in the second figure as the origin O, and let the length of the rod be $L$. Assume the rod is uniform with mass M. Now let us denote the direction of the rod and the direction of the forces as the positive $i-$ and $j-$ vectors respectively. We see that the net force on the system is $2Fj$ and the torque on the system is $\sum r_i \times F_i=Li \times Fj=LF k$. Now we see that $LFk \cdot 2Fj=0$, so the system simplifies to a single force acting through the COM of the rod with no net torque.
Assume we have two forces, $F_1$ and $F_2$, acting at points with position vectors $a(=\vec{OA})$ and $b(=\vec{OB})$ relative to some origin O respectively. Now the torque about the system is $\tau=a\times F_1 + b \times F_2$. Now if this system is a couple, $F_1=-F_2=F$ so $\tau= a \times F -b \times F=(a-b)\times F=\vec{BA} \times F$. This is why the torque of a couple does not depend on the point chosen to be the origin, since $\vec{BA}$ is the position vector of one point relative to the other,and O is not featured in the equation. However, in this case, with $F_1=F_2$, we have $\tau=a \times F+b\times F=(a+b)\times F=(\vec{OA}+\vec{OB})\times F$. Now we see that unlike previously, the torque of this system is actually dependent on the point chosen to be the position vector. However, as I have shown above, this system of forces simplifies to only a translational force; in fact, although I did it for a specific point, I believe it can be generalised to show that no matter what point you take as the origin, the system simplifies to a single force acting through the COM of the rod.