# What actually happens when a rod starts to rotate?

Consider a rod in the absence of gravity. If we were to give the left end of the rod a perpendicular nudge in the upwards direction, it spins about its centre of mass.

If we were to analyse what is happening on a molecular scale, we find that the molecules in the vicinity of the nudge had moved upwards, pulling the molecules down the rod in the same direction. This continues down the rod, resulting in the rod spinning in clockwise.

What I do not understand is why the upward motion of the molecules ceases when we reach the centre of gravity, and the molecules on the right side of the centre of gravity in turn move downwards. In order for the molecules to move downwards, mustn't there be a downward force exerted on the right side of the rod?

In this situation, shouldn't the effect of the force be transferred down the rod, resulting in the rod spinning about its right end? Why does the effect stop at the center of mass of the rod, and then reverse directions? I have tried looking for an answer but has yet to find one.

• You should correct it to "Centre of mass". In the absence of gravity, you cannot define the centre of gravity. – Mechanic7 Oct 22 '18 at 8:08
• You "can't understand it" because what you assumed is actually wrong. If you nudge one end of the rod upwards, the center of mass will move upwards, in exactly the same way as if you nudged the CM instead of the end. If you want the rod to rotate about its CM without moving the CM, you have to nudge one end upwards and the other end downwards. – alephzero Oct 22 '18 at 9:20
• The place to find "an answer" is a book or website on classical (Newtonian) mechanics, in particular the section on the motion of rigid bodies. – alephzero Oct 22 '18 at 9:23
• If the CM moves, then the object is undergoing translational motion. But since our force was directed perpendicular to the rod, shouldn't it only increase the angular momentum of the rod and not the linear momentum? – Luo Zeyuan Oct 22 '18 at 9:42
• I do actually have a couple of introductory texts, but these books cover rotation in a rather mathematical way. I am trying to get an intuitive sense of these mathematical equations. – Luo Zeyuan Oct 22 '18 at 9:49

## 1 Answer

When you apply a force on an end of the rod, the rod will simply undergo translational motion. To rotate the rod, you have to apply a couple. The torque on the rod is given by $$r\times F$$ where $$r$$ is the perpendicular distance between the two forces. So in this case, the total force on the rod is zero and hence the rod doesn't translate. But since a torque is acting on the rod, it increases its angular momentum and hence the rod rotates. When you start approaching the centre of mass, the torque keeps on decreasing till it becomes completely zero at the centre of mass as $$r = 0$$

• What if we consider a rod pivoted at the CM? If we were to nudge the rod on one side, then it will rotate and is no longer in equilibrium. In this case, only one force was needed to get the rod to rotate. Is this because the pivot exerts a normal force? So if we were to consider the rod in space, to get it rotating without any translation, we have to apply a force at one end and an equal and opposite force at the CM? – Luo Zeyuan Oct 22 '18 at 10:06
• @LuoZeyuan Yes, but in this case, the torque will be less. – Mechanic7 Oct 22 '18 at 10:10
• Lesser than if the force was applied at both ends of the rod? – Luo Zeyuan Oct 22 '18 at 10:12
• @LuoZeyuan Yes because the perpendicular distance between the forces has halved. – Mechanic7 Oct 22 '18 at 10:13
• I see. What if we apply an upward force of 5N on one end of the rod and a downward force of 10N on the other end (given that the rod is in space). Do we calculate the torque as 2(5*length of rod/2)? And does the remaining 5N of force on side in turn increase the linear momentum of the rod? – Luo Zeyuan Oct 22 '18 at 10:35