# Conservation of Angular Momentum about the Instantaneous centre of rotation

While solving some problems on rotational kinematics involving ,say, a cylinder rolling without slipping against a rough surface, I used conservation of angular momentum about the instantaneous centre of rotation.

This gives the same result as applying the equations of kinematics on the rigid body.

My question is, why can we do this over a finite time interval if the instantaneous centre keeps moving?

• Well what if you we moving along with the rolling object? Does anything change? Commented May 5, 2019 at 10:25

An applied torque about a point P is equal to the time rate of change of angular momentum about P under the following conditions:

a) P is at rest in an inertial frame.

b) P is the centre of mass of the object.

c) P is moving parallel to the centre of mass.

It is not true if P is moving in any other manner.

There are many different ways to see that this would work.

The simplest, sidestepping way, is to consider, say, a cylinder rolling down an inclined plane. You can write down the relevant energies, i.e. linear KE, rotational KE and GPE, and see that the time derivative of the sum of these three terms gives rise to a zero as long as the no-slipping condition is true. This means that you can have linear and rotational acceleration and yet there will be no problems involved in using conservation concepts.

But I am sure you are more interested in a direct explanation. So let us consider that you have a force applied tangentially upon a cylinder. At the centre of mass of the cylinder, add a zero by adding and subtracting a force, of same magnitude and direction to the force tangential to the surface of the cylinder. You are always allowed to add a zero any time you like, to any mathematical system whereby addition is possible. Now, there is a force parallel to the surface tangential force, but acting through the centre of mass (or any pivoting point, whichever is more convenient). This force thus generates a pure linear motion for the centre of mass, no rotation whatsoever. The leftover is a couple---two anti-parallel forces of the same magnitude, i.e. no net force, but separated by a perpendicular distance. A couple causes pure rotation, and you can see that it does not depend upon where you pick the pivot, its contribution to rotation will be the same.

This means that we have successfully decoupled the linear motion from the rotational motion, and thus whatever you can prove about the angular momentum's behaviour, will hold true in general. Then you can see that it is ok that the instantaneous centre of rotation is moving; the conserved quantity does not care about where that point actually is.

There is an ugly story about this topic, but I don't feel like sharing it right away. It should be a sufficient solution for you as this currently is.

See the thing is you actually conserved the momentum about a single point on the ground which made u think that u are conserving the momentum about the instantaneous centre on the ground. Just try to conserve the momentum about a single point on the ground then u will get a conservation of momentum equation which happened to be coincidentally your conservation of momentum equation about instantaneous centre of rotation that is why conserving the momentum about instantaneous centre of rotation is working in this case