Does Newton's Second Law for Torque Work for the Centre of Mass of an Object?

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SUMMARY

The discussion centers on the application of Newton's Second Law for torque to the center of mass of an object, specifically in the context of a rolling ball as presented in Tipler & Mosca's 5th edition. Participants confirm that the equation τnet ext = Iα is valid for the center of mass, as demonstrated through a proof involving the parallel axis theorem. The conversation clarifies that while the moment of inertia is calculated around a fixed axis, the center of mass remains the pivotal point for analyzing rotational motion, especially in frictionless scenarios like billiard balls on a table.

PREREQUISITES
  • Understanding of Newton's Second Law for rotation
  • Familiarity with moment of inertia and its calculation
  • Knowledge of the parallel axis theorem
  • Basic principles of rotational dynamics and rolling motion
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  • Study the derivation of the parallel axis theorem in detail
  • Explore the implications of frictionless surfaces on rolling motion
  • Learn about the differences in moment of inertia for fixed versus off-center axes
  • Investigate real-world applications of torque in rigid body dynamics
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Physics students, educators, and anyone interested in understanding the principles of rotational dynamics and torque, particularly in the context of objects rolling without slipping.

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I am reading Tipler & Mosca 5th edition. On pages 289-290 there is an example problem which seems to assume that Newton's second law for torque works just as well when we apply it to the centre of mass of an object. However, before this example problem was introduced, the authors did not state anything about NII for centre of mass and only derived it for torques in general. After the example problem however, they seem to be saying that NII for torques does indeed work for centres of mass (see yellow box).

Here is the example problem:
tipler 1.png
tipler 2.png


As you can see, in the problem they seem to assume that ##\tau_{net ext}=I\alpha## also holds as ##\tau_{cm}=I_{cm}\alpha##, at least for this particular problem. However, my problem is that they only state that it holds for centres of mass after the example problem, rather than before it. This makes me believe that I am missing something and maybe what they did is just a straightforward application of NII which I am not seeing? Maybe we do not need the theorem they state afterwards (in the yellow box) in order to solve the example problem as they solved it?

In any case, I think I have managed to prove that their application of NII-for-torques to centre of mass of the sphere is valid. Here is my proof:

If we apply NII without assuming it works for centre of mass, we have to use the fact that the ball rolls about an axis in contact with the ground (earlier in this chapter in the book it says that a rolling ball rolls about an axis on the ground). Since the radius of the ball is R, then applying NII for torques we get ##F(x+R)=I\frac{a}{R}## where ##a## is the tangential acceleration. Now by the parallel axis theorem, we have ##I=I_{cm}+MR^2## giving ##F(x+R)=(I_{cm}+MR^2)\frac{a}{R}##. Now by NII for linear motion, ##F=Ma## and so the last equation simplies to ##Fx=I_{cm}\alpha##, which is NII-for-torques applied to the centre of mass of the ball, as required.
 
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An object floating freely in space will rotate around its center of mass.

In your billiard ball example, the rotation is around the center of mass, too.

Newton's second law for rotation is

torque = moment of inertia × angular acceleration.

The moment of inertia is calculated around a fixed axis attached to the object. In your billiard ball example, the axis goes through the center of mass. That is why the book uses the moment of inertia calculated around the center of mass of the ball.

---

If you attach a fixed axis which does not go through the center of the ball, then the moment of inertia is different. You need to integrate to calculate the moment of inertia for an off-center axis.
 
Heikki Tuuri said:
An object floating freely in space will rotate around its center of mass.

In your billiard ball example, the rotation is around the center of mass, too.

Newton's second law for rotation is

torque = moment of inertia × angular acceleration.

The moment of inertia is calculated around a fixed axis attached to the object. In your billiard ball example, the axis goes through the center of mass. That is why the book uses the moment of inertia calculated around the center of mass of the ball.

---

If you attach a fixed axis which does not go through the center of the ball, then the moment of inertia is different. You need to integrate to calculate the moment of inertia for an off-center axis.
But in this book it says that a rolling wheel will rotate around an axis that is in contact with the ground. Is this only for a wheel and doesn't apply to a sphere? :

https://archive.org/details/PhysicsForScientistsAndEngineersExtendedEdition5TiplerMosca/page/n317
 
The axis is contact with the ground? No. The axis is, of course, at the center of the wheel.

Maybe you mean that the wheel will not slide on the ground?

In the textbook exercise, they look for a solution where the ball rotates at exactly the speed for which it does not slide. The translational speed matches the rotational speed of the edge of the ball.

They assume that the billiard table is frictionless, like ice. You have to hit the ball so that it gets exactly the right rotation. Friction does not help you.

Look at the wheel of a car. The rotation of the wheel exactly matches the translational velocity of the car, so that the rubber tire does not skid on the road. The wheel of a car gets the right rotation speed from friction.
 
The instantaneous motion of a rigid body can be divided into a translation and a rotation in many different ways. No single way is the true fact of the matter. Both rotation around the axle and rotation around the momentary point of contact with the roadway are correct.
 
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