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

In summary, the conversation discusses the application of Newton's second law for torque to the center of mass of an object, specifically in the example problem in Tipler & Mosca's 5th edition textbook. The book assumes that the law holds for the center of mass, but only explicitly states it after the example problem is solved. The conversation also delves into the concept of rotation around the center of mass and the need to integrate to calculate the moment of inertia for an off-center axis. It also mentions the difference between a wheel and a sphere in terms of rotation and friction.
  • #1
walking
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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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  • #2
Bump please
 
  • #3
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.
 
  • #4
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
 
  • #5
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.
 
  • #6
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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What is torque?

Torque is a measure of the twisting force that causes an object to rotate around an axis. It is calculated by multiplying the force applied to an object by the distance from the axis of rotation to the point where the force is applied.

How is torque related to centre of mass?

The centre of mass is the point at which the weight of an object is evenly distributed. When a force is applied to an object at its centre of mass, it will not rotate. However, if the force is applied at a different point, it will create a torque and cause the object to rotate around its centre of mass.

What is the difference between static and dynamic equilibrium?

In static equilibrium, an object is at rest and the forces acting on it are balanced. This means that the object is not rotating. In dynamic equilibrium, an object is in motion but the forces acting on it are still balanced, resulting in a constant speed and direction of motion.

How is centre of mass calculated?

The centre of mass can be calculated by finding the weighted average of the positions of all the particles that make up an object. This can be done by dividing the sum of the mass of each particle multiplied by its position by the total mass of the object.

Why is centre of mass important in physics?

The centre of mass is important in physics because it helps us understand the motion of objects. It allows us to predict how an object will move when forces are applied to it and to determine the stability of an object. It is also used in many engineering applications, such as designing structures and vehicles.

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