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Suppose we are studying the case of uniform circular motion. The analysis of such motion is usually done using Newton's second law as $ma=m v^2/r$. Even when the motion isn't purely circular, like the case for central force problems, we still use familiar Newton's second law in our analysis. Moreover, the object in question is treated as a point mass.

However, the motion of a body in an orbit can be considered to be a special case of rotational mechanics. Suppose we have a ball spinning about its axis. We analyze this using the rotational analog of Newton's second law - using torques and moment of inertia - $\tau=I\alpha=I d^2\theta/dt^2$. This equation basically tells us how fast would a body spin if we apply a torque to it.

When a body spin about an axis, the different mass elements that constitute the entire body, can be thought of as moving in an orbit about the axis. The necessary centripetal acceleration is provided by the forces that make sure that the body doesn't deform - electrostatic forces I suppose. The torque, therefore, is the manifestation of where we are applying the force on our extended body. Similarly, the angular momentum is describing the momentum of individual mass elements away from the axis.

Is it possible in principle, to analyze the rotational motion of an object purely using force, mass, momentum, etc, instead of torque, angular momentum, and moment of inertia? Something along the lines of finding the forces on each of these small particles, and then integrating over the entire body, and solving the equations of motion to describe how the body as a whole rotates ? I'm sure, during any such analysis, we would be forced to re-invent moment of inertia and torque somewhere in the derivation. However, in principle, is it possible to describe the rotation of a body by using (starting with) $F=ma$, or rather $F=mv^2/r$ ?

In the case of point particles, we do use Newton's laws directly to analyze motion. In the case of extended objects, we use the rotational analogs of our linear concepts. Is it theoretically possible (definitely impractical), to describe the motion of rigid extended bodies, by starting with $F=ma$, and then deriving all the rotational analogs from this starting point?

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    $\begingroup$ The answer is "yes", but that is not long enough to post as an answer. Not only is it possible, it is exactly what we do when we develop rotational mechanics. $\endgroup$ Nov 5, 2021 at 17:53
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    $\begingroup$ If you couldn't, it would seem to imply that rotational mechanics don't obey Newton's laws - there would be some aspect of physics that applies to rotational problems but not linear ones, or vice versa. $\endgroup$ Nov 5, 2021 at 20:51

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As pointed out in a comment: all expressions of rotational dynamics can be constructed from the expressions of linear dynamics.

The only real difference is that linear dynamics is possible with one degree of spatial freedom, whereas dynamics requires a minimum of two spatial dimensions.

For example, there is the linear mechanics classroom demonstration device called 'air track'. The motion is effectively constrained to one spatial dimension.

For demonstration of circumnavigating motion, around some point of attraction of repulsion, an air table is used.


In the case of circular motion there are two ways of decomposing that motion that have practical use. One way is to decompose the motion along the axes of a rectangular grid. Then circular motion can be represented as a linear combination of two harmonic oscillations, 90 degrees out of phase.

Of course the decomposition used most often is according to polar coordinates: radial distance to the center of attraction/repulsion, and angle with respect to some reference line.

In both cases the motion is decomposed in two components. The component motions are at 90 degrees to each other.


Angular momentum

The concept of angular momentum had a precursor: Kepler's law of areas.
The very first proposition in Newton's Principia is a demonstration that Kepler's law of Areas follows logically from the laws of motion as laid down in the Principia.

I represented that derivation in an answer here on stackexchange to a question titled 'Intuition for angular momentum' The elements that go into that derivation are all from linear mechanics.


Point masses versus extended objects.

In mechanics when the motion of a solid object is modeled with an equation what is being tracked is the motion of the center of mass. Whatever the shape of some object A is, if it isn't touching any other objects (hence the motion is not affected) then that object A is effectively treated as a point mass. That is how it is possible to describe an extended object with a single value for its inertia; you treat it as if it is a point mass.

In angular mechanics there is the concept of Moment of inertia. If you have a wheel you can use the moment of inertia of that wheel in a calculation.

That raises the question: there is no way to treat that wheel as a point mass, it is inherently an extended object. How can you assign a single value to the moment of inertia?

The entities of angular mechanics are constructed by capitalizing on rotational symmetry.

A wheel is (for the purpose of the calculation) rotationally symmetric. For simplicity we count only the mass of the rim (treating the mass of the spokes as negligable). For the magnitude of the moment of inertial we treat all of the mass of the rim as if concentrated in a point at some distance 'r' to the center of rotation. This is valid because all that counts for the magnitude of the moment of inertia is the distance to the center of rotation.

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Yes it is possible but very complicated as compared to when done by simple torque and moment of inertia and all. You can take a simple situation and use only newton's three laws of motion and some constraints like for a rigid body you have the constraint that the distances between any two particles will always be same throughout the motion then do some math you'll end up at relations same as that of torque equations. Do try this it will increase your mathematical skills and will give you a broader look at these simple things.

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
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    Nov 5, 2021 at 21:33

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