Torque and Angular Momentum - Origin Misconception

[Lost1ne]

Homework Statement

(Problems/diagrams referenced are attached as images.)

Homework Equations

Net torque about an origin = time derivative of the angular momentum vector about the same origin.

The Attempt at a Solution

I've solved these problems before, but I'm now looking back at them and questioning my understanding of them.

I've noticed how many problems are illustrated with what seems to be a cross-section of the object that we are analyzing in that we draw the problem such that our origin (that we self-select for calculating angular momentum and torque) and the center of mass of our object are in the same plane.

It seems that the reason for this is to make the problem simpler to solve, but is it always an accurate method to use? Should I view this as an assumption made to make the problem simpler to solve, particularly in the car problem where clearly a 3D vector would have to be used to illustrate the vector from the center of mass to the bottom of the tire where static friction is applied. (Solving this problem in a 2-D world allows you to use "2 feet", given in the problem, as the perpendicular component of this vector, making the torque calculation about the center of mass simpler. However, in 3-D, it seems that this perpendicular vector would be composed of two sub-components, one to get from the center of mass straight to the ground, and another to get from there to the same line where the static friction vector would be found on.)

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As it relates to the pulley image, can we agree that choosing our origin to be in a different plane would most certainly change our calculations? Is my way of thinking correct?: In this scenario, not only is the origin simply in the same plane as the pulley center of mass, but the origin is chosen as the center of mass. This eliminates the need to worry about an orbital angular momentum component which makes this problem (or similar problems where this approach is taken) easier to solve. Drawing and solving our problem in this 2-D manner is then encouraged.

If we chose our origin to be on this axis of rotation but not on the center of mass or not on this axis of rotation at all, we would get a more complex 3-D or 2-D (if the origin is still in the same plane as the center of mass) where we would then have to be concerned with an angular momentum component. (Although, however, in this specific pulley scenario, it seems that the orbital component will still be zero as the center of mass velocity of the pulley is constrained to be zero.)

Edit: The car problem also uses the origin as the center of mass as we elect to sum the torques about the center of mass. I guess I would like to know why we treat the center of mass as if it is in the same plane as the static friction and normal forces. Am I missing something conceptually here, or is this merely an assumption made to simplify the problem?

 

[haruspex]

However, in 3-D, it seems that this perpendicular vector would be composed of two sub-components, one to get from the center of mass straight to the ground, and another to get from there to the same line where the static friction vector would be found on.)

You can decompose the force in several ways, but the most useful is to take advantage of the symmetry of the problem.
It is evident that the forces from the two rear wheels (say) will be mirror images of each other about the central vertical plane through the length of the car. Thus, their torques about that axis (i.e., running up the slope) will cancel; likewise for the axis normal to the plane of the slope. At the same time, their torques about the axis horizontally across the slope, normal to the mirror plane, will be equal and combine. Therefore it suffices to consider forces projected onto that plane.

As it relates to the pulley image, can we agree that choosing our origin to be in a different plane would most certainly change our calculations?

The calculations, but not the answer.
I feel you are looking for a theorem that says all choices will lead to the same answer, therefore choose the most convenient.
Indeed, that can be proved. You might care to attempt it yourself.

 

[Lost1ne]

I feel you are looking for a theorem that says all choices will lead to the same answer, therefore choose the most convenient.
Indeed, that can be proved. You might care to attempt it yourself.

Oh, that the net torque summed about any and all origins is the same if the net force on the system is zero? I didn't think of referencing that. So, if my thinking is correct, this determines that the time-derivative of the angular momentum vector of the pulley analyzed about any origin leads to the same answer. In this pulley scenario, for an origin not at the center of mass, an orbital angular momentum component may be introduced but it will equal zero (as the velocity of the pulley's center of mass is zero), allowing us to simplify our equation as the time-derivative of only the spin angular momentum vector which will be found to be the same about any origin.

 

[haruspex]

the time-derivative of the angular momentum vector of the pulley analyzed about any origin leads to the same answer

The same answer to the underlying problem, yes. (Obviously not the same answer for the torque about the axis if a different axis is being chosen.)
E.g. for the pulley, if we were to choose an axis off to one side of its mass centre then we would have to include the torque due to the force supporting the pulley's axle. We know that force will balance the gravitational force on the pulley and the tensions in the rope, so we can write down what this torque is in terms of those. When the dust clears, the horizontal offset of the chosen axis cancels out and we are back to the original equation.

 

[Lost1ne]

The same answer to the underlying problem, yes. (Obviously not the same answer for the torque about the axis if a different axis is being chosen.)

This leads to another question that I have. When analyzing torques, should they be thought of torques about an axis or torques about a point in space? In the pulley example when the origin is chosen to be at the center of mass and the math simplifies to the net torque is equal to I*α (for I is constant), should I think of α as the angular acceleration of the body about our chosen origin or about the axis that runs through our chosen origin? I think that axis answer would make more intuitive sense, but this seems to contradict what my textbook teaches as torques and angular momentum being defined about single points (origins) in space.

Claiming that an object is rotating about a point in space doesn't seem very clear (rotating about it in what manner?) Defining an axis seems more appropriate. However, defining torques and angular momentum about an origin (point in space) also seems to be more clear when considering $\vec r$ x $\vec F$ and $\vec r$ x m$\vec v$ as $\vec r$ may be thought of as extending from our origin to some particle. If we attempt to define torque and angular momentum in terms of axes, it's unclear of how $\vec r$ should be defined (unless its perpendicular component is used, I guess).

 

[haruspex]

This leads to another question that I have. When analyzing torques, should they be thought of torques about an axis or torques about a point in space? In the pulley example when the origin is chosen to be at the center of mass and the math simplifies to the net torque is equal to I*α (for I is constant), should I think of α as the angular acceleration of the body about our chosen origin or about the axis that runs through our chosen origin. I think that axis answer would make more intuitive sense, but this seems to contradict what my textbook teaches as torques and angular momentum being defined about single points (origins) in space.

Claiming that an object is rotating about a point in space doesn't seem very clear (rotating about it in what manner?) Defining an axis seems more appropriate.

In 3D, a torque should in general be considered about a point. E.g. suppose a body is mounted on a rod attached to a universal joint, and several forces and torques act on the body. You do not know in advance which axis it will rotate around. You need to treat each torque as a 3D vector. Only when they are summed will the rotation axis emerge.
That said, in many cases an axis arises naturally, e.g. as an axle. The axle mounting will provide a torque normal to the axle (as a vector) as necessary to counter any such component of the net torque from the forces of interest. Thus it is only necessary to consider torques about the axle.