Given a force in 3d what is the resulting force and torque?

[SirSpence]

I'll preface this by saying I've done these sorts of problems before (though it was only in 2d) in my statics and dynamics class. I have no idea where my book went and the major problem I'm dealing with is vocabulary, so if I say something that doesn't make sense, please let me know, I've almost completely forgotten the specific terms. The units are also entirely arbitrary. I'm not trying to solve a specific problem so much as trying to create a program that will calculate the answer.

Given an object that is not experiencing gravity, if a force is applied (1,1,1) away from the COM and the force is a vector of (1,1,0.5) what is the resulting vector of force that the object experiences and how much torque does it experience along each axis. For ease of math, let's say the object has a mass of 1. I am assuming the object is a solid cube such that the COM is in the center of the cube and rotated such that each corner of the cube is at, for example (1,1,1) or (-1,1,1) No need to deal with the cube being rotated strangely.

I need to know this so that I can figure out how much the cube moves and rotates, I will need to convert the torque into rpms.

If I can get help even with just figuring out the name of the type of problem I am trying to solve, that would be great!

Otherwise, if someone wants to help me solve it, what information am I missing? (Other than the units, you can pick and choose whatever.)

Thanks!

 

[jbriggs444]

The "vector of force" that the object experiences is the full force. If the force is (1,1,0.5) then the rate of change of linear momentum of the block due to the force will be (1,1,0.5). There is no reduction in force due to distance of the point of application from the center of gravity.

In two dimensional formulations of angular momentum, torque depends on the applied force (in the plane) and on the perpendicular distance of the point of application from a chosen reference axis. Often one will choose the center of mass of a planar object to define the reference axis.

In the three dimensional formulation, torque depends on the applied force and upon a chosen reference point. It is the vector cross product of the position of the point of application (conventionally labelled r) and the force itself (conventionally labelled F).

$$\vec \tau = \vec r \times \vec F$$

This torque is a vector quantity. Technically it is a "pseudo-vector" due to technical details that we need not get in to. It points perpendicular to both the force and the displacement -- it is aligned with the axis about which rotation will tend to occur.

The three components of torque correspond to the twisting tendency about the three axes, x, y and z.

Torque also gives the rate of change of angular momentum. If you multiply a constant torque by the time for which it was applied you get the angular momentum imparted by that torque.

Angular momentum is a vector quantity. The three components roughly correspond to rotation about the three axes.

It would be nice to think that angular momentum is equal to rotation rate times a scalar moment of inertia. However, in three dimensions life is not that simple.

[here my training in three dimensional rotation leaves off. Now I begin relating what I think I know instead of what I actually do know]

In three dimensions, moment of inertia is a tensor, not a scalar. You can think of this as a consequence of the fact that the planar moment of inertia (which is a scalar) of an object depends on the axis you choose through that object.

Even knowing their angular momentum, three dimensional objects do not always rotate at well defined rates about each axis. There are some cool videos of "torque free rotation"

 

[DrClaude]

Given an object that is not experiencing gravity, if a force is applied (1,1,1) away from the COM and the force is a vector of (1,1,0.5) what is the resulting vector of force that the object experiences and how much torque does it experience along each axis. For ease of math, let's say the object has a mass of 1. I am assuming the object is a solid cube such that the COM is in the center of the cube and rotated such that each corner of the cube is at, for example (1,1,1) or (-1,1,1) No need to deal with the cube being rotated strangely.

If you are asking for a specific case, you will have to do this in the homework forum with an attempt at a solution.

Only the general principles of the physical situation can be discussed in this tread.

 

[SirSpence]

The "vector of force" that the object experiences is the full force. If the force is (1,1,0.5) then the rate of change of linear momentum of the block due to the force will be (1,1,0.5). There is no reduction in force due to distance of the point of application from the center of gravity.

Surely some force is transferred into angular rotation? If there was no rotation as a result of the force being in line with the COM, I would understand this.

In the three dimensional formulation, torque depends on the applied force and upon a chosen reference point. It is the vector cross product of the position of the point of application (conventionally labelled r) and the force itself (conventionally labelled F).

→τ=→r×→F​

\vec \tau = \vec r \times \vec F
This torque is a vector quantity. Technically it is a "pseudo-vector" due to technical details that we need not get in to. It points perpendicular to both the force and the displacement -- it is aligned with the axis about which rotation will tend to occur.

When I have some more time I will get back to you. As it stands, I need sleep desperately, which is a great combination when it comes to figuring out math.

I wrote out my confusion about cross products being applicable and then I realized, the value of the vector is rotation around an axis, not rotation in that direction. That makes what you wrote make sense.

That leaves my followup question.

Now that I have a vector for rotation, how do I convert that into rpms (or degrees per second) given I have the total mass of the object and the point at which the torque is being applied?

Thank you for your help!

If you are asking for a specific case, you will have to do this in the homework forum with an attempt at a solution.

Only the general principles of the physical situation can be discussed in this tread.

I wish! I could probably get away with using a calculator if that was the case. I was only providing that example in an attempt to help explain what I am trying to do. If I was more confident that I could explain it in a way that people would understand, I wouldn't have brought up the example at all.

 

[jbriggs444]

Surely some force is transferred into angular rotation?

The contribution to angular rotation does not deduct from the contribution to linear momentum. The two are independent.

Now that I have a vector for rotation, how do I convert that into rpms (or degrees per second) given I have the total mass of the object and the point at which the torque is being applied?

You apply force at a point. The offset from the reference point tells you how much torque that amounts to.

Torque does not have a point of application. You already factored the point of application in when you computed the torque.

If your torque vector ($\tau$) is constant, you can multiply by the time ($\Delta t$) for which it is applied to get the resulting change in angular momentum ($\Delta L$)
$$\Delta L = \tau \Delta t$$
If you make the [false for a cube but true for a uniform sphere] simplifying assumption that the moment of inertia of the object is the same about every axis then angular momentum ($L$) is given by moment of inertia ($I$) times angular rotation rate ($\omega$).
$$L=I \omega$$
The angular rotation rate is conventionally expressed in radians per unit time. That convention keeps factors of pi and 180 from showing up in the formulas.

 

[SirSpence]

Sweet! It appears I have it working.

I'm currently getting the cross product of the force and the point of application, (relative to the COM) dividing the result by the mass and then multiplying the result by the elapsed time. I am then able to use the resulting vector to calculate the change in rotation.

I am not looking forward to calculating the moment of inertia, fortunately the approximation is sufficient for my needs at the moment.

Thank you for your help!

 

[jbriggs444]

I am not looking forward to calculating the moment of inertia

Sad to say, without having calculated a moment of inertia, the magnitude of your result is meaningless. It has the wrong units.

 

[SirSpence]

Yeah, fortunately I can get away without doing more than treating it as a sphere for now.