Rigid body, torque and moment of inertia

In summary, the speaker is having trouble with the angular velocity in their rigid body simulator. They suspect they may have forgotten something basic and are using a long rectangular box with uniform density as their setup. They apply a force at one end of the box and use Euler integration to find the linear and angular velocities. They calculate the moment of inertia using the formula I = (1/12)m(w^2 + h^2) and find that for a long stick, the moment of inertia is much larger than the mass, resulting in a small angular acceleration. The speaker is unsure if their analysis is correct and is seeking clarification. They mention using tangential forces, but later correct themselves and say they meant orthogonal forces. The other speaker points out
  • #1
Lord Crc
343
47
Hi,

it's been quite some years since I took some entry-level mechanics at uni, and my book is no longer with me, so I'm struggling a bit here.

I'm making a very simple rigid body simulator just for fun, but the angular velocity is not working out right. I suspect I've forgotten something crucial and basic. My setup is as follows: I've got a long rectangular box (stick) with dimensions w x h x d, and which has uniform density. The center of mass of the box is [itex]\vec{x}[/itex], and the orientation is given by the rotation angle [itex]\phi[/itex]. I use simple Euler integration (for now) to find the linear velocity [itex]\vec{v}[/itex] and angular velocity [itex]\omega[/itex].

I apply a force [itex]\vec{F}[/itex] at some point [itex]\vec{r}[/itex] at one end of the stick. Here both [itex]\vec{F}[/itex] and [itex]\vec{r}[/itex] are in the local (unrotated) frame centered at the center of mass of the box.

For the linear part I find the rotated force vector [itex]\vec{F}'[/itex] by rotating [itex]\vec{F}[/itex] by [itex]\phi[/itex], and then integrate [itex]\vec{x}_{t + \Delta t} = \vec{x}_t + \Delta t \vec{v}_t[/itex] and [itex]\vec{p}_{t + \Delta t} = \vec{p}_t + \Delta t \vec{F}'[/itex], and recover the linear velocity [itex]\vec{v}_{t + \Delta t} = \frac{1}{m} \vec{p}_{t + \Delta t}[/itex].

To find the torque I first find [itex]\tau = \| \vec{r} \times \vec{F} \|[/itex]. Then I integrate [itex]\phi_{t + \Delta t} = \phi_t + \Delta t \omega_t[/itex] and [itex]L_{t + \Delta t} = L_t + \Delta t \tau[/itex]. I then recover the angular velocity [itex]\omega_{t + \Delta t} = I^{-1} L_{t + \Delta t}[/itex] for use in the next timestep.

I calculate the moment as inertia as [itex]I = \frac{1}{12}m(w^2 + h^2)[/itex], where the mass is given as [itex]m = \rho w h d[/itex] and [itex]\rho[/itex] is the density of the box.

The issue I have is that for a long stick the moment of inertia is so large compared to the mass (due to the [itex]w^2[/itex] [itex]h^2[/itex] terms) that even though I apply a constant tangential force at the end of the stick ([itex]\vec{F}[/itex] stays constant in the local, unrotated frame) the stick hardly rotates. Instead the linear velocity increases and the stick translates as if I had applied the force very close to the center of mass.

I assume I've done some silly above, and would really appreciate if someone could point it out.
 
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  • #2
Ok so stepping back a bit, assume I have a thin rod of uniform density and length [itex]\ell = 2r[/itex] (that is, [itex]r[/itex] is the distance from center of mass to either endpoint).

For the linear acceleration I have
[tex]\vec{v} = \frac{1}{m}\vec{p} \\
\frac{d\vec{v}}{dt} = \frac{1}{m}\frac{d\vec{p}}{dt} \\
\frac{d\vec{v}}{dt} = \frac{1}{m}\vec{F}[/tex]

Assuming the force is always tangential to the the length of the rod and that I apply the force at one end of the rod then [itex]\tau = rF[/itex] where [itex]F = \|\vec{F}\|[/itex]. The moment of inertia is [itex]I = \frac{1}{12}m\ell^2 = \frac{1}{3}mr^2[/itex]. Using this the angular acceleration becomes, as far as I can gather, the following:
[tex]\omega = I^{-1}L \\
\frac{d\omega}{dt} = I^{-1}\frac{dL}{dt} \\
\frac{d\omega}{dt} = I^{-1}\tau \\
\frac{d\omega}{dt} = I^{-1}rF \\
\frac{d\omega}{dt} = \frac{3}{mr^2}rF \\
\frac{d\omega}{dt} = \frac{3}{mr}F[/tex]

Then if the above is correct, this means for a fixed force [itex]F[/itex] the linear acceleration will go as [itex]1/r[/itex] as I increase [itex]r[/itex], but the angular acceleration will go as [itex]1/r^2[/itex]. Thus for a sufficiently long rod the angular acceleration will vanish. So, is this a correct analysis or am I doing it wrong?

The results feels very counter intuitive to me, but it wouldn't be the first time my gut feeling is wrong when it comes to physics :)
 
  • #3
A problem with your first post is that you assume you can describe the orientation of rigid body with one angle, and that the change of that one angle is angular velocity. That is not true motion is not planar; you need more angles and their relationship with angular velocity is quite complex.

In your second post, you fared better, except in two places: you said "tangential" where you should have said "orthogonal", and your assessments of acceleration as functions of ##1/r## are wrong.
 
  • #4
voko said:
A problem with your first post is that you assume you can describe the orientation of rigid body with one angle, and that the change of that one angle is angular velocity. That is not true motion is not planar; you need more angles and their relationship with angular velocity is quite complex.

Which other angles are you referring to? My external forces will be strictly in the xy plane.

voko said:
In your second post, you fared better, except in two places: you said "tangential" where you should have said "orthogonal", and your assessments of acceleration as functions of ##1/r## are wrong.

Use of tangential is probably due to my 3D graphics background, but yes I meant orthogonal to the length of the rod. I'd be very happy if you could shed some light on where I'm making the mistake w.r.t. the acceleration.

Cheers
 
Last edited:
  • #5
Lord Crc said:
Which other angles are you referring to? My external forces will be strictly in the xy plane.

Are you saying that your motion is essentially planar? That was not evident from your description. If that is the case, then you can use a single angle and all the simplicity coming with that.

Use of tangential is probably due to my 3D graphics background, but yes I meant orthogonal to the length of the rod. I'd be very happy if you could shed some light on where I'm making the mistake w.r.t. the acceleration.

Your final equation relates angular acceleration with force via ##1/r##, yet you spoke of ##1/r^2##, which is definitely wrong.
 
  • #6
voko said:
Are you saying that your motion is essentially planar? That was not evident from your description.

Sorry, my bad. I rewrote the post a couple of times, must have accidentally deleted that part :(

voko said:
Your final equation relates angular acceleration with force via ##1/r##, yet you spoke of ##1/r^2##, which is definitely wrong.
The rod has uniform density so the mass would increase with r as well, no? But that shouldn't really matter I guess as long as the angular acceleration has "one extra" [itex]1 / r[/itex] factor compared to linear acceleration?

Cheers
 
  • #7
Lets go back to your post #1. Since the problem is planar, I suggest that you stop treating your object as a 3D box, but treat it as a rectangular plate, and adjust your equations correspondingly.
 

1. What is a rigid body?

A rigid body is a theoretical model used in physics to describe an object that does not deform or change shape when subjected to external forces. It is made up of an infinite number of infinitesimal particles that are all fixed in position relative to each other.

2. What is torque?

Torque is a measure of the rotational force applied to a rigid body. It is calculated by multiplying the force applied to the body by the distance from the axis of rotation to the point where the force is applied.

3. How is torque related to angular acceleration?

According to Newton's second law of motion, the net torque applied to a rigid body is equal to the product of its moment of inertia and its angular acceleration. This means that the greater the torque applied to a body, the greater its angular acceleration will be.

4. What is moment of inertia?

Moment of inertia, also known as rotational inertia, is a measure of an object's resistance to changes in its rotational motion. It is calculated by summing the products of each particle's mass and its squared distance from the axis of rotation.

5. How is moment of inertia different from mass?

Moment of inertia is often referred to as the rotational equivalent of mass. While mass measures an object's resistance to linear motion, moment of inertia measures its resistance to rotational motion. Objects with a higher moment of inertia will require more torque to achieve the same angular acceleration as objects with a lower moment of inertia.

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