Torque <-> rotational velocity?

[Hnefi]

Hello. I'm having a bit of a problem. I need to calculate a change in rotational velocity on a rigid object given a force (actually an impulse, but nevermind) acting on that object.

I can calculate the torque without problem, by doing the cross product of the force vector and the offset vector, the offset being the difference between the point on which the force is applied and the centre of mass for the object. Fine. But I need to translate that into rotational velocity, or rather change in rotational velocity.

The object has a known mass and the rotation must be calculated in global Euler angles (as in it doesn't matter how the object is rotated or what its current angular velocity is). How do I do this? I tried projecting the force vector onto the cross product between the offset vector and the torque, but that didn't work. I've been searching the 'net, but for some reason the answer to this question is pretty darn hard to find.

Thanks in advance for any help.

 

[Hootenanny]

Hello. I'm having a bit of a problem. I need to calculate a change in rotational velocity on a rigid object given a force (actually an impulse, but nevermind) acting on that object.

I can calculate the torque without problem, by doing the cross product of the force vector and the offset vector, the offset being the difference between the point on which the force is applied and the centre of mass for the object. Fine. But I need to translate that into rotational velocity, or rather change in rotational velocity.

The object has a known mass and the rotation must be calculated in global Euler angles (as in it doesn't matter how the object is rotated or what its current angular velocity is). How do I do this? I tried projecting the force vector onto the cross product between the offset vector and the torque, but that didn't work. I've been searching the 'net, but for some reason the answer to this question is pretty darn hard to find.

Thanks in advance for any help.

Hi Hnefi and welcome to PF,

Am I missing something here, or isn't this just a simple application of,

\sum_i \boldmath{F}_i\times\boldmath{r}_i = \mathbb{I}_G\frac{d}{dt}\boldmath{\omega}

 

[Hnefi]

Ack, you're right. I overengineered the problem. Simply applying the torque directly did the trick. Thanks for your help.

 

[D H]

Hi Hnefi and welcome to PF,

Am I missing something here, or isn't this just a simple application of,

\sum_i \boldmath{F}_i\times\boldmath{r}_i = \mathbb{I}_G\frac{d}{dt}\boldmath{\omega}

Not quite that simple. In inertial coordinates, the correct expression is

\sum_i \mathbf{F}_i\times\mathbf{r}_i<br /> = \frac{d\mathbf L_I}{dt}<br /> = \frac{d}{dt}\left({\mathbb{I}}_I\,\mathbf{\omega}_I\right)

where the subscript I on the angular momentum \mathbf L, inertia tensor \mathbb{I} and angular velocity \boldmath{\omega} indicate that the quantities in question are to be expressed in terms of a non-rotating (i.e. inertial) frame. The problem is that the inertia tensor for a rigid body as observed from an inertial frame is not constant. The inertia tensor for a rigid body is constant in a body-fixed frame. It is much more convenient to do the calculations in the body-fixed (i.e., rotating) frame. However, this means that one must introduce a fictitious torque.

The transport theorem relates the time derivative of some vector quantity \mathbf q from the perspective of an inertial observer and a body-fixed observer:

\frac{d \mathbf q}{dt_I} = \frac{d \mathbf q}{dt_B} + \mathbf{\omega} \times \mathbf q

With \mathbf q = \mathbf L, this becomes

\frac{d \mathbf L}{dt_I} = \mathbb{I} \frac{d \mathbf{\omega}}{dt_B} + \mathbf{\omega} \times (\mathbb{I} \,\mathbf{\omega})

Combining with the first equation,

\frac{d \mathbf \omega_B}{dt}<br /> = \mathbb{I}_B^{\;-1}\left(\sum_i \mathbf{F}_{i_B}\times \mathbf{r}_{i_B} - \mathbf{\omega}_B \times (\mathbb{I}_B\, \mathbf{\omega}_B)\right)

Here the subscripts B denote that the quantities in question, including external forces, are to be expressed in body-fixed coordinates.

 

[Hootenanny]

I stand corrected DH. In all honesty I thought the OP mentioned body angles, but I see that I was mistaken.

I doff my cap to you sir, nice post!