How to get the linear and angular acceleration generated by a force vector field?

I am working on a physics simulation and I have to calculate the angular acceleration in degrees per seconds squared around the point on the object located relatively to the center of a vector field (0,0) and the linear acceleration in Meters per seconds squared. Every vector in the vector field represents a force in Newton pointing in the direction of the vector. This is based in a side viewed 2-D world. Hope you have a way to do it.

Let exist a generic 2D space, and there is a 2D object in that space, called $S$. The point $\mathbf r = (x, y, z)$ is a generic point of the space, while $\mathbf r' = (x', y', z')$ is a generic point inside $S$. Let a force field $\mathbf F(\mathbf r, t)$, which means, at each point $\mathbf r$, in a given time $t$, there is a force $\mathbf F$, given by $\mathbf F = \mathbf F(\mathbf r, t)$. Since it is 2D, our vectors lies inside a plan. In a given point $\mathbf r'$ the object $S$ has a density $\sigma = \sigma(\mathbf r')$.

To make things more general, let's think in a real field $\mathbf E(\mathbf r, t)$. The field is linearly related to the force by parameter $q$: $$\mathbf F = q\mathbf E = m\frac{d^2\mathbf r}{dt^2}$$

Where $q$ is a field parameter inherent to the object $S$. For example, if $\mathbf E$ is an electric field, then $q$ will be the total charge of the object $S$. If $\mathbf E$ is an gravitational field, then $q$ is the total mass of the object $S$. The density $\sigma$ is surface density of the field parameter. Hence, if $\mathbf E$ is an electric field, $\sigma$ is the surface charge density of object $S$ in a given point $\mathbf r'$. If $\mathbf E$ is an gravitational field, $\sigma$ is the surface mass density of object $S$ in a given point $\mathbf r'$.

Notice: $$q = \iint_S dq = \iint_S \sigma(\mathbf r')dS$$

Where $q$ is the total charge, or the total mass, etc, of the object.

Linear Dynamics

Now, let's go to the dynamics. The linear motion: $\mathbf F = q\mathbf E$. Therefore, the force of the whole object, is an integration: $$\mathbf F = \iint_S\mathbf Edq = \iint_S\mathbf E(\mathbf r', t)\sigma(\mathbf r')dS$$

The force then, can be related to its linear acceleration $\mathbf a$ (m/s$^2$) using: $$\mathbf F = m\mathbf a = m\frac{d^2\mathbf r}{dt^2}$$

Rotational Dynamics

It will rotate around their center of mass, which can be computed with an integration around the whole object $S$: $$\mathbf R_{cm} = \frac{1}{q}\iint_S\mathbf r'\sigma(\mathbf r')dS$$

Since $S$ is two-dimensional, then this will be a surface integral. Same way, the torque of the object when rotated around the center of mass: $$\mathbf \tau = \iint_S (\mathbf r' - \mathbf R_{cm})\times\mathbf E(\mathbf r', t)\sigma(\mathbf r')dS$$

If you know the shape of the object and its density, you can do the integration manually and then plug the result in the computer. I recommend this method. If you don't know the objects shape, you can do the integration numerically (which will take time, and often not suited for a real time simulation). The torque and the angular aceleration $\alpha$ (rad / sec$^2$) are related. So as the angular momentum and the angular speed $\omega$ (rad / sec). $$\tau = \frac{dL}{dt},\quad\quad\quad \tau = I\alpha,\quad\quad\quad L = I\omega, \quad\quad\quad \alpha = \frac{d\omega}{dt}$$

where, $I$ is the inertia tensor of the object. Since we are all in 2D here, the angular momentum and the torque will be in a single direction: The direction perpendicular to the 2D plan everything happen. And the inertia tensor will collapse to a single number: The moment of inertia when rotation axis is at the center of mass is calculated: $$I = \iint_S\mathbf (\mathbf r' - \mathbf R_{cm})^2\sigma(\mathbf r')dS$$

• may I know how to get the torque in degrees per second too? I have an unclear idea how to find it. How do I do it if the component is made of smaller parts with different densitys? – Wicpar Dec 23 '14 at 15:06
• $\omega$ is the measure what you want, in rad per second, which you can convert to degree per second. – Physicist137 Dec 23 '14 at 15:10
• wow! thanks for the update, you just saved me tons of work... I was trying to get ω but now you give me α all is way easyer because I do not have to reverse derivate τ... – Wicpar Dec 25 '14 at 11:54
• May I know a bit more about some variables, especially in F(r',t) wich I do not exactly understand, are they geometrical coordinates? – Wicpar Dec 25 '14 at 11:57
• I have edited to bring more details about $\mathbf F(\mathbf r', t)$. And to organize a little bit more =). And made a review. – Physicist137 Dec 25 '14 at 14:28