# Rigid Body Dynamics for Dummies

Hopefully this writeup can help demystify rigid body dynamics for those who have a hard time remembing their first undergraduate M.E. class.

## Time derivative of a rotation matrix

Let’s start with the basics, the same vector expressed in two coordinate frames. We’ll call x_b the position of a point in some body frame and x_i that same point but expressed in an inertial frame. They are related by a matrix which converts body coordinates to inertial ones, R_b2i

x_i = R_b2i * x_b


Okay, nothing too crazy yet. Let’s also assume x_b is a fixed point on our rigid body, which is to say it is not moving in the body frame.

Now, what happens when we take a time derivative of x_i?

x_i_dot = R_b2i * x_b_dot + (R_b2i)_dot * x_b


But remember, x_b does not move in the body frame. This means x_b_dot = 0.

eq1

x_i_dot = (R_b2i)_dot * x_b


Now we just need to figure out what to make of the time derivative of a rotation matrix, (R_b2i)_dot.

Recall the simple relationship between linear and angular velocity:

v = omega X r


We can use this equation to find the inertial velocity of x_b expressed in the body frame:

x_b_dot_inertial = omega_b X x_b


Where omega_b is the angular velocity of the body frame expressed in the body frame. Now to convert x_b_dot_inertial into x_i is a simple matter:

x_i_dot = R_b2i * x_b_dot_inertial = R_b2i * omega_b X x_b


We can further take advantage of the skew symmetric matrix and define a matrix omega_b_cross which has the property:

omega_b_cross * r = omega_b X r


Putting this all together we have:

eq2

x_i = R_b2i * omega_b_cross * x_b


Now we can combine eq1 and eq2 to solve for (R_b2i)_dot

eq3

(R_b2i)_dot = R_b2i * omega_b_cross


## Newton-Euler equations

The Newton-Euler equations describe the rotation and translation of a rigid body under the influence of a force and moment. With the math described in the last section figuring them out is a simple matter.

We must first remember that momentum (angular and linear) is only conserved in an inertial frame. If we were interested in equations that worked in an inertial frame we’d be done and could resort to the simple equations everyone learns in highschool physics

F = m * a
tau = I * omega_dot


Unfortunately the inertial frame is often a very inconvenient place to work in. For example, the moment of inertia tensor I will very likely be changing as our rigid body spins in an inertial frame. Let’s see if we can express the same type of equations but in the more convenient body frame.

Let’s begin with conservation of momentum using body frame velocities which are rotated into the inertial frame by R_b2i

p_i = R_b2i * m * v_b
h_i = R_b2i * I * omega_b


Now we can take a time derivative and use the fact that the time derivative of linear and angular momentum are defined as force and torque, respectively

F_i = R_b2i * m * a + (R_b2i)_dot * m * v_b
tau_i = R_b2i * I * omega_b_dot + (R_b2i)_dot * I * omega_b


And using eq3

F_i = R_b2i * m * a + R_b2i * omega_b_cross * m * v_b
tau_i = R_b2i * I * omega_b_dot + R_b2i * omega_b_cross * I * omega_b



Now we simply multiply both sets of equations by R_i2b

eq4

F_b = m * a + m * omega_b X v_b
tau_b = I * omega_b_dot + omega_b X I * omega_b


And there you have it, the equations which relate linear and angular acceleration expressed in the body frame with forces and torques which are also expressed in the body frame.

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Written by David Friedman on 05 March 2017