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How to convert angular velocity to rotation matrix?

  1. Oct 17, 2013 #1

    I have two questions related to angular velocity:

    1. According to rotational damper, Torque = Viscous Damping Coefficient * Angular Velocity. This equation gives the unit of Angular Velocity as meter square per second. How is it equivalent to rad per second?

    2. If I have an angular velocity with it's three components ωx, ωy and ωz, how can I get the rotation matrix?

    Thank you.
  2. jcsd
  3. Oct 17, 2013 #2
    I'm not understanding where m^2/s comes from. Angular momentum has units of 1/s, torque has units of N*m. Have you checked your units for the viscous damping coefficient? And where is the moment of inertia? The left hand side should ordinarily be torque/moi.
  4. Oct 17, 2013 #3

    D H

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    1. The rotary damping coefficient is not the same as the linear damping coefficient. The rotary damping coefficient has units of mass*length2/time, or in SI units, newtons*meters*seconds.

    2. This is akin to asking if I have velocity how to I get position, except the rotational analog is much tougher than the relation between velocity and position.

    One approach is to use quaternions. The time derivative of a quaternion is half of the quaternion product of the rotation quaternion and angular velocity as a pure quaternion, possibly negated, and possibly the reverse order (angular velocity times quaternion as opposed to quaternion times angular velocity). Which is which depends on the conventions you adopt.

    the problem becomes a bit closer to that of velocity and position once you have the quaternion derivative. There's still a catch because numerical integration is inevitably going to make your integrated quaternion be something other than a unit quaternion. There are kludges galore for dealing with this problem. A simple kludge is to normalize the integrated quaternion after each numerical integration step. A much, much better approach is to use Lie group integration techniques. That, however, is a rather advanced topic.
  5. Oct 17, 2013 #4
    That sounds very complicated... I would have said find the axis of rotation and perform a coordinate transform to a cylindrical polar system where the z axis is aligned with this axis of rotation. The rotation matrix would then be an addition to the theta angle... so [0,0,0;0,(1+k),0;0,0,0] where k is the fractional rotation. Then perform the inverse transform to find the form in the Cartesian axes.

    Maybe a brute force?
  6. Oct 17, 2013 #5

    D H

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    Doesn't work. Rotation is complicated because, well, rotation is complicated. The rotation group SO(3) is a non-commutative Lie group.
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