question on spin physics in space
Author
I don't know if i'm even smart enough to ask the right question... but here goes
can a body spin along multiple axis in space or will the two spins cancel out and merge through gyroscopic forces?
modeling asteroids, if they are spinning x+3/sec and get imparted with a y+5/sec spin through a collision, how do those 2 spins react to each other? will it find one axis to rotate on instantly or will it wobble around as the two other spins degrade?
if that doesn't make any sense than i don't know enough to ask what i want to know ;P
Euler's equations of rotational motion of a rigid body are of the form:
Where the I is the moment of inertia, N is torque, omega is the angular velocity, and epsilon is the Levi-Civita density.
[Edited by - Mastaba on August 19, 2006 9:58:43 PM]
Where the I is the moment of inertia, N is torque, omega is the angular velocity, and epsilon is the Levi-Civita density.
[Edited by - Mastaba on August 19, 2006 9:58:43 PM]
Author
Wow!... im not sure i can even read that...
;P
That'll get me started in reaserching this online... Thanks.
;P
That'll get me started in reaserching this online... Thanks.
Rotations on one axis do not affect the rotations on a perpendicular axis. So, rotating 50 degrees around the x axis, won't change that 75 degree rotation about the y axis, or the 0 degree rotation about the z axis, for example. Also, an rotation about any axis can be defined as rotations about the xyz axes. So, a 50 degree rotation around the (0.7,0.7,0) axis could be defined as two rotations, one about the x axis and one about the y axis.
As the converse to Ezbez's reply (and, I think, the answer to your question), any number of rotations (including the three pseudo-linearly-independent rotations about x-, y- and z- axes) can be expressed as a single rotation about a some axis.
To picture this, imagine the tangential velocity of a point on a sphere, fixed in world coordinates. Suppose this point isn't on any of the six poles, to avoid gimbal clashes. Applying a rotation about one axis will give rise to a constant velocity at this point, in some fixed direction. Now applying a second rotation about another axis will produce another velocity in a different direction. But since these velocities are tangential, the velocities combine just as they do in the euclidean plane: A single new velocity is created (in a new direction), the linear combination of the first two.
Contrived as this example is, it shows that angular velocities combine linearly just as planar velocities do.
I don't like the idea of decomposing rotations into x, y and z components as it causes much confusion for those who aren't confident with linear algebra. Rotation in three dimensions actually has only two degrees of freedom (which is why using your 2D mouse, you can make your first-person camera look in any direction). So although any set of x-, y- and z-axis rotations spans the vector space, they do not form a basis. One consequence of this is that your new coordinates are not unique, which confuses many people.
I much prefer to use a spherical polar coordinate system where 'r' is fixed, equal to one. This way, your remaining coordinates are unique and the whole system is much easier to picture (and hence, debug). [/rant]
Regards
Admiral
To picture this, imagine the tangential velocity of a point on a sphere, fixed in world coordinates. Suppose this point isn't on any of the six poles, to avoid gimbal clashes. Applying a rotation about one axis will give rise to a constant velocity at this point, in some fixed direction. Now applying a second rotation about another axis will produce another velocity in a different direction. But since these velocities are tangential, the velocities combine just as they do in the euclidean plane: A single new velocity is created (in a new direction), the linear combination of the first two.
Contrived as this example is, it shows that angular velocities combine linearly just as planar velocities do.
I don't like the idea of decomposing rotations into x, y and z components as it causes much confusion for those who aren't confident with linear algebra. Rotation in three dimensions actually has only two degrees of freedom (which is why using your 2D mouse, you can make your first-person camera look in any direction). So although any set of x-, y- and z-axis rotations spans the vector space, they do not form a basis. One consequence of this is that your new coordinates are not unique, which confuses many people.
I much prefer to use a spherical polar coordinate system where 'r' is fixed, equal to one. This way, your remaining coordinates are unique and the whole system is much easier to picture (and hence, debug). [/rant]
Regards
Admiral
In addition to the above post, note also that while there is always one axis around which the body instantaneously rotates, that axis is not necessarily constant with respect to time, even when no torque is applied.
I think the way you're going about this is wrong. A collision does not "impart" angular velocity, it changes the angular momentum, which in general is not related (by a scalar) to the angular velocity.
[Edited by - mooserman352 on August 20, 2006 11:51:22 AM]
I think the way you're going about this is wrong. A collision does not "impart" angular velocity, it changes the angular momentum, which in general is not related (by a scalar) to the angular velocity.
[Edited by - mooserman352 on August 20, 2006 11:51:22 AM]
Given an object that is spinning about some axis, how the object responds to a torque about some other axis is dependent on the details of the principal moments of inertia. If for a normal ball, then all the I's would be equal and the Euler equation above simplifies to the rotational form of Newton's second law, and all the components of rotation are independent. That's not the case if the object is more complex in shape, and has one or more principal moments of inertia that are unique.
[Edited by - Mastaba on August 20, 2006 2:55:04 PM]
[Edited by - Mastaba on August 20, 2006 2:55:04 PM]
This topic is closed to new replies.
Advertisement
Popular Topics
Advertisement
