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Over the last few days, I've been trying to understand how applying forces to a system of point masses will effect the system's linear and angular velocity in 3 dimensions. I came across this answer that described a solution for a rigid bar in 2 dimensions. From this I believe I could figure out the velocities that I need, however with my relatively limited knowledge of physics, I'm having difficulties translating the solution into 3 dimensions. How could I go about doing so?

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For linear acceleration you can always use Newton's second law, assuming relativistic effects can be neglected,

$$ \sum \vec{F}_\textrm{external} = m\,\vec{a}. \tag{1} $$

For the rotation you could use Euler's equation of rotation,

$$ \dot{\vec{\omega}} = I^{-1}\left(\sum M_\textrm{external} - \vec{\omega}\times(I\cdot\vec{\omega})\right), \tag{2} $$

with $I$ the moments of inertia tensor, $\vec{\omega}$ the angular velocity vector and each moment is calculated around the center of mass of the (rigid) body.

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  • $\begingroup$ How would I differentiate the torque from the linear force? Would the torque just be the component of the force that's perpendicular to the vector going from the center of mass to the point, and then the linear force would be whatever of that force is left? $\endgroup$ – Juddily Oct 3 '16 at 14:52
  • $\begingroup$ @Juddily The linear external force is just literally the sum of all external forces, so you do not need to subtract any perpendicular component. Each external torque can be found by taking the cross product between an external force and the (position) vector between the center of mass and the point at which the force is applied. The reason why you do not need to reduce for example the (linear) force is because as the body also starts rotating the distance over which this force has done work also increases and would explain the additional form of (rotational) kinetic energy. $\endgroup$ – fibonatic Oct 3 '16 at 16:43
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So you mean, you are having your masses lying on a three dimensional space on one 3-d position vector $r$. Easy. Just break up the vector into its components using trigonometry, break up the masses, or rather, break up the effective inertia of the total mass, and then you solve for any variable you want. In case you're wondering how the heck angular momentum gets broken, remember it is just a perpendicular vector and it can be broken up into components perpendicular to the position vector's components easily.

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  • $\begingroup$ If this is wrong , please correct. I have just written what was roughly in my mind. $\endgroup$ – TESLAGEN Oct 3 '16 at 9:18

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