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apologies if this isn't worded great: I don't know all the technical terms for what I'm describing.

Given a 3D object in space (no gravity or air resistance or anything), let's say its center of mass is at $(0,0,0)$. We have a force vector $\vec{F}=<0,1,0>$ acting on it at $(0,0,1)$ like in the following image:an image of the mentioned scenario

How would I go about calculating the object's linear movement and angular movement? Conceptually I feel like the object would both move in the $+y$ direction and also begin to pitch down below the $y$ axis. How do I break up the resulting movement into its angular and linear components? What would the process be for multiple force vectors all around the object?

Here's what I was thinking so far, although I'm not sure if it would work:

  1. You can find the resulting force vector by taking the vector summation of all force vectors acting on the object, and treat it as a linear force acting on the center of mass.
  2. You can find the "resultant moment"(?) by taking the sum of the vector cross product of each force vector and it's corresponding position vector from a point $O$ (the center of mass in this case?). If the magnitude is zero, then there are is no angular movement. Otherwise, you could break that resulting "moment" down into it's three axes and calculate angular movement that way.

Thank you in advance for any help!

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  • $\begingroup$ Looks like you’re on the right track $\endgroup$
    – Bob D
    Commented Sep 5 at 18:45
  • $\begingroup$ Alright, thank you. $\endgroup$ Commented Sep 5 at 19:28

1 Answer 1

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Let us give this body a time in which all of these events are happening as $t$ The will accelerate both linearly and angularly.

The linear moment is

$$\mathbf{p}(t) = \int \mathbf{F}(t) \, dt$$ And as your force is constant and not changing, it simplifies to : $$p = \mathbf{F} \cdot t$$

So your linear moment is $[0\hat{i}+1\hat{j}+0\hat{k}]\cdot t$ Let us name this value as $\mathbf{p_{o}}$ Your angular moment can be calculated using this formula: $$\mathbf{L} = \mathbf{r} \times \mathbf{p}$$ So $\mathbf{L}$ is $[0\hat{i} + 0\hat{j} + 1\hat{k}] \times ([0\hat{i} + 1\hat{j} + 0\hat{k}] \cdot t)$ is your answer...

After all the calculation, the answer simplifies to $[-t\hat{i}+ 0\hat{j} + 0 \hat{k}]$

I hope this clears your doubt!

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