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Let me set up the following problem for a rectangle floating in space:

  • We know its dimensions.
  • We know its mass.
  • There's a force pushing it for a known amount of time - we know the angle & magnitude of the force.
  • We know the point on the rectangle the force is being applied.

Here's a picture I made of the situation: enter image description here

I can use the torque & moment of inertia equations to determine the angular acceleration this rectangle will experience.

But I would also imagine this rectangle will experience some "straight" acceleration.

For example in this picture, I can see the rectangle rotating counterclockwise, but also moving in an up-left direction.

My Question: I would use my torque & moment of inertia equations to determine the angular acceleration of the rectangle. But that's only part of the motion it would experience. How would I calculate the "straight" acceleration it has?

My best guess is that the "straight" acceleration is just going to be f/m, but since the rectangle is also rotating while the force is being applied, then the force vector keeps changing and this will make for some difficult computations?

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  • $\begingroup$ Is the force always at the same orientation, or does it rotate with the body? Also is the time of force application small compared to the motion or not? BTW linear acceleration is governed by $\vec{F} = m \vec{a}_{cm}$ $\endgroup$ – ja72 Dec 10 '15 at 16:17
  • $\begingroup$ This video explains this question: m.youtube.com/watch?v=vWVZ6APXM4w $\endgroup$ – Michal Jun 21 '16 at 10:48
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I'm not sure what you're referring to by 'straight' acceleration. If that angle is the angle at which a force, $F$, is being applied, then the horizontal force is: $$ F_x=F\cdot cos(\theta)\\ F_y=F\cdot sin(\theta) $$ These will get the forces along the X and Y axis respectively. Assuming width, $w$, length, $l$, height, $h$, mass, $m$ and $\text{'some distance'}$, $d$, then we can find the torque by: $$ \tau = F \space sin(\theta)\space (d - \frac{width}{2}) $$ From there you should be able to figure out any other rotational force or what have you. I'm not entirely sure as to what you're asking, but let me know if I'm on the right track!

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  • $\begingroup$ What I mean by "straight" acceleration is the acceleration you would get from the basic formula a=fnet\m, it's not going to cause any rotation, it will be translation. $\endgroup$ – Aaron Oct 17 '14 at 10:16
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It seems you have no issue with determining angular acceleration in force systems. For 2D:

$$\dot \omega = \tau / J = F r / J$$

where $\dot \omega$ is the angular acceleration, $\tau$ is the net torque applied, and $r$ is the distance between the centre of mass and the line of action of the force.

As for translational motion (i.e. straight motion), you suspect correctly:

$$\vec a = \vec F / m$$

It is the direction and magnitude, not the distance from the centre of mass, of the force that influences the translational acceleration. This relationship holds true for any particular instance in time, even if the force changes due to rotation, i.e.:

$$\vec a(t) = \vec F(t) / m$$

So, for example, if the force could be described over time with:

$$\vec F(t) = (t^2 + 7)\vec i + (\cos{t}) \vec j$$

then translational acceleration is:

$$\vec a(t) = [(t^2 + 7)\vec i + (\cos{t}) \vec j]/m $$

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Although you haven't explicit said it, I get the impression that you are wondering if the force is somehow split between providing rotational vs. linear acceleration. That does not happen.

It's a both/and situation, not an either/or situation. A force can cause both accelerations simultaneously.

The linear will be (instantaneously) $$\vec{a}=\frac{F_{net}}{m}$$ and $$\vec{\alpha}=\frac{\Sigma (\vec{R}\times\vec{F})}{\mathcal{I}},$$ where $\vec{R}$ is the vector pointing from the point of rotational interest to the application point of each force in the sum. For your problem, you only have one force. The difficulty will become how does that force change over time, and how do you describe it mathmetically.

But the force isn't split into different roles. It plays both roles.

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  • $\begingroup$ Using vectors somewhat implies 3D motion, and thus the equation of motion is $\sum ( \vec{R} \times \vec{F}) = \mathcal{I} \vec{\alpha} + \vec{\omega} \times \mathcal{I} \vec{\omega}$. $\endgroup$ – ja72 Dec 10 '15 at 16:31
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If you look at the planar motion, with the force $\vec{F}=(F_x,F_y)$ going through a point $\vec{r}=(x,y)$ applied to a rigid body, then the equations of motion are:

$$ \begin{align} F_x & = m \ddot{x}_C \\ F_y & = m \ddot{y}_C \\ (x-x_C) F_y - (y-y_C) F_x & = I_C \ddot{\theta} \end{align} $$

where $(x_C,y_C)$ is the location of the center of mass and $I_C$ is the mass moment of inertia about the center of mass.

Note that $\ddot{x}_C$, $\ddot{y}_C$ and $\ddot{\theta}$ are the linear and angular accelerations of the center of mass.

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