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In statics, the net force and the net torque must be both zero. The net torque is the vector sum of all the involved torques (first moment of each force) about any arbitrary point $P$. The point $P$ about which we calculate each torque is arbitrary. That is static.

In the case of dynamics, where objects are moving and acceleration, which point do we choose to calculate the torques about? I think any arbitrary point would work. But that will change the value (magnitude and direction) of the net torque....Choosing the center of mass CM is just an arbitrary choice.

When we solve a problem, we first define the reference frame and choose the suitable coordinate system. The we should, I guess, select the point $P$ about which to calculate the torques since that decision seems to after the dynamical behavior of the system...

Any inputs?

Thanks!

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  • $\begingroup$ If you take a car, how is the centre of mass a point you choose? $\endgroup$
    – Solar Mike
    Dec 24, 2021 at 17:13
  • $\begingroup$ In any situation, the force and torque change but the mass does not, unless it is a deformable body that the shape is not constant, thus the mass center. $\endgroup$
    – r13
    Dec 24, 2021 at 20:56

2 Answers 2

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Any point can be used for the calculation of the moments, however there are more considerations compared to the static case. This is easier with an example:

Assume the case of cylindrical mass which has a wire around it and its left to drop. In that case

enter image description here

The left is the Free body diagram, and right is the kinetic diagram (which has the accelerations).

If you write the equations of equilibrium for x, and y (just for completeness) you get:

$$ x:\qquad 0 = m \cdot a_x \tag{eq.1} $$ $$ y:\qquad T- W = m \cdot a_y \tag{eq.2}$$

This does not change for any point that is selected for the moments. However for the moments you can do one of the following things (there are probably more):

  1. select the center of mass, and use the mass moment of inertia about the center of mass G (this is the simplest).
  2. select another point, and use the mass moment of inertia about the center of mass G. in that case you need to consider the inertial forces.
  3. select another point (let's call it P), and use the mass moment of inertia about that point P. in that case you the inertial forces are consider around that point.

All of the above are (almost miraculously) equivalent.

1. moment about G

$$\sum M_G = I_G \cdot \alpha\Rightarrow -T\cdot r = I_G \cdot \alpha$$

where:

  • $I_G$ is the mass moment of inertia about G (and its equal to $\frac{1}{2}mr^2$
  • $\alpha$ is angular acceleration

$$-T\cdot r = \frac{1}{2} mr^2 \cdot \alpha \tag{eq.3}$$

2. moment about P using $I_G$ mass moment about G

Assume the following point P

enter image description here

In that case:

$$- W\cdot r = I_G\cdot \alpha +\color{red}{m\cdot a_y} \cdot r \\Rightarrow $$ $$- W\cdot r - \color{red}{m\cdot a_y} \cdot r = I_G\cdot $$ $$ - (W + m\cdot a_y) \cdot r = I_G\cdot \alpha $$

However from eq.2: $ T = m \cdot a_y + W$. So we end up with

$$ - T \cdot r = I_G\cdot \alpha \tag{eq.4}$$

Which is the same as eq.3.

3. moment about P using $I_P$ (mass moment about P)

if the take the point P and the mass moment about P, then the inertial forces are drawn around P. In that case:

$$- W\cdot r = I_P\cdot \alpha \\Rightarrow $$

(i.e. we don't need to consider the inertial forces like in case 2). However in that case:

  • $I_P$ is equal to $I_P = \frac{1}{2}m r^2 + m*r^2$ (parallel axis theorem).

So: $$- W\cdot r = (\color{red}{\frac{1}{2}m r^2} + m\cdot r^2)\cdot \alpha \\Rightarrow $$ $$- W\cdot r = (\color{red}{I_G} + m*r^2)\cdot \alpha \\Rightarrow $$ $$- W\cdot r = \color{red}{I_G}\cdot \alpha + \color{green}{m\cdot r^2\cdot \alpha} \tag{eq:5.a}$$

With a little rearranging the term $\color{green}{m\cdot r^2\cdot \alpha}$, becomes:

$\color{green}{m\cdot r^2\cdot \alpha} =(m\cdot r\cdot \alpha)\cdot r = m\cdot (r\cdot \alpha)\cdot r \tag{eq:6a}$

However because there are also kinematic constraints. e.g. in this particular problem the relationship between $\alpha$ and $a_y$ is $$\alpha \cdot r = a_y$$

therefore, eq.6a becomes: $$m\cdot a_y \cdot r \tag{eq:6b}$$

By substituting eq. 6b in eq.5a we obtain:

$$- W\cdot r = \color{red}{I_G}\cdot \alpha + m\cdot a_y \cdot r $$

$$- W\cdot r - m\cdot a_y \cdot r =I_G\cdot \alpha + $$

$$- T\cdot r = I_G\cdot \alpha \tag{eq:5.b}$$

which is equivalent to equation eq.3.

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  • $\begingroup$ Thank you. I am processing. But, let's say, there is a single force $F_1$ acting on a body. Depending on which point we choose to calculate the moment of the force about, the moment $\tau = f \times r$ will be different because the vector $r$ will be different. So same body, same physical situation but different torque just because we pick different pivot points. And different torque magnitudes will produce different rotational behaviors... $\endgroup$ Dec 24, 2021 at 19:59
  • $\begingroup$ even if you have a single external force, the methodology is the same (you still have the inertial terms) . $\endgroup$
    – NMech
    Dec 24, 2021 at 20:43
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An object will turn around its center of mass either with constant speed or accelerate around it by an angular acceleration $\alpha$, in space.

$$\alpha=\frac{\tau}{I}$$

And we know $I$ is the second moment of the area with respect to the center of mass.

So it is required to take the center of mass as the reference point, or else we can pick any random point but then we would have to transfer the resultant torque by parallel axis transfer to the center of mass.

Edit

If we use a solid bar with length, L, and a mass, m, at rest as an example and apply a force F at distance x from its Cm, it will accelerate and rotate with these accelerations:

$$\alpha_{linear}=\frac{F}{m} \ and \quad \alpha_{ angular}= \frac{F*x}{mL^2/12}$$

The interesting thing is the bigger the x, the more work is done by the same force

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