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I know it was a question but when I asked last time I did not know the answer. It has been suggested that the question is not clear, it's badly written. Now I know the answer and I can introduce them here if you let me. I can prove that the questions are correct.

The question is how to calculate force vectors acting on points in these rigid bodies?

enter image description here enter image description here enter image description here

I need some time to translate. I will write answers one by one if I have the opportunity to do this or someone will answer before me.

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closed as off-topic by Emilio Pisanty, Chris, GiorgioP, John Rennie, Carl Brannen Apr 16 at 1:32

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I do not understand why this question was closed? Question about the physical mechanism of action on the points. Full answer based on book patterns and complete calculations and despite such strong arguments, knowledge is considered heresy. If can not deny these calculations, they should be hidden somewhere deeply.

Eli answer showed us how to calculate the resultant of forces acting on the point, I have called this to the forces of bonds. This force is perpendicular to the axis of rotation, but in our cases it is not a central force. There are two options bonds force is a component of the central (centripetal) force or central force si a component of the bonds force (diagrams in the previous answer). Because in books it is written that centripetal forces are always central forces, I bet for a long time that the first option is real but it turned out that the second option is true.

I have a lot of calculations that's why I share them. Today I will count the points of the first diagram on numbers, because it is easier to understand this for the first time. for the points we have

$$ {\omega}(-2,0,0) ; \alpha=30 ; \beta=60$$ $$ m_1=m_2=1$$ $$ m_3=m_4=2$$ $$ \vec r_1 (\frac 1 2, \frac {\sqrt3}2,0 ) $$ $$ \vec r_2 ( -\frac 1 2,\frac {-\sqrt3}2,0 ) $$ $$ \vec r_3 ( \frac {-\sqrt3}2,\frac 1 2,0 ) $$ $$ \vec r_4 ( \frac {\sqrt3}2,-\frac 1 2,0 ) $$

velocity vector for $m_3$

$$ \vec v_1 = \vec \omega \times \vec r_1 = (0,0,-\sqrt3) $$ $$ \vec v_2 = (0,0,\sqrt3) $$ $$ \vec v_3 = (0,0,-1) $$ $$ \vec v_4 = (0,0,1) $$

centripetal acceleration resulting from the angular velocity vector

$$ \vec a_1 = \vec \omega \times \vec v_1 = (0,-2 \sqrt3,0) $$ $$ \vec a_2 = (0,2 \sqrt3,0) $$ $$ \vec a_3 = (0,-2,0) $$ $$ \vec a_4 = (0,2,0) $$

forces $\vec F = m \vec a$ $$ \vec F_1 (0,-2 \sqrt3,0) $$ $$ \vec F_2 (0,2 \sqrt3,0) $$ $$ \vec F_3 (0,-4,0) $$ $$ \vec F_4 (0,4,0) $$

you can see the sum of the resultant forces is equal to zero.

Because the value of the all position vector $ \| \vec r \| = 1$ we can calculate the perpendicular component angular velocity vector $\vec \omega`_p $ for $\vec r $ from the inverse of a vector product

$$ \vec \omega`_{p1} = \vec r_1 \times \vec v_1 = ( -\frac 3 2, \frac {\sqrt 3} 2,0) $$ $$ \vec \omega`_{p2} = ( -\frac 3 2, \frac {\sqrt 3} 2,0) $$ $$ \vec \omega`_{p3} = (- \frac 1 2, -\frac {\sqrt 3} 2,0) $$ $$ \vec \omega`_{p4} = (- \frac 1 2, -\frac {\sqrt 3} 2,0) $$

As we can see the angular velocity of the body consists of a sum of angular velocities on perpendicular axes. It works for two-dimensional space for three-dimensional space it's different but there is no place to describe it.

We can now calculate the central component of bond forces

$$ \vec F_{1c} =m \vec \omega`_{p1} \times \vec v_1 = (- \frac 3 2, -\frac{3 \sqrt3} 2,0) $$ $$ \vec F_{2c} = ( \frac 3 2, \frac{3 \sqrt3} 2,0) $$ $$ \vec F_{3c} = (\sqrt3 , -1 ,0) $$ $$ \vec F_{4c} = (-\sqrt3 , 1 ,0) $$

the central forces are zeroing, but the component of the force that stays gives a non-zero moment of force $ \vec F_{1f} = \vec F_1 - \vec F_{1c}$

$$ \vec F_{1f} = ( \frac 3 2, - \frac{\sqrt3} 2,0) $$ $$ \vec F_{2f} = ( -\frac 3 2, \frac{\sqrt3} 2,0) $$ $$ \vec F_{3f} = ( -\sqrt3, -3,0) $$ $$ \vec F_{4f} = ( \sqrt3, 3,0) $$

we count the moment of forces

$$ \vec M_1 = \vec r_1 \times \vec F_{1f} = (0, 0, - \sqrt3) $$ $$ \vec M_2 = (0, 0, -\sqrt3) $$ $$ \vec M_3 = (0, 0, 2 \sqrt3) $$ $$ \vec M_4 = (0, 0, 2 \sqrt3) $$

sum of moments of forces $$ M=(0, 0, 2 \sqrt3) $$

Now use Euler's equations for free rotational motion of rigid bodies.

$$ I_x \frac {d \omega_x``} {dt} = I_y \omega_y`` \omega_z`` - I_z \omega_z`` \omega_y`` $$ $$ I_y \frac {d \omega_y``} {dt} = I_z \omega_z`` \omega_x`` - I_x \omega_x`` \omega_z`` $$ $$ I_z \frac {d \omega_z``} {dt} = I_x \omega_x`` \omega_y`` - I_y \omega_y`` \omega_x`` $$

We must change the reference system to the arrangement of the main axes of inertia.

$$ \vec r`_1 (1,0,0 ) $$ $$ \vec r`_2 (-1,0,0 ) $$ $$ \vec r`_3 (0,1,0 ) $$ $$ \vec r`_4 (0,-1,0 ) $$

main moments of inertia

$$ I_x =4 ; I_y=2; I_z=6 $$

We already counted the angular velocity on the main axes

$$ \vec \omega_y`` = \| \vec \omega_{p1}`\|\ = \sqrt 3 $$ $$ \vec \omega_x`` = \| \vec \omega_{p3}`\|\ = 1 $$ $$ \vec \omega`` = (-1, \sqrt 3 ,0)$$

The first two Euler equations are zero and third is $$ I_z \frac {d \omega_z``} {dt} = M_z = 4 \sqrt 3 – 2 \sqrt 3 $$

That is the same as the sum of our moments of force that we have calculated. This way works always if our position vectors are equal $ \| \vec r \| = 1$ .

SECOND SCHEME

Each of the three examples that are in the question are different, to see if my method works in different situations. In books it is written that: the ellipsoid of inertia rotates and the distribution of masses (points) does not matter. In the second example, we have three points whose position vectors are not perpendicular to each other but the sum of their inertia on the X and Y axes is the same. So we count the main moments of inertia and create a model with arms perpendicular to each other withd the same ellipsoid of inertia.

enter image description here

So we have

$$ \vec r`_1 =(-r`_x, 0, 0) $$ $$ \vec r`_2 =(0, r`_y, 0) $$ $$ \vec r`_3 =(0, -r`_y, 0) $$ $$ \vec r`_4 =(r`_x, 0, 0) $$ $$m`= \frac 3 4$$

$$ \Omega (\omega_x,0,0); I_x=2mr`_y^2= \frac 3 2; I_y=2mr`_x^2=\frac 3 2; I_z=2mr`_x^2+2mr`_y^2=3 $$

substitute $w_x=-1$

$m`_1$ i $m`_4$ they are on the axis of rotation, so their velocity vectors are zero, we can omit them in calculations.

angular velocity for $m`_2$ i $m`_3$

$$ \vec v`_2 = \vec \omega \times \vec r`_y = (0,0,\omega_x r`_y)= (0,0,-1)$$ $$ \vec v`_3 = (0,0,-\omega_x r`_y)= (0,0,1) $$

we now count the force of bonds

$$\vec F_d = m (\vec \omega \times \vec v)$$ $$ \vec F`_{d2} = (0,-m`\omega_x^2 r`_y,0) = (0, -\frac 3 4,0)$$ $$ \vec F`_{d3} = (0,m`\omega_x^2 r`_y,0) = (0, \frac 3 4,0)$$

They are central forces, the moments of force are zero.

Let's check how the force of bonds acting on the point in the second diagram used my method. $m_1$ is on the axis of rotation and we omit. Remembering that in the calculations the position vector must equal one, stays only

$$ \vec r_2 =(r_x, r_y, 0) = ( \frac 1 2,\frac {\sqrt3}2,0 ) $$ $$ \vec r_3 =(r_x, -r_y, 0)= ( \frac 1 2,-\frac {\sqrt3}2,0 ) $$ $$m_1=m_2=m_3=1$$

velocity vectors for points

$$ \vec v_2 = \vec \omega \times \vec r_y = (0,0,\omega_x r_y)= (0,0,-\frac {\sqrt3}2)$$ $$ \vec v_3 = (0,0,-\omega_x r_y)= (0,0,\frac {\sqrt3}2) $$

We now count the component angular velocity perpendicular to the position vector $$ \vec \omega`_p = \vec r \times \vec v $$ $$ \vec \omega`_{p2} = (\omega_x r_y^2,-\omega_x r_y r_x,0) =( -\frac 3 4, \frac {\sqrt 3} 4,0) $$ $$ \vec \omega`_{p3} = (\omega_x r_y^2,\omega_x r_y r_x,0) =( -\frac 3 4, - \frac {\sqrt 3} 4,0) $$

now the components of the angular velocity parallel to the position vector $$ \vec \omega`_{ii} = \vec \omega - \vec \omega`_p $$ $$ \vec \omega`_{ii2} = (\omega_x - \omega_x r_y^2 , \omega_x r_y r_x , 0) =( -\frac 1 4, -\frac {\sqrt 3} 4,0) $$ $$ \vec \omega`_{ii3} = (\omega_x - \omega_x r_y^2 , -\omega_x r_y r_x , 0) =( -\frac 1 4, \frac {\sqrt 3} 4,0) $$

The matter is complicated components of angular velocity perpendicular to one position vector it is not parallel to another position vector.

enter image description hereTo find a solution I had to understand that, that the source of the moment of force is the angular velocity parallel to the position vector relative to the center of mass, consistent with the formula for centripetal acceleration $\vec a_c = \vec \omega \times \vec v $ enter image description here

we count acceleration vectors. First from perpendicular angular velocity. $$ \vec a_{p2} = (-\omega_x^2 r_y^2 r_x,-\omega_x^2 r_y^3,0) =( -\frac 3 8, -\frac {3 \sqrt 3} 8,0) $$ $$ \vec a_{p3} = (-\omega_x^2 r_y^2 r_x,\omega_x^2 r_y^3,0) =( -\frac 3 8, \frac {3 \sqrt 3} 8,0) $$

Now the acceleration of the angular velocity parallel to the position vector $$ \vec a_{ii2} = (\omega_x^2 r_y^2 r_x, \omega_x^2 r_y^3 -\omega_x^2 r_y ,0) =( \frac 3 8, -\frac { \sqrt 3} 8,0) $$ $$ \vec a_{ii3} = (\omega_x^2 r_y^2 r_x, -\omega_x^2 r_y^3 + \omega_x^2 r_y ,0) =( \frac 3 8, \frac { \sqrt 3} 8,0) $$

adding together the components of acceleration vectors we get $$ \vec a_{p2} + \vec a_{ii2} = (0,-\omega_x^2 r_y ,0) = - (\vec a_p3 + \vec a_ii3) $$

So the resultant forces acting on points are perpendicular to the axis of rotation, and the moments of forces after adding become zero, as expected.

THIRD DIAGRAM

I finished writing down the rotation mechanics of point in rigid body. There are answers, described mechanisms and deates of this movement. If anyone is interested please contact me. Unfortunately, only the Polish version is currently available.

the third diagram $$ {\omega}( -\frac 1 2, \frac 1 2 ,0) $$ $$ m_1 = 3 ; m_2=4 ; m_3 =2 $$ $$ \vec r_1 (-1, 0,0 ) $$ $$ \vec r_2 ( \frac 1 4,\frac 1 4,0 ) $$ $$ \vec r_3 ( 1,-\frac 1 2,0 ) $$

main moments of inertia

$$ I_x= m_2 r_{2y}^2 + m_3 r_{3y}^2 = \frac 3 4 $$ $$ I_y= m_1 r_{1x}^2 + m_2 r_{2x}^2 + m_3 r_{3x}^2 = \frac {21} 4 $$ $$ I_z= m_1 r_{1x}^2 + m_2 (r_{2x}^2 + r_{2y}^2) + m_3 (r_{3x}^2 + r_{3y}^2) = 6$$

because $\omega_z=0 $ only the third Euler equation is non-zero

$$ I_z \frac {d \omega_z} {dt} = I_x \omega_x \omega_y - I_y \omega_y \omega_x $$

so we have

$$ I_z \frac {d\omega_z} {dt} = -\frac 9 8 \tag 1 $$

we calculate the velocity vector of points

$$ \vec v_1 = \vec \omega \times \vec r_1 = (0,0, \frac 1 2) $$ $$ \vec v_2 = (0,0, -\frac 1 4) $$ $$ \vec v_3 = (0,0, -\frac 1 4) $$

components of the angular velocity perpendicular to the position vector

$$ \vec \omega`_{p1} = \vec r_1 \times \vec v_1 = ( 0, \frac 1 2,0) $$ $$ \vec \omega`_{p2} = ( -\frac 1 2, \frac 1 2,0) $$ $$ \vec \omega`_{p3} = ( \frac 1 {10}, \frac 1 5,0) $$

components of the angular velocity parallel to the position vector $\omega`_r = \omega - \omega`_p $

$$ \vec \omega`_{r1} = ( - \frac 1 2,0,0) $$ $$ \vec \omega`_{r2} = ( 0, 0,0) $$ $$ \vec \omega`_{r3} = ( - \frac 6 {10}, \frac 4 {10},0) $$

Central accelerations

$$ \vec a_{c1} = \vec \omega`_{p1} \times \vec v_1 = ( \frac 1 4, 0,0) $$ $$ \vec a_{c2} = (- \frac 1 8,- \frac 1 8,0) $$ $$ \vec a_{c3} = (- \frac 1 {20},\frac 1 {40},0) $$

other accelerations

$$ \vec a_{s1} = \vec \omega`_{r1} \times \vec v_1 = ( 0, \frac 1 4,0) $$ $$ \vec a_{s2} = (0,0,0) $$ $$ \vec a_{s3} = (- \frac 3 {40},-\frac 6 {40},0) $$

vectors of forces acting on points $$ \vec F_{c1} = m_1 \vec a_{c1} = ( \frac 3 4, 0,0) $$ $$ \vec F_{c2} = (- \frac 1 2,- \frac 1 2,0) $$ $$ \vec F_{c3} = (- \frac 2 {20},\frac 1 {20},0) $$

and

$$ \vec F_{s1} = m_1 \vec a_{s1} = ( 0, \frac 3 4, 0) $$ $$ \vec F_{s2} = (0,0,0) $$ $$ \vec F_{s3} = (- \frac 3 {20},-\frac 6 {20},0) $$

the sum of all forces equals zero

$$ \sum \vec F_{c1} + \sum \vec F_{s1} = 0 $$

but the moments of force from forces are uncentral is

$$ \vec M_1 = \vec r_1 \times \vec F_{s1} = (0, 0, -\frac 3 4) $$ $$ \vec M_2 = (0, 0, 0) $$ $$ \vec M_3 = (0, 0, - \frac {15} {40}) $$

the sum of these moments of force is $$ \sum M = - \frac 9 8 $$

exactly as much as we expected (1).

I do not know why I was banned again and I can not ask more questions. Who is afraid of these questions? Who is afraid of new ideas and new knowledge?

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The centripetal force is:

$$\vec{F_c}=m\,\left[\vec{\Omega}\times \left(\vec{\Omega}\times \vec{R}\right)\right]\tag 1$$

where in your case :

$$ \vec{\Omega}=\left[ \begin {array}{c} \omega_{{x}}\\ \omega_{{y} }\\ 0\end {array} \right] $$

and

$$\vec{R}=\left[ \begin {array}{c} r_{{x}}\\ r_{{y}} \\ 0\end {array} \right] $$ $ \Rightarrow$ equation (1)

$$\vec{F_c}=m\,\left[ \begin {array}{c} \omega_{{y}}\omega_{{x}}r_{{y}}-{\omega_{{y} }}^{2}r_{{x}}\\ -{\omega_{{x}}}^{2}r_{{y}}+\omega_{{ x}}\omega_{{y}}r_{{x}}\\ 0\end {array} \right]$$

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  • $\begingroup$ Good way but not complete. First example m1, from your solution the centripetal force has only y component, Fx=0. Your centripetal force is not directed to the center of mas. Correct but incomplete result because the force which is the result is not a central force. $\endgroup$ – Sylwester L Mar 3 at 15:53

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