motion.ar

studiot, the reference point is the origin O of the reference frame which makes measurements of: [latex]\mathbf{r_{a_o}}, \mathbf{v_{a_o}}, \mathbf{a_{a_o}}, \mathbf{r_{b_o}}, \mathbf{v_{b_o}}, \mathbf{a_{b_o}}, [/latex] etc.

In the equation of the topic # 1, a particle A is related with another particle B. In the above example, it is possible that a particle D exerts a force [latex]\mathbf{F}_a[/latex] on a particle A, and a particle E exerts a force [latex]\mathbf{F}_b[/latex] on a particle B, where [latex]\mathbf{F}_a=\mathbf{F}_b[/latex] and [latex]m_a=m_b[/latex] In the above example, it is possible that the particle A does not exert any force on the particle B, and the particle B does not exert any force on particle A.

For example, through an imaginary experiment, this is, exposing an exercise. In your imaginary experiment you should determine the initial conditions.

swansont, for example, if we consider a single particle A in a uniform gravitational field [latex]g[/latex] then the mechanical energy is constant. [latex]\frac{1}{2} m_a \mathbf{v}_a^2 + m_a \, \mathbf{g} \cdot \mathbf{r}_a = constant[/latex] This constant depends on the value of the mass and of the initial values of [latex]\mathbf{v}_a[/latex] and [latex]\mathbf{r}_a[/latex] Similarly, the constant of the posted #3 depends on the values of the masses and of the initial values of [latex]\mathbf{v}_a, \mathbf{v}_b, \mathbf{a}_a, \mathbf{a}_b, \mathbf{r}_a, [/latex] and [latex]\mathbf{r}_b[/latex]

Ok, swansont, but on the right side there is an indefinite integral. Therefore, in the previous equation, it follows that [latex]\frac{1}{2}\,m_am_b\left[(\mathbf{v}_a-\mathbf{v}_b)^{2}+(\mathbf{a}_a-\mathbf{a}_b)\cdot(\mathbf{r}_a-\mathbf{r}_b)\right]=constant[/latex]

In classical mechanics, this topic presents a scalar equation of motion, which can be applied in any reference frame (rotating or non-rotating) (inertial or non-inertial) without the necessity of introducing fictitious forces. If we consider two particles A and B of mass [latex]m_a[/latex] and [latex]m_b[/latex] respectively, then the scalar equation of motion, is given by: [latex]\frac{1}{2}\,m_am_b\left[(\mathbf{v}_a-\mathbf{v}_b)^{2}+(\mathbf{a}_a-\mathbf{a}_b)\cdot(\mathbf{r}_a-\mathbf{r}_b)\right]=\frac{1}{2}\,m_am_b\left[2\int\left(\frac{\mathbf{F}_a}{m_a}-\frac{\mathbf{F}_b}{m_b}\right){\cdot}\;d(\mathbf{r}_a-\mathbf{r}_b)+\left(\frac{\mathbf{F}_a}{m_a}-\frac{\mathbf{F}_b}{m_b}\right)\cdot(\mathbf{r}_a-\mathbf{r}_b)\right][/latex] where [latex]\mathbf{v}_a[/latex] and [latex]\mathbf{v}_b[/latex] are the velocities of particles A and B, [latex]\mathbf{a}_a[/latex] and [latex]\mathbf{a}_b[/latex] are the accelerations of particles A and B, [latex]\mathbf{r}_a[/latex] and [latex]\mathbf{r}_b[/latex] are the positions of particles A and B, and [latex]\mathbf{F}_a[/latex] and [latex]\mathbf{F}_b[/latex] are the net forces acting on particles A and B. This scalar equation of motion is invariant under transformations between reference frames. In addition, this scalar equation of motion would be valid even if Newton's three laws of motion were false.