contestada

verify that the positions of two particles can be written in terms of CM the CM and relative positions as r1 = R + m2r/M and r2 = R − m1r/M. Hence confirm that the total KE of the two particles can be expressed as T = 1 2MR˙ 2 + 1 2 µr˙ 2 , where µ denotes the reduced mass µ = m1m2/M.

Respuesta :

cjmejiab

To solve the problem it is necessary to resort to the concepts of kinetic energy of the bodies.

Kinetic energy in vector form can be expressed as

[tex]KE = \frac{1}{2}m\vec{R}^2[/tex]

According to the description given we have to

[tex]r_1 = R+ m_2r/M[/tex]

[tex]r_2 = R-m_1 r/M[/tex]

Equating both equation we have that

[tex]R = \frac{m_1r_1+m_2r_2}{m_1+m_2}[/tex]

[tex]R = \frac{m_1r_1+m_2r_2}{M}[/tex]

The kinetic energy of the two particles would be given by

[tex]T = \frac{1}{2} (m_1\vec{r_1}^2+m_2\vec{r_2}^2)[/tex]

[tex]T = \frac{1}{2} (m_1(\vec{R}+\frac{m_2}{M}\vec{r})^2+m_2(\vec{R}-\frac{m_1}{M}\vec{r})^2)[/tex]

[tex]T = \frac{1}{2} (m\vec{R}^2+\frac{m_1m_2}{M}\vec{r}^2)[/tex]

We have the consideration that

[tex]\mu = \frac{m_1m_2}{M}[/tex]

Then replacing,

[tex]T = \frac{1}{2}(m\vec{R}^2+\mu\vec{r^2})[/tex]

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