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Eli
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you start always with the position vector to the masses .

Case I

$$\vec P_m=\begin{bmatrix} X \\ Y \\ \end{bmatrix}=\left[ \begin {array}{c} \left( l+x \right) \sin \left( \phi \right) \\ - \left( l+x \right) \cos \left( \phi \right) \end {array} \right] $$ the generalized coordinates are $~x(t)~,\phi(t)~$

from here the velocity $$v^2=\dot X^2+\dot Y^2 $$ and the kinetic energy $$T=\frac m2\,v^2+\frac k2\,(l+x)^2$$

Case II

Position to $~m_1$

$$\vec P_1=\begin{bmatrix} X_1 \\ Y_1 \\ \end{bmatrix}=\left[ \begin {array}{c} r\sin \left( \phi_{{1}} \right) \\ r\cos \left( \phi_{{1}} \right) \end {array} \right] $$

Position to $~m_2$ $$\vec P_2=\begin{bmatrix} X_2 \\ Y_2 \\ \end{bmatrix}=2\,\vec P_1+\left[ \begin {array}{c} r\sin \left( \phi_{{2}} \right) \\ r\cos \left( \phi_{{2}} \right) \end {array} \right] $$

the generalized coordinates are $~\phi_1(t)~,\phi_2(t)~$

thus the velocities are

$$v_1^2=\dot X_1^2+\dot Y_1^2 \\ v_2^2=\dot X_2^2+\dot Y_2^2$$

and the kinetic energy

$$T=\frac m2\,(v_1^2+v_2^2)$$

What is the general principle that determines whether an independent variable should be treated as a parameter or explicitly as a variable in such problems?

the only independent variable are the generalized coordinates , thusthose are implicit time dependent.

you start always with the position vector to the masses .

Case I

$$\vec P_m=\begin{bmatrix} X \\ Y \\ \end{bmatrix}=\left[ \begin {array}{c} \left( l+x \right) \sin \left( \phi \right) \\ - \left( l+x \right) \cos \left( \phi \right) \end {array} \right] $$ the generalized coordinates are $~x(t)~,\phi(t)~$

from here the velocity $$v^2=\dot X^2+\dot Y^2 $$ and the kinetic energy $$T=\frac m2\,v^2+\frac k2\,(l+x)^2$$

Case II

Position to $~m_1$

$$\vec P_1=\begin{bmatrix} X_1 \\ Y_1 \\ \end{bmatrix}=\left[ \begin {array}{c} r\sin \left( \phi_{{1}} \right) \\ r\cos \left( \phi_{{1}} \right) \end {array} \right] $$

Position to $~m_2$ $$\vec P_2=\begin{bmatrix} X_2 \\ Y_2 \\ \end{bmatrix}=2\,\vec P_1+\left[ \begin {array}{c} r\sin \left( \phi_{{2}} \right) \\ r\cos \left( \phi_{{2}} \right) \end {array} \right] $$

the generalized coordinates are $~\phi_1(t)~,\phi_2(t)~$

thus the velocities are

$$v_1^2=\dot X_1^2+\dot Y_1^2 \\ v_2^2=\dot X_2^2+\dot Y_2^2$$

and the kinetic energy

$$T=\frac m2\,(v_1^2+v_2^2)$$

What is the general principle that determines whether an independent variable should be treated as a parameter or explicitly as a variable in such problems?

the only independent variable are the generalized coordinates , thus are implicit time dependent.

you start always with the position vector to the masses .

Case I

$$\vec P_m=\begin{bmatrix} X \\ Y \\ \end{bmatrix}=\left[ \begin {array}{c} \left( l+x \right) \sin \left( \phi \right) \\ - \left( l+x \right) \cos \left( \phi \right) \end {array} \right] $$ the generalized coordinates are $~x(t)~,\phi(t)~$

from here the velocity $$v^2=\dot X^2+\dot Y^2 $$ and the kinetic energy $$T=\frac m2\,v^2+\frac k2\,(l+x)^2$$

Case II

Position to $~m_1$

$$\vec P_1=\begin{bmatrix} X_1 \\ Y_1 \\ \end{bmatrix}=\left[ \begin {array}{c} r\sin \left( \phi_{{1}} \right) \\ r\cos \left( \phi_{{1}} \right) \end {array} \right] $$

Position to $~m_2$ $$\vec P_2=\begin{bmatrix} X_2 \\ Y_2 \\ \end{bmatrix}=2\,\vec P_1+\left[ \begin {array}{c} r\sin \left( \phi_{{2}} \right) \\ r\cos \left( \phi_{{2}} \right) \end {array} \right] $$

the generalized coordinates are $~\phi_1(t)~,\phi_2(t)~$

thus the velocities are

$$v_1^2=\dot X_1^2+\dot Y_1^2 \\ v_2^2=\dot X_2^2+\dot Y_2^2$$

and the kinetic energy

$$T=\frac m2\,(v_1^2+v_2^2)$$

What is the general principle that determines whether an independent variable should be treated as a parameter or explicitly as a variable in such problems?

the only independent variable are the generalized coordinates , those are implicit time dependent.

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Eli
  • 12.9k
  • 2
  • 11
  • 31

you start always with the position vector to the masses .

Case I

$$\vec P_m=\begin{bmatrix} X \\ Y \\ \end{bmatrix}=\left[ \begin {array}{c} \left( l+x \right) \sin \left( \phi \right) \\ - \left( l+x \right) \cos \left( \phi \right) \end {array} \right] $$ the generalized coordinates are $~x(t)~,\phi(t)~$

from here the velocity $$v^2=\dot X^2+\dot Y^2 $$ and the kinetic energy $$T=\frac m2\,v^2+\frac k2\,(l+x)^2$$

Case II

Position to $~m_1$

$$\vec P_1=\begin{bmatrix} X_1 \\ Y_1 \\ \end{bmatrix}=\left[ \begin {array}{c} r\sin \left( \phi_{{1}} \right) \\ r\cos \left( \phi_{{1}} \right) \end {array} \right] $$

Position to $~m_2$ $$\vec P_2=\begin{bmatrix} X_2 \\ Y_2 \\ \end{bmatrix}=2\,\vec P_1+\left[ \begin {array}{c} r\sin \left( \phi_{{2}} \right) \\ r\cos \left( \phi_{{2}} \right) \end {array} \right] $$

the generalized coordinates are $~\phi_1(t)~,\phi_2(t)~$

thus the velocities are

$$v_1^2=\dot X_1^2+\dot Y_1^2 \\ v_2^2=\dot X_2^2+\dot Y_2^2$$

and the kinetic energy

$$T=\frac m2\,(v_1^2+v_2^2)$$

What is the general principle that determines whether an independent variable should be treated as a parameter or explicitly as a variable in such problems?

the only independent variable are the generalized coordinates , thus are implicit time dependent.

you start always with the position vector to the masses .

Case I

$$\vec P_m=\begin{bmatrix} X \\ Y \\ \end{bmatrix}=\left[ \begin {array}{c} \left( l+x \right) \sin \left( \phi \right) \\ - \left( l+x \right) \cos \left( \phi \right) \end {array} \right] $$ the generalized coordinates are $~x(t)~,\phi(t)~$

from here the velocity $$v^2=\dot X^2+\dot Y^2 $$ and the kinetic energy $$T=\frac m2\,v^2+\frac k2\,(l+x)^2$$

Case II

Position to $~m_1$

$$\vec P_1=\begin{bmatrix} X_1 \\ Y_1 \\ \end{bmatrix}=\left[ \begin {array}{c} r\sin \left( \phi_{{1}} \right) \\ r\cos \left( \phi_{{1}} \right) \end {array} \right] $$

Position to $~m_2$ $$\vec P_2=\begin{bmatrix} X_2 \\ Y_2 \\ \end{bmatrix}=2\,\vec P_1+\left[ \begin {array}{c} r\sin \left( \phi_{{2}} \right) \\ r\cos \left( \phi_{{2}} \right) \end {array} \right] $$

the generalized coordinates are $~\phi_1(t)~,\phi_2(t)~$

thus the velocities are

$$v_1^2=\dot X_1^2+\dot Y_1^2 \\ v_2^2=\dot X_2^2+\dot Y_2^2$$

and the kinetic energy

$$T=\frac m2\,(v_1^2+v_2^2)$$

you start always with the position vector to the masses .

Case I

$$\vec P_m=\begin{bmatrix} X \\ Y \\ \end{bmatrix}=\left[ \begin {array}{c} \left( l+x \right) \sin \left( \phi \right) \\ - \left( l+x \right) \cos \left( \phi \right) \end {array} \right] $$ the generalized coordinates are $~x(t)~,\phi(t)~$

from here the velocity $$v^2=\dot X^2+\dot Y^2 $$ and the kinetic energy $$T=\frac m2\,v^2+\frac k2\,(l+x)^2$$

Case II

Position to $~m_1$

$$\vec P_1=\begin{bmatrix} X_1 \\ Y_1 \\ \end{bmatrix}=\left[ \begin {array}{c} r\sin \left( \phi_{{1}} \right) \\ r\cos \left( \phi_{{1}} \right) \end {array} \right] $$

Position to $~m_2$ $$\vec P_2=\begin{bmatrix} X_2 \\ Y_2 \\ \end{bmatrix}=2\,\vec P_1+\left[ \begin {array}{c} r\sin \left( \phi_{{2}} \right) \\ r\cos \left( \phi_{{2}} \right) \end {array} \right] $$

the generalized coordinates are $~\phi_1(t)~,\phi_2(t)~$

thus the velocities are

$$v_1^2=\dot X_1^2+\dot Y_1^2 \\ v_2^2=\dot X_2^2+\dot Y_2^2$$

and the kinetic energy

$$T=\frac m2\,(v_1^2+v_2^2)$$

What is the general principle that determines whether an independent variable should be treated as a parameter or explicitly as a variable in such problems?

the only independent variable are the generalized coordinates , thus are implicit time dependent.

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Eli
  • 12.9k
  • 2
  • 11
  • 31

you start always with the position vector to the masses .

Case I

$$\vec P_m=\begin{bmatrix} X \\ Y \\ \end{bmatrix}=\left[ \begin {array}{c} \left( l+x \right) \sin \left( \phi \right) \\ - \left( l+x \right) \cos \left( \phi \right) \end {array} \right] $$ the generalized coordinates are $~x(t)~,\phi(t)~$

from here the velocity $$v^2=\dot X^2+\dot Y^2 $$ and the kinetic energy $$T=\frac m2\,v^2+\frac k2\,(l+x)^2$$

Case II

Position to $~m_1$

$$\vec P_1=\begin{bmatrix} X_1 \\ Y_1 \\ \end{bmatrix}=\left[ \begin {array}{c} r\sin \left( \phi_{{1}} \right) \\ r\cos \left( \phi_{{1}} \right) \end {array} \right] $$

Position to $~m_2$ $$\vec P_2=\begin{bmatrix} X_2 \\ Y_2 \\ \end{bmatrix}=\vec P_1+\left[ \begin {array}{c} r\sin \left( \phi_{{2}} \right) \\ r\cos \left( \phi_{{2}} \right) \end {array} \right] $$$$\vec P_2=\begin{bmatrix} X_2 \\ Y_2 \\ \end{bmatrix}=2\,\vec P_1+\left[ \begin {array}{c} r\sin \left( \phi_{{2}} \right) \\ r\cos \left( \phi_{{2}} \right) \end {array} \right] $$

the generalized coordinates are $~\phi_1(t)~,\phi_2(t)~$

thus the velocities are

$$v_1^2=\dot X_1^2+\dot Y_1^2 \\ v_2^2=\dot X_2^2+\dot Y_2^2$$

and the kinetic energy

$$T=\frac m2\,(v_1^2+v_2^2)$$

you start always with the position vector to the masses .

Case I

$$\vec P_m=\begin{bmatrix} X \\ Y \\ \end{bmatrix}=\left[ \begin {array}{c} \left( l+x \right) \sin \left( \phi \right) \\ - \left( l+x \right) \cos \left( \phi \right) \end {array} \right] $$ the generalized coordinates are $~x(t)~,\phi(t)~$

from here the velocity $$v^2=\dot X^2+\dot Y^2 $$ and the kinetic energy $$T=\frac m2\,v^2+\frac k2\,(l+x)^2$$

Case II

Position to $~m_1$

$$\vec P_1=\begin{bmatrix} X_1 \\ Y_1 \\ \end{bmatrix}=\left[ \begin {array}{c} r\sin \left( \phi_{{1}} \right) \\ r\cos \left( \phi_{{1}} \right) \end {array} \right] $$

Position to $~m_2$ $$\vec P_2=\begin{bmatrix} X_2 \\ Y_2 \\ \end{bmatrix}=\vec P_1+\left[ \begin {array}{c} r\sin \left( \phi_{{2}} \right) \\ r\cos \left( \phi_{{2}} \right) \end {array} \right] $$

the generalized coordinates are $~\phi_1(t)~,\phi_2(t)~$

thus the velocities are

$$v_1^2=\dot X_1^2+\dot Y_1^2 \\ v_2^2=\dot X_2^2+\dot Y_2^2$$

and the kinetic energy

$$T=\frac m2\,(v_1^2+v_2^2)$$

you start always with the position vector to the masses .

Case I

$$\vec P_m=\begin{bmatrix} X \\ Y \\ \end{bmatrix}=\left[ \begin {array}{c} \left( l+x \right) \sin \left( \phi \right) \\ - \left( l+x \right) \cos \left( \phi \right) \end {array} \right] $$ the generalized coordinates are $~x(t)~,\phi(t)~$

from here the velocity $$v^2=\dot X^2+\dot Y^2 $$ and the kinetic energy $$T=\frac m2\,v^2+\frac k2\,(l+x)^2$$

Case II

Position to $~m_1$

$$\vec P_1=\begin{bmatrix} X_1 \\ Y_1 \\ \end{bmatrix}=\left[ \begin {array}{c} r\sin \left( \phi_{{1}} \right) \\ r\cos \left( \phi_{{1}} \right) \end {array} \right] $$

Position to $~m_2$ $$\vec P_2=\begin{bmatrix} X_2 \\ Y_2 \\ \end{bmatrix}=2\,\vec P_1+\left[ \begin {array}{c} r\sin \left( \phi_{{2}} \right) \\ r\cos \left( \phi_{{2}} \right) \end {array} \right] $$

the generalized coordinates are $~\phi_1(t)~,\phi_2(t)~$

thus the velocities are

$$v_1^2=\dot X_1^2+\dot Y_1^2 \\ v_2^2=\dot X_2^2+\dot Y_2^2$$

and the kinetic energy

$$T=\frac m2\,(v_1^2+v_2^2)$$

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Eli
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  • 31
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