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## Homework Statement

The first block with mass [itex] m_1 [/itex] slides without friction on a wedge which has an incline of angle [itex]\alpha[/itex]. The wedged shaped block has a mass [itex] m_2 [/itex]. The second block is also allowed to slide on a flat frictionless line. Find Lagrange's equations of motion.

## Homework Equations

[itex] L = T-U [/itex]

[itex] \frac{∂L}{∂q_j}-\frac{d}{dt}\frac{∂L}{∂\dot{q_j}}=0 [/itex]

## The Attempt at a Solution

First I'll define a vector [itex] \vec{R} [/itex] that points to the center of mass of the system. Also [itex] \vec{r_1} [/itex] is the vector that points to [itex] m_1 [/itex] and [itex] \vec{r_2} [/itex] is the vector that points to [itex] m_2 [/itex]. Additionally [itex] \vec{R}=\frac{1}{M}[m_1\vec{r_1}+m_2\vec{r_2}][/itex]. Finally let [itex] \vec{r} = \vec{r_2}-\vec{r_1} [/itex].

If I combine these equations I'll get [itex] \vec{r_1} = \vec{R}+\frac{m_2}{M}\vec{r} [/itex] and [itex] \vec{r_2} = \vec{R}-\frac{m_1}{M}\vec{r} [/itex]

Also [itex] \dot{\vec{r_1}} = \dot{\vec{R}}+\frac{m_2}{M}\dot{\vec{r}} [/itex] and [itex]\dot{\vec{r_2}} = \dot{\vec{R}}-\frac{m_1}{M}\dot{\vec{r}}[/itex]

Now I need the total kinetic energy of the system: [itex] T [/itex]

[itex] T = \frac{M}{2}(\dot{\vec{r_1}}+\dot{\vec{r_2}})^2 [/itex] Now I'll check this expression for accuracy when [itex] \dot{\vec{r_2}} = \vec{0} [/itex]. When the second block is stationary that implies that [itex] \dot{\vec{R}} = \frac{m_1}{M}\dot{\vec{r}} [/itex] Which if I substitute back into [itex] T = \frac{M}{2}(\dot{\vec{r_1}})^2 [/itex] gives [itex] \frac{M}{2}(\frac{m_1+m_2}{M}\dot{\vec{r}})^2[/itex] The kinetic energy of the second block will be zero since it is at rest so setting [itex] m_2=0 [/itex] gives [itex] \frac{m_1}{2}(\dot{\vec{r}})^2 [/itex] So in the limit that the velocity of the second block goes to zero I get back to the familiar [itex] T [/itex] of just one particle.

Now [itex] \dot{\vec{R}} = (\dot{X},\dot{Y}) [/itex] and [itex] \dot{\vec{r}} = (\dot{x},\dot{y}) [/itex]

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

[itex] T=\frac{M}{2}[4\dot{\vec{R}}\cdot\dot{\vec{R}}+(\frac{m_2-m_1}{M})^2\dot{\vec{r}}\cdot\dot{\vec{r}}+4\frac{m_2-m_1}{M}\dot{\vec{R}}\cdot\dot{\vec{r}}] [/itex]

[itex] T = \frac{M}{2}[4(\dot{X}^2+\dot{Y}^2)+(\frac{m_2-m_1}{M})^2(\dot{x}^2+\dot{y}^2)+4\frac{m_2-m_1}{M}(\dot{X}\dot{x}+\dot{Y}\dot{y})][/itex]

Okay thats [itex] T [/itex] Now I need [itex] U [/itex]

[itex] U = M(\vec{g}\cdot\vec{r_1}+\vec{g}\cdot\vec{r_2})[/itex]

where [itex] \vec{g} = (0,-g) [/itex], [itex] \vec{r_1} = (X+\frac{m_2}{M}x,Y+\frac{m_2}{M}y) [/itex], and [itex] \vec{r_2} = (X-\frac{m_1}{M}x,Y-\frac{m_1}{M}y)[/itex]

Okay so now I need to apply my constraints which are now: [itex]\frac{Y}{X}=tan(\alpha)[/itex] and [itex] \frac{y}{x}=tan(\alpha) [/itex] so plugging in to both [itex] T [/itex] and [itex] U [/itex]

[itex] T=\frac{M}{2}[4(1+tan^2(\alpha))\dot{X}^2+(\frac{m_2-m_1}{M})^2(1+tan^2(\alpha))\dot{x}^2+4\frac{m_2-m_1}{M}(1+tan^2(\alpha))\dot{X}\dot{x}] [/itex]

[itex] T =\frac{Msec^2(\alpha)}{2}[4\dot{X}^2+(\frac{m_2-m_1}{M})^2\dot{x}^2+4\frac{m_2-m_1}{M}\dot{X}\dot{x}] [/itex]

and

[itex] U = M(\vec{g}\cdot(\vec{r_1}+\vec{r_2})) = M[-2gtan(\alpha)X-g\frac{m_2-m_1}{M}tan(\alpha)x] [/itex]

[itex] U = Mgtan(\alpha)[-2X-\frac{m_2-m_1}{M}x] [/itex]

Then [itex] L= T-U [/itex] and [itex] H = T+U [/itex]

[itex] L = T -U = \frac{Msec^2(\alpha)}{2}[4\dot{X}^2+(\frac{m_2-m_1}{M})^2\dot{x}^2+4\frac{m_2-m_1}{M}\dot{X}\dot{x}]-Mgtan(\alpha)[-2X-\frac{m_2-m_1}{M}x] [/itex]

Now the momenta are: [itex] p_X = \frac{∂L}{∂\dot{X}} = \frac{Msec^2(\alpha)}{2}[8\dot{X}+4\frac{m_2-m_1}{M}\dot{x}] [/itex] and [itex] p_x =\frac{∂L}{∂\dot{x}}= Msec^2(\alpha)[(\frac{m_2-m_1}{M})^2\dot{x}+2\frac{m_2-m_1}{M}\dot{X}] [/itex]

Also: [itex] \frac{∂L}{∂X} = 2Mgtan(\alpha) [/itex] and [itex] \frac{∂L}{∂x} = (m_2-m_1)gtan(\alpha) [/itex]

Then the Lagrange equations are:

[itex]x[/itex]: [itex] (m_2-m_1)gtan(\alpha)-Msec^2(\alpha)[(\frac{m_2-m_1}{M})^2\ddot{x}+2\frac{m_2-m_1}{M}\ddot{X}]=0 [/itex]

[itex]X[/itex]: [itex] 2Mgtan(\alpha)-Msec^2(\alpha)[4\ddot{X}+2\frac{m_2-m_1}{M}\ddot{x}]=0 [/itex]