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- Homework Statement
- Find the lagrangian of the double pendulum, kinetic energy

- Relevant Equations
- L=T-U

This is from Taylor's classical mechanichs, 11.4, example of finding the Lagrangian of the double pendulum

Relevant figure attached below

Angle between the two velocities of second mass is

$$\phi_2-\phi_1$$

Potential energy

$$U_1=m_1gL_1$$

$$U_2=m_2g[L_1\cos(1-\phi_1)+L_2(1-\phi_2)]$$

$$(U\phi_1,phi_2)=m_1gL_1+m_2g[L_1\cos(1-\phi_1)+L_2(1-\phi_2)]$$

Kinetic energy of the second mass in the double pendulum

$$T_1=\frac{1}{2}m_1L_1^2\dot\phi_1^2$$

$$T_2=\frac{1}{2} m_2[L_1^2\dot\phi_1^2+2L_1L_2\dot\phi_2\dot\phi_1\cos(\phi_1-\phi_2)+L_2^2\dot\phi_2^2]$$

Where $$ T_2=\frac{1}{2}m_2(v_1+L_2\dot\phi_2)^2$$

I am trouble understanding the cos term here. Does it come from the unit vector? Can someone explain why it's involved, because I could not follow why it's here after squaring the velocity for the second kinetic energy

$$2L_1L_2\dot\phi_2\dot\phi_1\cos(\phi_1-\phi_2)$$

Relevant figure attached below

Angle between the two velocities of second mass is

$$\phi_2-\phi_1$$

Potential energy

$$U_1=m_1gL_1$$

$$U_2=m_2g[L_1\cos(1-\phi_1)+L_2(1-\phi_2)]$$

$$(U\phi_1,phi_2)=m_1gL_1+m_2g[L_1\cos(1-\phi_1)+L_2(1-\phi_2)]$$

Kinetic energy of the second mass in the double pendulum

$$T_1=\frac{1}{2}m_1L_1^2\dot\phi_1^2$$

$$T_2=\frac{1}{2} m_2[L_1^2\dot\phi_1^2+2L_1L_2\dot\phi_2\dot\phi_1\cos(\phi_1-\phi_2)+L_2^2\dot\phi_2^2]$$

Where $$ T_2=\frac{1}{2}m_2(v_1+L_2\dot\phi_2)^2$$

I am trouble understanding the cos term here. Does it come from the unit vector? Can someone explain why it's involved, because I could not follow why it's here after squaring the velocity for the second kinetic energy

$$2L_1L_2\dot\phi_2\dot\phi_1\cos(\phi_1-\phi_2)$$

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