MHB Double pendulum: energy function

[mathmari]

Hey!

I am looking at the following:

Find the energy function for the double pendulum, with vertical constant and uniform gravity field

• for motion restricted at vertical plane
• for motion in space

We have that the energy function is the sum of the potential and the kinetic energy, right?

The potential energy of the system is given by $PE=m_1gy_1+m_2gy_2$.

The kinetic energy is given by $KE=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$.

Therefore, the energy function is equal to $$E=PE+KE=m_1gy_1+m_2gy_2+\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ Is this correct?

How can we distinct between motion restricted at vertical plane and motion in space? And what does a uniform gravitational field imply?

(Wondering)

 

[I like Serena]

How can we distinct between motion restricted at vertical plane and motion in space?

Hey mathmari! (Smile)

In a vertical plane we only have $x$ and $y$.
So $v_1^2 = \dot x_1^2 + \dot y_1^2$.

For motion in space we also have a 3rd coordinate (perhaps we should pick the vertical coordinate to be $z$ instead of $y$ to avoid confusion.) (Thinking)

And what does a uniform gravitational field imply?

That we have constant vertical acceleration by gravity $g$, which is what you already have.
That is as opposed to how gravity actually works, which is $\frac{GM}{r^2}$, where $M$ is the mass of the earth, and $r$ is the distance to the center of the earth. At the surface we can treat it as a uniform field though. (Thinking)

 

[mathmari]

Hey mathmari! (Smile)

In a vertical plane we only have $x$ and $y$.
So $v_1^2 = \dot x_1^2 + \dot y_1^2$.

For motion in space we also have a 3rd coordinate (perhaps we should pick the vertical coordinate to be $z$ instead of $y$ to avoid confusion.) (Thinking)

So, to difference between the two motions is at the kinetic energy?

Do we get for the first case:

The potential energy of the system is given by $PE=m_1gy_1+m_2gy_2$.
The kinetic energy is given by $KE=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}m_1\left (\dot x_1^2 + \dot y_1^2\right )+\frac{1}{2}m_2\left (\dot x_2^2 + \dot y_2^2\right )$.

and for the second case:

The potential energy of the system is given by $PE=m_1gy_1+m_2gy_2$.
Here do we have to use also $z_1,z_2$ ? But how?
The kinetic energy is given by $KE=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}m_1\left (\dot x_1^2 + \dot y_1^2+\dot z_1^2\right )+\frac{1}{2}m_2\left (\dot x_2^2 + \dot y_2^2+\dot z_2^2\right )$.

? Or do we have to use for example polar coordinates? (Wondering)

 

[I like Serena]

Yep. (Nod)

We can write the same thing in polar coordinates, but there is no need.
The only reason to do so, is if it helps us to solve a follow up question, and we can do it then.

 

[mathmari]

Yep. (Nod)

So, we do we get the following energy function?

1. for motion restricted at vertical plane
   
   The potential energy of the system is given by $PE=m_1gy_1+m_2gy_2$.
   The kinetic energy is given by $KE=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}m_1\left (\dot x_1^2 + \dot y_1^2\right )+\frac{1}{2}m_2\left (\dot x_2^2 + \dot y_2^2\right )$.
   
   So, the total energy is given by $$E=PE+KE=m_1gy_1+m_2gy_2+\frac{1}{2}m_1\left (\dot x_1^2 + \dot y_1^2\right )+\frac{1}{2}m_2\left (\dot x_2^2 + \dot y_2^2\right )$$
   
2. for motion in space
   
   The potential energy of the system is given by $PE=m_1gy_1+m_2gy_2$.
   The kinetic energy is given by $KE=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}m_1\left (\dot x_1^2 + \dot y_1^2+\dot z_1^2\right )+\frac{1}{2}m_2\left (\dot x_2^2 + \dot y_2^2+\dot z_2^2\right )$.
   
   So, the total energy is given by $$E=PE+KE=m_1gy_1+m_2gy_2+\frac{1}{2}m_1\left (\dot x_1^2 + \dot y_1^2+\dot z_1^2\right )+\frac{1}{2}m_2\left (\dot x_2^2 + \dot y_2^2+\dot z_2^2\right )$$

Is the potential energy in both cases the same? (Wondering)

 

[I like Serena]

Yes and yes. (Mmm)

 

[mathmari]

Yes and yes. (Mmm)

Ok! Thank you very much! (Smile)