# Double pendulum: energy function

• MHB
• mathmari
In summary, the potential energy is the same for both cases, and the kinetic energy is also the same for both cases.
mathmari
Gold Member
MHB
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)

mathmari said:
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)

mathmari said:
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)

I like Serena said:
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)

Last edited by a moderator:
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.

I like Serena said:
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)

Yes and yes. (Mmm)

I like Serena said:
Yes and yes. (Mmm)

Ok! Thank you very much! (Smile)

## 1. What is a double pendulum?

A double pendulum is a physical system consisting of two pendulums connected by a joint. The first pendulum is attached to a fixed point, while the second pendulum hangs from the end of the first pendulum. When the system is in motion, the two pendulums swing and interact with each other.

## 2. How is the energy of a double pendulum calculated?

The energy of a double pendulum can be calculated using its kinetic and potential energy. The kinetic energy is determined by the mass and velocity of each pendulum, while the potential energy is determined by the height and angle of each pendulum.

## 3. What is the energy function of a double pendulum?

The energy function of a double pendulum is a mathematical equation that represents the total energy of the system at any given point in time. It takes into account the kinetic and potential energy of both pendulums and can be used to analyze the behavior of the system.

## 4. How does the energy function affect the motion of a double pendulum?

The energy function has a direct impact on the motion of a double pendulum. As the pendulums swing back and forth, the energy function changes, causing fluctuations in the system's total energy. This can result in chaotic and unpredictable motion.

## 5. What are some real-world applications of the double pendulum?

The double pendulum has applications in various fields, such as physics, engineering, and mathematics. It is often used as a model to study chaotic systems and can also be found in certain types of clocks and toys. Additionally, the double pendulum is used in sports science to analyze the movements of athletes during complex motions.

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