Classical Mechanics: Lagrangian of a falling yo-yo

In summary, the problem discusses a yo-yo with a string wrapped around it, moving in a vertical straight line. The string is attached to a support that has upward displacement h(t) at time t. The rotation angle φ of the yo-yo is taken as a generalized coordinate and Lagrange's equation is found for this system. The acceleration of the yo-yo is determined and the support's acceleration h(t) is calculated in order for the center of the yo-yo to remain at rest. The total energy of the system at time t is also discussed. There is a discussion about the Lagrange's equation being independent of h(t), and a link to a forum discussion is provided for more information.
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NahsiN
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Homework Statement


A uniform circular cylinder (a yo-yo) of radius a and mass M has a string wrapped around
it that can unwind without slipping. The yo-yo moves in a vertical straight line and the
straight part of the string is vertical as well. The other end of the string is fastened to a
support that has upward displacement h(t) at time t. Here h(t) is a prescribed function, not
a degree of freedom.
a. Take the rotation angle φ of the yo-yo as a generalized coordinate and find Lagrange’s equation.
b. Find the acceleration of the yo-yo. What must the support’s acceleration h(t) be so
that the centre of the yo-yo can remain at rest?
c. Suppose the system starts from rest. Find an expression for the total energy E = T+V at time t, in terms of h and h.

Homework Equations


L=T-V
T=[tex]\frac{1}{2}[/tex]M[tex]\dot{y}[/tex][tex]^{2}[/tex]+[tex]\frac{1}{2}[/tex]I[tex]\dot{\varphi}[/tex][tex]^{2}[/tex]
Defining downwards as my positive direction: V=-Mgy
Rolling without slipping ---> a[tex]\dot{\varphi}[/tex]=[tex]\dot{y}[/tex]
Since the support is moving upwards given by (h(t)) ---> y=[tex]\widetilde{y}[/tex]-h(t) where [tex]\widetilde{y}[/tex]= the distance between the yo-yo and the support at time t. So, [tex]\dot{y}[/tex]=[tex]\dot{\widetilde{y}}[/tex]-[tex]\dot{h(t)}[/tex]

The Attempt at a Solution


With these equations and treating [tex]\varphi[/tex] as my generalized coordinate, it's easy to obtain the Lagrange's equation of motion for this system. I get [tex]\dot{}\dot{\varphi}[/tex]=2/3*g/a

Now for part b) of the question, I have [tex]\dot{}\dot{y}[/tex]=[tex]\dot{}\dot{h}[/tex]. If this is correct, I can figure this out using the without slipping equation above.
From here on before starting part c), I am wondering why the Lagrange's equation I got is independent of h(t) (i.e its the same as if the support was fixed). And if its wrong, it has to do something with the relation between y and h(t). Any help is apprecaited
 
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Related to Classical Mechanics: Lagrangian of a falling yo-yo

1. What is the Lagrangian of a falling yo-yo?

The Lagrangian of a falling yo-yo is a mathematical expression that describes the total kinetic and potential energy of the yo-yo as it falls. It is represented by the symbol L and is equal to the difference between the kinetic energy and potential energy of the yo-yo.

2. How is the Lagrangian of a falling yo-yo calculated?

The Lagrangian of a falling yo-yo is calculated using the equation L = T - V, where T represents the kinetic energy of the yo-yo and V represents the potential energy. The kinetic energy of the yo-yo can be calculated using its mass, velocity, and moment of inertia, while the potential energy is calculated based on the height and gravitational acceleration.

3. Why is the Lagrangian an important concept in classical mechanics?

The Lagrangian is an important concept in classical mechanics because it provides a more efficient and comprehensive way to describe the motion of a system. It allows for the use of generalized coordinates, which simplifies the equations of motion and makes it easier to solve complex problems. It also takes into account the conservation of energy and momentum, making it a powerful tool for analyzing physical systems.

4. How is the Lagrangian of a falling yo-yo different from that of a simple pendulum?

While both the falling yo-yo and simple pendulum have a similar mathematical expression for their Lagrangian, the main difference lies in the moment of inertia. The yo-yo has a variable moment of inertia as it falls due to the changing radius of rotation, while the simple pendulum has a constant moment of inertia. This results in a more complex and non-linear equation for the yo-yo's Lagrangian.

5. Can the Lagrangian be used to predict the motion of a falling yo-yo?

Yes, the Lagrangian can be used to predict the motion of a falling yo-yo. By using the Euler-Lagrange equation, which relates the Lagrangian to the equations of motion, one can calculate the trajectory of the yo-yo as it falls. However, this method may not be as straightforward as using other equations such as Newton's laws, and may require more advanced mathematical techniques.

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