Skip to main content
added 4 characters in body; edited tags
Source Link
Qmechanic
  • 206.6k
  • 48
  • 566
  • 2.3k

I have been trying to figure out the equation of motion of a rotating disc with an added mass fixed on the surface of the disc at a certain distance r$r$ from the disc's centre point.

The blue dot is the added mass

I have approached the problem in a few ways. Using the Euler-Lagrange equation with the following Lagrangian: $$L=\frac{1}{2}I\omega ^2 - mgr \sin\theta $$ where I is the total moment of inertia, m$m$ is the added mass. I get $$\ddot{\theta}=\frac{-mgr\cos\theta}{I}$$ How would one carry on from here?

Another approach I tried was to look at the kinetic energy of the system, to find the angular velocity as a function of position but I got stuck there too. $$\omega (\theta) = \sqrt{\omega^2 - \frac{2mgr\sin\theta}{I}}$$ I end up with the integral: $$ t(\theta)=\int_{}^{} \frac{1}{\sqrt{\overline{\omega}^2-\frac{2mgr\sin\theta}{I}}} d\theta$$ Which I thought could be inversed to find $\theta (t)$.

I have with some help been able to write this as an incomplete elliptic integral of the first kind by using the substitutions $a=\overline{\omega}^2$, $b=-\frac{2mgr}{I}$ and $\phi = \frac{\pi}{4}-\frac{\theta}{2}$ into $$t(\theta)=-2\sqrt{a-b}\int\frac{d\phi}{\sqrt{1-\frac{2b}{b-a}\sin^2\phi}}$$

I don't really know how I am supposed to think about this integral.

I have also solved the problem numerically resulting in the following graph, but it was hard to find a good approximation of the function:

enter image description here

The numerical method made me think that the function should be a function of the integral boundaries, to find the time. I think that it should look something like this:

$$t(\theta)=-2\sqrt{a-b}\int_0^\theta\frac{d\phi}{\sqrt{1-\frac{2b}{b-a}\sin^2\phi}}$$

How could I use this find an approximation?

I have been trying to figure out the equation of motion of a rotating disc with an added mass fixed on the surface of the disc at a certain distance r from the disc's centre point.

The blue dot is the added mass

I have approached the problem in a few ways. Using the Euler-Lagrange equation with the following Lagrangian: $$L=\frac{1}{2}I\omega ^2 - mgr \sin\theta $$ where I is the total moment of inertia, m is the added mass. I get $$\ddot{\theta}=\frac{-mgr\cos\theta}{I}$$ How would one carry on from here?

Another approach I tried was to look at the kinetic energy of the system, to find the angular velocity as a function of position but I got stuck there too. $$\omega (\theta) = \sqrt{\omega^2 - \frac{2mgr\sin\theta}{I}}$$ I end up with the integral: $$ t(\theta)=\int_{}^{} \frac{1}{\sqrt{\overline{\omega}^2-\frac{2mgr\sin\theta}{I}}} d\theta$$ Which I thought could be inversed to find $\theta (t)$.

I have with some help been able to write this as an incomplete elliptic integral of the first kind by using the substitutions $a=\overline{\omega}^2$, $b=-\frac{2mgr}{I}$ and $\phi = \frac{\pi}{4}-\frac{\theta}{2}$ into $$t(\theta)=-2\sqrt{a-b}\int\frac{d\phi}{\sqrt{1-\frac{2b}{b-a}\sin^2\phi}}$$

I don't really know how I am supposed to think about this integral.

I have also solved the problem numerically resulting in the following graph, but it was hard to find a good approximation of the function:

enter image description here

The numerical method made me think that the function should be a function of the integral boundaries, to find the time. I think that it should look something like this:

$$t(\theta)=-2\sqrt{a-b}\int_0^\theta\frac{d\phi}{\sqrt{1-\frac{2b}{b-a}\sin^2\phi}}$$

How could I use this find an approximation?

I have been trying to figure out the equation of motion of a rotating disc with an added mass fixed on the surface of the disc at a certain distance $r$ from the disc's centre point.

The blue dot is the added mass

I have approached the problem in a few ways. Using the Euler-Lagrange equation with the following Lagrangian: $$L=\frac{1}{2}I\omega ^2 - mgr \sin\theta $$ where I is the total moment of inertia, $m$ is the added mass. I get $$\ddot{\theta}=\frac{-mgr\cos\theta}{I}$$ How would one carry on from here?

Another approach I tried was to look at the kinetic energy of the system, to find the angular velocity as a function of position but I got stuck there too. $$\omega (\theta) = \sqrt{\omega^2 - \frac{2mgr\sin\theta}{I}}$$ I end up with the integral: $$ t(\theta)=\int_{}^{} \frac{1}{\sqrt{\overline{\omega}^2-\frac{2mgr\sin\theta}{I}}} d\theta$$ Which I thought could be inversed to find $\theta (t)$.

I have with some help been able to write this as an incomplete elliptic integral of the first kind by using the substitutions $a=\overline{\omega}^2$, $b=-\frac{2mgr}{I}$ and $\phi = \frac{\pi}{4}-\frac{\theta}{2}$ into $$t(\theta)=-2\sqrt{a-b}\int\frac{d\phi}{\sqrt{1-\frac{2b}{b-a}\sin^2\phi}}$$

I don't really know how I am supposed to think about this integral.

I have also solved the problem numerically resulting in the following graph, but it was hard to find a good approximation of the function:

enter image description here

The numerical method made me think that the function should be a function of the integral boundaries, to find the time. I think that it should look something like this:

$$t(\theta)=-2\sqrt{a-b}\int_0^\theta\frac{d\phi}{\sqrt{1-\frac{2b}{b-a}\sin^2\phi}}$$

How could I use this find an approximation?

added 109 characters in body
Source Link
Loading
added 294 characters in body
Source Link
Loading
added 506 characters in body
Source Link
Loading
deleted 3 characters in body; edited tags
Source Link
Qmechanic
  • 206.6k
  • 48
  • 566
  • 2.3k
Loading
Source Link
Loading