- #1
dracobook
- 23
- 0
Homework Statement
I apologize if this is not the right place to put this. If it is not please redirect me for future reference.
11. Consider a uniform thin disk that rolls without slipping on a horizontal plane. A horizontal force is applied to the center of the disk and in a direction parallel to the plane of the disk.
a. Derive Lagrange's equations and find the generalized force.
Homework Equations
[tex] \mathbb{L} = T- U
[/tex]
[tex] \frac{d}{dt}( \frac{\partial T}{\partial \dot q_j}) - \frac{\partial T}{\partial q_j} = Q_j
[/tex]
[tex] \frac{d}{dt}(\frac{\partial \dot L}{\partial \dot q_j}) - \frac{\partial L}{\partial q_j} = 0
[/tex]
[tex] Q_j[/tex] is the generalized forces
The Attempt at a Solution
[tex] \dot x = \dot\theta R[/tex], where R is radius of disk.
[tex] T = \frac{1}{2} m v^2 + \frac{1}{2}I\dot\theta^2 \text{, }V=0\text{, } I=\frac{mR^2}{2} \Rightarrow \mathbb{L} = \frac{3}{4}m \dot x^2[/tex]
I would assume then that I would just need to plug in [tex] \mathbb{L} [/tex] into the euler Lagrange equation...however it seems that
[tex]
\frac{d}{dt}(\frac{\partial \dot L}{\partial \dot q_j}) - \frac{\partial L}{\partial q_j} = Q \neq 0
[/tex]
So my question is:
1) Why is this so? Is this because the potential is zero? Hence the generalized force cannot be derivable from a scalar potential? and hence we can only reduce it to
[tex] \frac{d}{dt}(\frac{\partial \dot T}{\partial \dot q_j}) - \frac{\partial T}{\partial q_j} = Q_j [/tex]
And so does that mean that whenever potential is zero we can only reduce the problem with the equation above?
2) Where does [tex] \frac{1}{2}I\dot\theta^2 [/tex] come from in the equation for kinetic energy??
Thanks in advance!
Last edited: