- #1
Arjan82
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- 442
This could also be posted in the Math / differential equations sub, but it also involves the derivation which is classical physics. So I was doubting .
So, I'm dusting off my dynamics a bit and found this problem of a thin beam on a frictionless surface in a different forum and decided to have a crack at it:
The surface is frictionless, so the end of the beam on the surface will slide to the right when it is released. What is the normal force / reaction force ##N##? This is the derivation of the differential equations:
First the relation between ##\theta(t)## and ##y(t)##, the latter is measured at the middle of the beam:
$$ y = \frac{l}{2}\cos(\theta) $$
$$\dot{y} = - \frac{l}{2}\sin(\theta) \dot{\theta}$$
$$\ddot{y} = - \frac{l}{2}\left(\dot{\theta}^2\cos(\theta) + \ddot{\theta}\sin(\theta)\right)$$
We have a coupled system for vertical force and momentum (positive CCW), so we need ##\Sigma F_y = ma_y## and ##\Sigma M = I\alpha##:
$$ \Sigma F_y = N - mg = m\ddot{y} = - m\frac{l}{2}\left(\dot{\theta}^2\cos(\theta) + \ddot{\theta}\sin(\theta)\right)$$
For the sum of the moments we take the point at the middle of the beam such that gravity falls out. Also, then we can take the moment of inertia around the center, ##I_m = 1/12ml^2##, so in this case:
$$ \Sigma M = \frac{l}{2}N\sin(\theta) = I_m\ddot\theta = \frac{1}{12}ml^2\ddot{\theta}$$
And here is our system of equations describing the beam, two equations (##\Sigma F_y## and ##\Sigma M##) with two unknowns (##N(t)## and ##\theta(t)##). Now solving these coupled non-linear differential equations analytically is hard if not impossible (I guess?!?).
Some questions:
So, I'm dusting off my dynamics a bit and found this problem of a thin beam on a frictionless surface in a different forum and decided to have a crack at it:
The surface is frictionless, so the end of the beam on the surface will slide to the right when it is released. What is the normal force / reaction force ##N##? This is the derivation of the differential equations:
First the relation between ##\theta(t)## and ##y(t)##, the latter is measured at the middle of the beam:
$$ y = \frac{l}{2}\cos(\theta) $$
$$\dot{y} = - \frac{l}{2}\sin(\theta) \dot{\theta}$$
$$\ddot{y} = - \frac{l}{2}\left(\dot{\theta}^2\cos(\theta) + \ddot{\theta}\sin(\theta)\right)$$
We have a coupled system for vertical force and momentum (positive CCW), so we need ##\Sigma F_y = ma_y## and ##\Sigma M = I\alpha##:
$$ \Sigma F_y = N - mg = m\ddot{y} = - m\frac{l}{2}\left(\dot{\theta}^2\cos(\theta) + \ddot{\theta}\sin(\theta)\right)$$
For the sum of the moments we take the point at the middle of the beam such that gravity falls out. Also, then we can take the moment of inertia around the center, ##I_m = 1/12ml^2##, so in this case:
$$ \Sigma M = \frac{l}{2}N\sin(\theta) = I_m\ddot\theta = \frac{1}{12}ml^2\ddot{\theta}$$
And here is our system of equations describing the beam, two equations (##\Sigma F_y## and ##\Sigma M##) with two unknowns (##N(t)## and ##\theta(t)##). Now solving these coupled non-linear differential equations analytically is hard if not impossible (I guess?!?).
Some questions:
- You can solve the ##\Sigma M## equation for ##\ddot{\theta}## and substitute in the ##\Sigma F_y## equation. If you then also take as initial condition ##\dot{\theta} = 0## and ##\theta = \theta_0## you end up with: $$N = \frac{mg}{1+3\sin^2(\theta_0)}$$ This would suggest that the initial condition for ##N## is not free to chose because otherwise it would lead to inconsistencies. But I would guess that to solve this equation you need an initial condition for ##N(t)##, ##\theta(t)##, ##\dot{\theta}(t)## and ##\ddot{\theta}(t)## (in fact, I want to solve this in Matlab for which I need to rewrite everything in a system of 4 equations which need 4 initial conditions...). This is paradoxical to me, what am I missing? (In fact, in the other forum (Dutch) the OP asked for the normal force ##N## directly after the beam is released and got this answer, which would suggest it is fixed)
- I'm guessing this is not the only or perhaps not even the easiest way to do this? Are there better ways? With energy conservation maybe? I seem to remember this was also possible.
- Is there some opensource software you can model this in and get the answers as a check? I've found HOTINT but I'm not yet able to get the reaction force out of this (only kinematics, i.e. speeds and accelerations). I've tried to solve this in Matlab (the ode45 solver) but hopelessly failed thus far...