Velocity of end points of the rod

[Nathanael]

Okay, now I've shown that (after the rod loses contact with the wall) it will not become airborne.

I've shown this via the inequality
4.5\cos^2(\theta)\sin(\theta)+25.5\sin(\theta)-31.5<0
which is always true.

If we are talking about after the rod loses contact with the wall, then \frac{d\omega}{dt}=0 and the inequality in my last post becomes g<\omega ^2\sin(\theta) which can be written as:

\frac{mgL}{12\sin(\theta)}<\frac{I\omega^2}{2} ... This is the condition for the rod to lose contact.

Now we can replace the right side of the inequality (rotational energy) with the conservation of energy equation. (I will also flip the inequality, so that we get the condition for the rod not to lose contact.) Then we get:

\frac{mgL}{12\sin(\theta)}>\frac{mgL}{2}(1-\sin\theta)-\frac{mgL}{18}-\frac{m}{2}V_y^2
(This only represents the rotational energy after it loses contact because \frac{mgL}{18} is the horizontal kinetic energy after losing contact.)

So then we just need to determine V_y(\theta) the vertical velocity of the center of mass as a function of the angle (which only needs to be true after it loses contact with the wall).

I determined that V_y^2=\frac{gL(\frac{8}{9}-\sin\theta)}{1+\frac{1}{3\cos^2\theta}}
(This equation should only be true after it loses contact with the wall, i.e. \theta<\arcsin(2/3))
Now, we just do a whole lot of manipulations to get to my final inequality (which turns out to be always true).Some how I feel like haruspex has a better way...

 

[ehild]

I determined that V_y^2=\frac{gL(\frac{8}{9}-\sin\theta)}{1+\frac{1}{3\cos^2\theta}}
(This equation should only be true after it loses contact with the wall, i.e. \theta<\arcsin(2/3))

Very nice!
Instead of finding when dv_y/dt can be equal to g, It is easier to find the condition when $\ddot \theta = 0$. If the rod loses contact with the floor, the torque becomes zero, and the angular acceleration is zero.
From conservation of energy, you get that $(\dot \theta )^2 = \frac {\frac{32}{9}-4 \sin(\theta)}{ \cos^2+\frac{1}{3}}$ which is equivalent with your formula for V_y² (I hope) .
Derive both sides. You get an equation for $\ddot\theta$ in terms of theta. It can be rewritten as a quadratic equation in sin(theta), which has no real solutions.

 

[Nathanael]

It is easier to find the condition when $\ddot \theta = 0$.

Ahh, of course!
I just used $\ddot \theta = 0$ as one piece of my method to simplify it... But the whole time I had a feeling, "there must be a simpler way..."

 

[Satvik Pandey]

I think you made a mistake with the signs (the accelerations are in opposite directions)

I am confused here. Acceleration due to gravity $g$ is in downward direction and the vertical velocity of the CoM is in downward direction so differentiating that wrt time gives acceleration (in downward direction).

I determined that V_y^2=\frac{gL(\frac{8}{9}-\sin\theta)}{1+\frac{1}{3\cos^2\theta}}
(This equation should only be true after it loses contact with the wall, i.e. \theta<\arcsin(2/3))

Now, we just do a whole lot of manipulations to get to my final inequality (which turns out to be always true).Some how I feel like haruspex has a better way...

Got that! Thanks! :)

Very nice!
Instead of finding when dv_y/dt can be equal to g, It is easier to find the condition when $\ddot \theta = 0$. If the rod loses contact with the floor, the torque becomes zero, and the angular acceleration is zero.
From conservation of energy, you get that $(\dot \theta )^2 = \frac {\frac{32}{9}-4 \sin(\theta)}{ \cos^2+\frac{1}{3}}$ which is equivalent with your formula for V_y² (I hope) .
Derive both sides. You get an equation for $\ddot\theta$ in terms of theta. It can be rewritten as a quadratic equation in sin(theta), which has no real solutions.

I think this is an easier way. $\ddot \theta = 0$-- I didn't think about that.

Thank you guys.:)