# Velocity of end points of the rod

• Satvik Pandey
In summary, the rigid uniform rod 'AB' of mass 'M' and length 'L' is pulled slightly (Gently) at the bottom at time t=0 when it just reaches the horizontal ground, and then finds the velocities of its end points.

## Homework Statement

Rigid uniform Rod 'AB' of mass 'M' and Length 'L' is pulled slightly ( Gently ) at the bottom at time t=0 when it just reaches the horizontal ground, at that time find the Velocities of End Points of the rod.

All the surfaces are friction less.

## The Attempt at a Solution

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I first tried to find the value of ##\theta## at which the rod left contact with the vertical wall.

From the figure

##{ V }_{ BC }=\frac { l\omega }{ 2 }##

##{ V }_{ BC }=\frac { l\omega }{ 2 } sin\theta \hat { i } +\frac { l\omega }{ 2 } cos\theta \hat { j } ##

and $${ V }_{ AC }=-\frac { l\omega }{ 2 } sin\theta \hat { i } -\frac { l\omega }{ 2 } cos\theta \hat { j }$$

As ##{ V }_{ BC }={ V }_{ B }-{ V }_{ C }## and ##{ V }_{ AC }={ V }_{ A }-{ V }_{ C }##

or ##{ V }_{ C }={ V }_{ B }-{ V }_{ BC }##

So ##{ V }_{ C }\hat { j } =-\frac { l\omega }{ 2 } cos\theta \hat { j } ## (considering velocity in vertical direction only. We are doing this because ##{ V }_{ B }\hat { j } =0##)

Similarly we can get

##{ V }_{ C }\hat { i } =\frac { l\omega }{ 2 } sin\theta \hat { i } ##...(4)

So ##{ V }_{ C }=\frac { l\omega }{ 2 } sin\theta \hat { i } -\frac { l\omega }{ 2 } cos\theta \hat { j } ##

Also from conservation of energy

##mg\frac { l }{ 2 } (1-sin\theta )=\frac { 1 }{ 2 } \left( m{ v }^{ 2 }+{ I }_{ C }{ \omega }^{ 2 } \right) ##......(1)

Also ##{ V }_{ C }={ \left( \frac { l\omega }{ 2 } \right) }^{ 2 }## (we can get this from eq(4))

Putting this in eq-1 we get

##{ \omega }^{ 2 }=\frac { 3g }{ l } (1-sin\theta )##...(2)

When the rod looses contact with the vertical rod the Normal force acting on Rod becomes 0. So no horizontal force act on the rod after the moment the rod left contact with vertical wall. So acceleration of the CoM of the rod in horizontal direction is 0.

So ##{ \frac { d }{ dt } (V }_{ C }\hat { i } )=\frac { d }{ dt } \left( \frac { l\omega }{ 2 } sin\theta \hat { i } \right) =0##

So ##\frac { d }{ d\theta } \left( \sqrt { \frac { 3g }{ l } (1-sin\theta ) } sin\theta \right) \frac { d\theta }{ dt } =0\\ ##

or ##\frac { d }{ d\theta } \left( \sqrt { (1-sin\theta ) } sin\theta \right) =0##

or ##cos\theta \sqrt { (1-sin\theta ) } =\frac { sin\theta cos\theta }{ 2\sqrt { (1-sin\theta ) } } ##

So ##sin\theta =\frac { 2 }{ 3 } ##

So at ## \theta=arcsin(2/3)## ##\omega=\sqrt{\frac{g}{L}}##

So ##{ V }_{ C }=\frac{\sqrt{gl}}{3} { i } -\frac{\sqrt{5gl}}{6} { j } ##

After loosing contact with the vertical wall no horizontal force acts on the CoM of the rod so the vertical component of the rod will not change.

So ##V_{A}=\frac{\sqrt{gl}}{3}##. Where ##V_{A}## is the velocity of the lower end of the rod when the rod becomes horizontal.

Let the rod make an angle ##\phi## withe the horizontal after loosing contact.

So ##-\frac { Nlcos\theta }{ 2 } =\tau ##

or ##\frac { Nlcos\phi }{ 2 } =-I\frac { d\omega }{ d\phi } \omega ##

or ##\frac { Nlsin\phi }{ 2 } =-I\frac { { \omega }^{ 2 } }{ 2 } +C##

At ##sin\theta=2/3 {\omega}^{2}=\frac{g}{l}##

So ##C=\frac { l(8N+mg) }{ 24 } ##

So ##\frac { Nlsin\phi }{ 2 } =-I\frac { { \omega }^{ 2 } }{ 2 } +\frac { l(8N+mg) }{ 24 }##

When the rod becomes horizontal ##\theta=0##
So ##N=\frac { m(l{ \omega }^{ 2 }-g) }{ 8 } ##

Now the acceleration of the CoM of the rod(##a##) is ##\frac{mg-N}{m}##

vertical velocity of CoM when rod left contact with the wall is ##\frac{\sqrt{5gl}}{2}##

distance covered by the CoM =##\frac{l}{3}##

So vertical velocity of the CoM of the rod when it becomes horizontal(##v##) is

## v^{2}=\frac { 5gl }{ 36 } +\frac { 2l(mg-N) }{ 3m } ##

Also when the rod becomes horizontal ##\frac{l\omega}{2}=v##. As the vertical velocity of point A is 0.

So on putting values of ##N## and ##\omega##

I got ##4v^{2}=\frac{13gl}{6}##

Also the ##V_{B}=2v##

So ##V_{B}=\sqrt{\frac{13gl}{6}}##.

Is my approach correct? Are my answers correct?

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Satvik Pandey said:
At $\sin \theta = 2/3 {\omega}^{2} = \frac {g}{l}$
I guess you mean
At $\sin \phi = \frac 23$, $\omega^2=\frac gl$
Satvik Pandey said:
So ##C=\frac { l(8N+mg) }{ 24 }##
But that N is the normal force at the moment it leaves contact with the wall, no?

haruspex said:
I guess you mean
At $\sin \phi = \frac 23$, $\omega^2=\frac gl$

No. ##\theta## is the angle(between rod and horizontal) at which the rod leaves contact with the vertical wall. Where as ##\phi## is a general angle(between rod and horizontal) after the rod leaves contact with the vertical wall.

So $\sin \theta = \frac 23$, $\omega^2=\frac gl$

haruspex said:
But that N is the normal force at the moment it leaves contact with the wall, no?
The rod leaves contact with the vertical wall but it remains in contact with the horizontal wall. In the FBD in #post1 I have mentioned it.

Satvik Pandey said:
No. ##\theta## is the angle(between rod and horizontal) at which the rod leaves contact with the vertical wall. Where as ##\phi## is a general angle(between rod and horizontal) after the rod leaves contact with the vertical wall.

So $\sin \theta = \frac 23$, $\omega^2=\frac gl$
Sure, but you wrote "At sinθ =2/3" in the context of a general equation involving ϕ. So I think what you intend is along the lines of "At ϕ = ϕ0 = θ". There are a couple of other places where it seems to me you wrote θ where you meant ϕ. E.g.
Satvik Pandey said:
When the rod becomes horizontal θ=0

Satvik Pandey said:
The rod leaves contact with the vertical wall but it remains in contact with the horizontal wall. In the FBD in #post1 I have mentioned it.
Yes, I understand N is the normal force from the ground here. You use N to substitute for a constant C in a general equation for the period after losing contact with the wall, and the basis for the substitution is the instant of losing contact with the wall. For that to be valid, the N substituted must be its value at that instant. Maybe I'm misreading your algebra, but you appear to treat this N later as the normal force at any later time.
It might help clarify things if you make more use of suffixes, like ϕ0 and N0 for the values at the instant of losing wall contact. Indeed, with that you could have used theta throughout, but with different suffixes/embellishments for the critical values.

I do have another concern I forgot to mention before. The question says "when it just reaches the horizontal ground". You calculated on the basis of its becoming horizontal, but are you sure it will be on the ground at that time? Maybe it becomes airborne at some point? If so, the question is not well posed.

Satvik Pandey
haruspex said:
Sure, but you wrote "At sinθ =2/3" in the context of a general equation involving ϕ. So I think what you intend is along the lines of "At ϕ = ϕ0 = θ". There are a couple of other places where it seems to me you wrote θ where you meant ϕ. E.g.

Yes, I understand N is the normal force from the ground here. You use N to substitute for a constant C in a general equation for the period after losing contact with the wall, and the basis for the substitution is the instant of losing contact with the wall. For that to be valid, the N substituted must be its value at that instant. Maybe I'm misreading your algebra, but you appear to treat this N later as the normal force at any later time.
It might help clarify things if you make more use of suffixes, like ϕ0 and N0 for the values at the instant of losing wall contact. Indeed, with that you could have used theta throughout, but with different suffixes/embellishments for the critical values.

I do have another concern I forgot to mention before. The question says "when it just reaches the horizontal ground". You calculated on the basis of its becoming horizontal, but are you sure it will be on the ground at that time? Maybe it becomes airborne at some point? If so, the question is not well posed.

Oh! I missed that. I assumed ##N## to be constant but it is a function of ##\phi##. How should I proceed?
I tried to find out the ##\theta## at which the lower end of the rod leaves contact with the ground.

The moment at which the rod leaves contact with ground the acceleration of the CoM of the rod will be ##g##.

So ##\frac { d }{ dt } \left( \frac { l\omega cos\theta }{ 2 } \right) =g##

or ##\theta =arcsin\left( \frac { \sqrt { 129 } +3 }{ 12 } \right) ##
But the value of number inside the brackets is coming out to be greater the 1. So, can I assume that the rod will not loose contact with the horizontal ground?

How should I proceed to solve the question?

Satvik Pandey said:
Oh! I missed that. I assumed ##N## to be constant but it is a function of ##\phi##. How should I proceed?
Once it leaves the wall, horizontal speed of the mass centre is fixed. So put that to one side and treat it as the CoM descending vertically at speed ##v = v(t) = (l/2)\omega \cos \phi##.
What energy equation do you now have?
Satvik Pandey said:
I tried to find out the ##\theta## at which the lower end of the rod leaves contact with the ground.

The moment at which the rod leaves contact with ground the acceleration of the CoM of the rod will be ##g##.

So ##\frac { d }{ dt } \left( \frac { l\omega cos\theta }{ 2 } \right) =g##

or ##\theta =arcsin\left( \frac { \sqrt { 129 } +3 }{ 12 } \right) ##
I don't know how you got that last equation, but I confirm that it will not become airborne.

Satvik Pandey
haruspex said:
Once it leaves the wall, horizontal speed of the mass centre is fixed. So put that to one side and treat it as the CoM descending vertically at speed ##v = v(t) = (l/2)\omega \cos \phi##.
What energy equation do you now have?

The initial velocity of the rod is ##\frac { l\omega cos{ (\phi }_{ 0 }) }{ 2 } =\frac { \sqrt { 5gl } }{ 6 } ##
where ##\phi_{0}## is the angle b/w rod and horizontal at the moment the rod looses contact with the vertical wall.

Initial potential of the rod is ##\frac{lsin\phi_{0}}{2}=\frac{l}{3}##

Let ##v'## be the velocity of the CoM of the rod when it becomes horizontal.

So by the conservation of mechanical energy---

##\frac { m }{ 2 } { \left( \frac { \sqrt { 5gl } }{ 6 } \right) }^{ 2 }+\frac { mgl }{ 3 } =\frac { m{ v' }^{ 2 } }{ 2 } ##

##v' =\sqrt { \frac { 29gl }{ 36 } } ##

So we have found the velocity of CoM of the rod when it becomes horizontal. When the rod becomes horizontal then the vertical velocity of point A(the point which was in contact with the ground) will be zero.

So ##\frac{l\omega}{2}=v'##. So can we find the velocity of CoM and the angular velocity of the rod about the CoM in this way?
Then velocity of the points A and B can be find by adding the velocity of the CoM and velocity about the CoM. right?

haruspex said:
I don't know how you got that last equation, but I confirm that it will not become airborne.

The 1st equation in #post5 can be written as --

##\frac { d }{ d\theta } \left( \frac { l{ \omega }^{ 2 }cos\theta }{ 2 } \right) =g##

Substituting the value of ##\omega## I got

or ##\frac { 3 }{ 2 } \frac { d }{ d\theta } (1−sinθ)cos\theta =1##

or ## 2{ sin }^{ 2 }\theta -sin\theta -1=\frac { 2 }{ 3 } ##

on solving this I got ##\theta =arcsin\left( \frac { 3\pm \sqrt { 129 } }{ 12 } \right) ##

Thanks for helping haruspex.:)

Satvik Pandey said:
When the rod looses contact with the vertical rod the Normal force acting on Rod becomes 0. So no horizontal force act on the rod after the moment the rod left contact with vertical wall. So acceleration of the CoM of the rod in horizontal direction is 0.
.....

So ##sin\theta =\frac { 2 }{ 3 } ##

So at ## \theta=arcsin(2/3)## ##\omega=\sqrt{\frac{g}{L}}##

So ##{ V }_{ C }=\frac{\sqrt{gl}}{3} { i } -\frac{\sqrt{5gl}}{6} { j } ##

It is correct so far.

Satvik Pandey said:
After loosing contact with the vertical wall no horizontal force acts on the CoM of the rod so the vertical component of the rod will not change.

So ##V_{A}=\frac{\sqrt{gl}}{3}##. Where ##V_{A}## is the velocity of the lower end of the rod when the rod becomes horizontal.

You meant that the horizontal component of the velocity of the CoM will not change. But VA will. XA=XC+L/2 cosθ, and θ changes with time. A is accelerating.
If you solve rotational motion of a rigid body, do it with respect to a fixed axis or with respect to the CoM. And apply conversation of energy. The frame of reference fixed to A is not inertial.

Satvik Pandey
Satvik Pandey said:
So by the conservation of mechanical energy---

##\frac { m }{ 2 } { \left( \frac { \sqrt { 5gl } }{ 6 } \right) }^{ 2 }+\frac { mgl }{ 3 } =\frac { m{ v' }^{ 2 } }{ 2 }##
You've left out rotational energy (both sides).

haruspex said:
You've left out rotational energy (both sides).
Sorry.:s

So
##\frac { { I }_{ c }{ \omega }_{ 0 }^{ 2 } }{ 2 } +\frac { m }{ 2 } { \left( \frac { \sqrt { 5gl } }{ 6 } \right) }^{ 2 }+\frac { mgl }{ 3 } =\frac { m{ v' }^{ 2 } }{ 2 } +\frac { { I }_{ c }{ \omega }_{ f }^{ 2 } }{ 2 } ##
where ##\omega_{ 0 }## is the angular velocity of the rod when it looses contact with the vertical wall and ##\omega_{ f }## is the angular velocity of the rod when it become horizontal.

or ##\frac { m{ l }^{ 2 } }{ 12 } \times \frac { { \omega }_{ 0 }^{ 2 } }{ 2 } +\frac { m }{ 2 } { \left( \frac { \sqrt { 5gl } }{ 6 } \right) }^{ 2 }+\frac { mgl }{ 3 } =\frac { m{ v' }^{ 2 } }{ 2 } +\frac { m{ l }^{ 2 } }{ 12 } \times \frac { { \omega }_{ f }^{ 2 } }{ 2 } ##

or ##\frac { 32gl }{ 36 } ={ v' }^{ 2 }+\frac { { l }^{ 2 }{ \omega }_{ f }^{ 2 } }{ 12 } ##.

When the rod becomes horizontal then the vertical component of velocity of point A(the point which was in contact with the ground) will be zero.

So is ##\frac { l{ \omega }_{ f }^{ 2 } }{ 2 } ={ v' }##?

ehild said:
You meant that the horizontal component of the velocity of the CoM will not change. But VA will. XA=XC+L/2 cosθ, and θ changes with time. A is accelerating.
If you solve rotational motion of a rigid body, do it with respect to a fixed axis or with respect to the CoM. And apply conversation of energy. The frame of reference fixed to A is not inertial.

Oh! yes I meant horizontal component.:s

Here let red arrow represent vertical and horizontal component of the velocity of the CoM and green arrow represent the angular velocity of the point A about CoM. Total velocity of point A is vector sum of these three vectors.
When the rod becomes horizontal point A will not have any vertical velocity. right?
So vertical component of velocity of CoM is equal to angular velocity of point A about CoM. right?
So being in opposite direction they cancel out. So remaining vector is the horizontal velocity of CoM. So isn't velocity of pointA equal to the component of the horizontal velocity of the CoM?

Satvik Pandey said:
So being in opposite direction they cancel out. So remaining vector is the horizontal velocity of CoM. So isn't velocity of point A equal to the component of the horizontal velocity of the CoM?

The horizontal component of the velocity of the CoM is equal to VA if the rod is horizontal.

Satvik Pandey
ehild said:
The horizontal component of the velocity of the CoM is equal to VA if the rod is horizontal.
So, is my expression of ##V_{A}## in post 1 correct?

What are your comments on my post #11. Should I find ##v'## and ##\omega_{f}## from the two equations?

then velocity of point B (when the rod becomes horizontal) is ##\frac { l{ \omega }_{ f } }{ 2 } +{ V }_{ CoM }##
where Vcom is the total velocity of the CoM of the rod when it becomes horizontal.

Satvik Pandey said:
Sorry.:s

So
##\frac { { I }_{ c }{ \omega }_{ 0 }^{ 2 } }{ 2 } +\frac { m }{ 2 } { \left( \frac { \sqrt { 5gl } }{ 6 } \right) }^{ 2 }+\frac { mgl }{ 3 } =\frac { m{ v' }^{ 2 } }{ 2 } +\frac { { I }_{ c }{ \omega }_{ f }^{ 2 } }{ 2 } ##
where ##\omega_{ 0 }## is the angular velocity of the rod when it looses contact with the vertical wall and ##\omega_{ f }## is the angular velocity of the rod when it become horizontal.

or ##\frac { m{ l }^{ 2 } }{ 12 } \times \frac { { \omega }_{ 0 }^{ 2 } }{ 2 } +\frac { m }{ 2 } { \left( \frac { \sqrt { 5gl } }{ 6 } \right) }^{ 2 }+\frac { mgl }{ 3 } =\frac { m{ v' }^{ 2 } }{ 2 } +\frac { m{ l }^{ 2 } }{ 12 } \times \frac { { \omega }_{ f }^{ 2 } }{ 2 } ##

or ##\frac { 32gl }{ 36 } ={ v' }^{ 2 }+\frac { { l }^{ 2 }{ \omega }_{ f }^{ 2 } }{ 12 } ##.

When the rod becomes horizontal then the vertical component of velocity of point A(the point which was in contact with the ground) will be zero.

So is ##\frac { l{ \omega }_{ f }^{ 2 } }{ 2 } ={ v' }##?
What is v'? is not it the vertical velocity of the CoM?

Satvik Pandey said:
So is ##\frac { l{ \omega }_{ f }^{ 2 } }{ 2 } ={ v' }##?
Almost... I think you left in something by mistake.

ehild said:
What is v'? is not it the vertical velocity of the CoM?

Yes it is a vertical velocity of CoM of the rod when it becomes horizontal.

haruspex said:
Almost... I think you left in something by mistake.

Did I commit mistake in the conservation of energy equation?

How are ω and v' related? And what is v' finally?

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ehild said:
How are ω and v' related? And what is v' finally?
You meant ωf and v'. right?
0.5lωf=v'.
I have mentioned that at the end of #post 11 but by mistake I typed ω2. I think haruspex was pointing that in post#16.

Is velocity of point B ##\sqrt { \frac { 25gl }{ 9 } } ##?

Yes, if B is the falling end :), but it is the speed, not the velocity. What is the velocity?

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Satvik Pandey said:
Is velocity of point B ##\sqrt { \frac { 25gl }{ 9 } } ##?
I get 26, not 25.

haruspex said:
I get 26, not 25.
How?

ehild said:
How?
No, I found my error. I agree 25 now.

ehild said:
Yes, if B is the falling end :), but it is the speed, not the velocity. What is the velocity?

Oh! yes it is speed.
Velocity will be ##\sqrt { \frac { gl }{ 9 } } \hat { i } -\sqrt { \frac { 8gl }{ 3 } } \hat { j } ##. right?

It is right :)

ehild said:
It is right :)

Thank you ehild for helping me again. You have helped me in my every thread.:) You are great. Thank you haruspex too.:)

Can you guys tell me that why point A does not left contact with the horizontal ground and become airborne?

I tried this

The moment at which the rod leaves contact with ground the acceleration of the CoM of the rod will be ##g##.

So ##\frac { d }{ dt } \left( \frac { l\omega cos\theta }{ 2 } \right) =g##

But ##\omega## is also a function of ##\theta##.
I am stuck here.

Satvik Pandey said:
So ##\frac { d }{ dt } \left( \frac { l\omega cos\theta }{ 2 } \right) =g##
I think you made a mistake with the signs (the accelerations are in opposite directions)

$\frac{d}{dt}\left( \frac{-L\omega \cos(\theta)}{2} \right)=\frac{L}{2}\left(\omega ^2\sin(\theta)-\frac{d\omega}{dt}\cos(\theta)\right)>g$

At best, $\frac{d\omega}{dt}$ is zero.
At worst, the rod is still in contact with the wall and so the net torque is in the "wrong direction" (helping point A stay on the ground).

I rearranged it into an inequality involving energy but I didn't quite show that the rod stays on the ground (although I thought I did!)

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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...

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Satvik Pandey
Nathanael said:
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 dvy/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 Vy2 (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.

Satvik Pandey and Nathanael
ehild said:
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..."

Nathanael said:
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).

Nathanael said:
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! :)

ehild said:
Very nice!
Instead of finding when dvy/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 Vy2 (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.:)