At what height does this plank leave the wall

[TSny]

You are asking for justification because $\vec L_p = I_p \omega$ is valid only for fixed-axis rotation. Right?

It is also valid in certain other cases. It is valid if point $p$ is the instantaneous center of rotation. But it is not generally true that $d L_p/dt = I_p \alpha$ when $p$ is the instantaneous center of rotation. So, for the plank problem, I believe that $d L_p/dt = I_p \alpha$ needs to be justified. (I think it can be justified.)

 

[Chestermiller]

@haruspex I did an analysis to see if I could derive a general relationship for quantifying the location of the instantaneous center of rotation, given the velocity of the center of mass and the rate of rotation of the object. I probably "reinvented the wheel," but the results I obtained were as follows: If $v_x$ and $v_y$ are the cartesian components of the velocity of the center of mass, and $\omega$ is the counter-clockwise rate of rotation of the object, the coordinates of the center of rotation are given by the simple equations:$$x-x_c=-\frac{v_y}{\omega}$$ $$y-y_c=\frac{v_x}{\omega}$$. Is this consistent with your experience?

Chet

 

[Chestermiller]

Another attempt :
View attachment 209130
Center ūf mass motion gives,
$mg(-\hat y ) + N \hat y = m\ddot y \hat y ~~~~~~~~~~~~~~~~~ \ddot y <0. ~~~~~~~~~~~~~~(1)$

$y = l \sin \theta \\ \ddot y = l\{\ddot \theta \cos \theta - \sin \theta {\dot \theta }^2\}$
I can't decide here whether $\ddot \theta$ is positive or negative.
Since the torque about both center of mass C and pivot P is anti - clockwise,
considering $\vec \tau = I \vec \alpha : \vec \alpha = \ddot \theta \hat z,$ I decide $\ddot >0. ~~~~~~~~~~~~~~~~(2)$
But this method is valid only if $\vec \tau = I \vec \alpha$ is valid only for fixed axis rotation.
Using (1) and (2) ,
$N- mg = ml[ \cos \theta ~\ddot \theta - \sin \theta ~ {\dot \theta }^2 ] ~~~~~~~~~~~~~~~~~~~~~~(1.1)$

Similarly,
$\ddot x =0 ~ and~ x = l \cos \theta$ gives,

$\ddot \theta = \cot \theta {\dot \theta}^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

(1.1) and (3) gives,
$N - mg = \frac {ml {\dot \theta }^2 \cos {2 \theta} } {\sin \theta } ~~~~~~~~~~~~~~~(4)$

Now, considering rotational motion about center of mass,
Torque about center of mass,
$Nl\cos \theta = I_{cen} \alpha ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(5)$
Now, can I take $\ddot \theta = \alpha$?

Is this correct so far?

Your vertical force balance is correct. You also need to write the equation for the horizontal force balance, including the horizontal normal force. Your moment balance omits the contribution of the horizontal normal force, and it must be included. You are going to need to combine these three equations in order to solve for the angular acceleration in terms of the angle $\theta$. You can then integrate that equation to get the angular velocity as a function of the initial angle and the current angle.

More precisely, what you can do is solve the horizontal and vertical force balance equations for the two normal forces, and then substitute these into the moment balance equation. This will give you a relationship for the angular acceleration.

 

[haruspex]

Now, how to get to your statement mathematically?

My difficulty is that we are dealing with the instant that contact is lost. Deriving to get the velocity and acceleration up until that instant is fine, and of course the velocity obtained will also be correct at that instant, but can we be sure the acceleration obtained is correct? Maybe it is, and I'm being overcautious.

 

[haruspex]

@haruspex I did an analysis to see if I could derive a general relationship for quantifying the location of the instantaneous center of rotation, given the velocity of the center of mass and the rate of rotation of the object. I probably "reinvented the wheel," but the results I obtained were as follows: If $v_x$ and $v_y$ are the cartesian components of the velocity of the center of mass, and $\omega$ is the counter-clockwise rate of rotation of the object, the coordinates of the center of rotation are given by the simple equations:$$x-x_c=-\frac{v_y}{\omega}$$ $$y-y_c=\frac{v_x}{\omega}$$. Is this consistent with your experience?

Chet

Yes, that looks right. It just says $\vec v=\vec x\times\vec\omega$.
But in this case the easiest way to find it is by taking the normals to the known velocity vectors.

 

[Chestermiller]

My difficulty is that we are dealing with the instant that contact is lost. Deriving to get the velocity and acceleration up until that instant is fine, and of course the velocity obtained will also be correct at that instant, but can we be sure the acceleration obtained is correct? Maybe it is, and I'm being overcautious.

I don't think you are being overcautious. He can't just set the horizontal acceleration to zero, because the horizontal force history prior to loss of contact is important in determining what is happening at the time of contact. It is necessary to characterize the entire time interval between the initial state and the time that contact is lost (or to apply the energy balance at least at the two end points).

 

[Chestermiller]

Yes, that looks right. It just says $\vec v=\vec x\times\vec\omega$.
But in this case the easiest way to find it is by taking the normals to the known velocity vectors.

Yes. That's exactly what I did.

Thanks. Now, I'd like to learn more as to how this factors into the balance of moments.

 

[J Hann]

An interesting discussion.
I arrived at a solution using minimal calculus and basic physics and wanted to check my result.
I found a solution using Lagrangian mechanics at the following web page:

http://physics.columbia.edu/files/physics/content/Quals2010Sec1.pdf

and my result agreed with the one found there.

 

[Chestermiller]

An interesting discussion.
I arrived at a solution using minimal calculus and basic physics and wanted to check my result.
I found a solution using Lagrangian mechanics at the following web page:

http://physics.columbia.edu/files/physics/content/Quals2010Sec1.pdf

and my result agreed with the one found there.

In my judgment, there was quite a bit of calculus in there.

 

[J Hann]

In my judgment, there was quite a bit of calculus in there.

I didn't use advanced mechanics.
I only used introductory Physics and a smattering of rudimentary calculus.