Angular momentum about ICOR of a rod

[Krushnaraj Pandya]

Homework Statement

A rod (mass M, length L) is placed vertically on a smooth horizontal surface. Rod is released and after some time velocity of COM is v downwards and at this moment rod makes 60 degrees with horizontal. Find angular momentum of rod about Instantaneous center of rotation.

Homework Equations

All equations pertaining to rotational mechanics

The Attempt at a Solution

from energy conservation I wrote (Mg√3L)/4=1/2 Mv^2 + 1/2 ml^2/12 w^2=1/2 I(about ICOR)*w^2
I intuitively guessed ICOR must be on the lower right part of the rod and wrote xw=v where x is distance of ICOR from COM...not sure how to proceed further

 

[Nathanael]

The Attempt at a Solution

from energy conservation I wrote (Mg√3L)/4= ...

Mg√3L/4 is not the energy released by gravity.

I intuitively guessed ICOR must be on the lower right part of the rod and wrote xw=v where x is distance of ICOR from COM...not sure how to proceed further

That’s not the correct center. The correct center has every point on the rod moving perpendicularly. But in the absence of horizontal forces, the center moves straight down, which is not perpendicular to the point you chose.

I think we could systematically find the center of rotation based on the property I just mentioned (remember it need not be on the rod) but I also think we don’t have to.

You put the same w for rotation about the CoM as for rotation about the CoR. Was this on purpose?

Let me give a hint as to how we can avoid finding the CoR...
We are asked to find L_CoR = I_CoRw_CoR
We know that E_total = 0.5I_CoR(w_CoR)²
Thus we can write L in terms of E instead of I.

 

[Krushnaraj Pandya]

Mg√3L/4 is not the energy released by gravity.

yes, I forgot to subtract it from L/2

 

[Krushnaraj Pandya]

That’s not the correct center. The correct center has every point on the rod moving perpendicularly. But in the absence of horizontal forces, the center moves straight down, which is not perpendicular to the point you chose.

so it must be on the horizontal axis passing through COM, correct? and xw(ICOR)=v; x is dist. between ICOR and COM

 

[Krushnaraj Pandya]

Thus we can write L in terms of E instead of I.

2E/w(icor) = L, I can put the potential energy lost i.e Mg(2-√3)L/4 as E but we still don't know w(icor)

 

[Krushnaraj Pandya]

I think we could systematically find the center of rotation

I would like to do that too...I want the concepts I learn to be perfect and I should know how to solve this problem in all possible ways to a reasonable extent

 

[Nathanael]

Yes you have the right idea for finding the center of rotation. We can also choose another point we know the motion of, then the CoR is the intersection of the lines normal to their motion. (The contact point is a nice choice, as it only moves horizontally.)

About my question about w... did you mean for w_CoR to be the same as w_CoM? What about the rate of change of the angle that the rod makes with the ground, is that related to w? Can you maybe express v in terms of w as well?

 

[Krushnaraj Pandya]

We can also choose another point we know the motion of, then the CoR is the intersection of the lines normal to their motion. (The contact point is a nice choice, as it only moves horizontally.)

we don't know w yet, so can't define the velocity of any other point...if I could then I'd find CoR easily by the method you mentioned.

 

[Krushnaraj Pandya]

Yes you have the right idea for finding the center of rotation. We can also choose another point we know the motion of, then the CoR is the intersection of the lines normal to their motion. (The contact point is a nice choice, as it only moves horizontally.)

About my question about w... did you mean for w_CoR to be the same as w_CoM? What about the rate of change of the angle that the rod makes with the ground, is that related to w? Can you maybe express v in terms of w as well?

I meant to answer that...I did not do that on purpose, I just forgot to mention w(CoR) and just typed w

 

[Krushnaraj Pandya]

What about the rate of change of the angle that the rod makes with the ground, is that related to w?

dθ/dt is the definition of w, so the rate of change of angle w.r.t to ground would be w about that point on the ground (which point is this exactly though?)

 

[Krushnaraj Pandya]

Can you maybe express v in terms of w as well?

very easily if we knew w about a point and the location of that point, its just xw(CoR)

 

[Nathanael]

we don't know w yet, so can't define the velocity of any other point...if I could then I'd find CoR easily by the method you mentioned.

We don’t know the velocity of any point yet, but we know the direction of those two points. Isn't that enough to find the instantaneous CoR?

dθ/dt is the definition of w, so the rate of change of angle w.r.t to ground would be w about that point on the ground (which point is this exactly though?)

It’s not “about that point on the ground” in the sense of pure rotation; its just the rate of change of orientation.

Anyway I thought you defined “w” to be about the center of mass, because that’s how you used it in your energy equation

I meant to answer that...I did not do that on purpose, I just forgot to mention w(CoR) and just typed w

Oh ok. Well what do you think the relationship is? There’s a simple relationship between w_CoM, w_CoR, and dθ/dt

very easily if we knew w about a point and the location of that point, its just xw(CoR)

We can do it without knowing that. Think about the frame of reference in which the contact point is stationary. Is w the same in this frame? It is an accelerated frame so there are fictional forces, but is the vertical direction affected? What is the vertical speed in this frame?

 

[Krushnaraj Pandya]

We don’t know the velocity of any point yet, but we know the direction of those two points. Isn't that enough to find the instantaneous CoR?

the velocity of the highest point of rod at this instant would be pointing generally in the south-west direction but till we know the exact inclination of the velocity vector, can we mathematically find CoR? that's what I ultimately want to solve as one half of this question

 

[Nathanael]

the velocity of the highest point of rod at this instant would be pointing generally in the south-west direction but till we know the exact inclination of the velocity vector, can we mathematically find CoR? that's what I ultimately want to solve as one half of this question

The two points I’m speaking of are the center of mass and the contact point. Are you implying you don’t know the exact inclination of these velocities...?

 

[Krushnaraj Pandya]

The two points I’m speaking of are the center of mass and the contact point. Are you implying you don’t know the exact inclination of these velocities...?

oh, right! sorry, it's really late here and I'm half dreaming...but I need to solve this before I sleep, I got your point (pun intended). let me think about the other things you said for a second...I'm trying to solve this again and see where I'm getting stuck

 

[Krushnaraj Pandya]

The two points I’m speaking of are the center of mass and the contact point. Are you implying you don’t know the exact inclination of these velocities...?

so CoR is somewhere above the lower half of the rod (outside the body), correct?
My brain is going really slow right now

 

[Krushnaraj Pandya]

There’s a simple relationship between w_CoM, w_CoR, and dθ/dt

I am not sure but upon visualizing it seems to be w_CoR=w_CoM + dθ/dt

 

[Krushnaraj Pandya]

We can do it without knowing that. Think about the frame of reference in which the contact point is stationary. Is w the same in this frame? It is an accelerated frame so there are fictional forces, but is the vertical direction affected? What is the vertical speed in this frame?

In this frame dθ/dt is w- the vertical direction is not affected, so the vertical speed is still v, I am visualizing this as a pure rotation about the contact point, I could write dθ/dt(L/2) as the velocity of COM

 

[Nathanael]

so CoR is somewhere above the lower half of the rod (outside the body), correct?

Yes... but you can be more precise. You should know the exact CoR from that reasoning.

oh, right! sorry, it's really late here and I'm half dreaming...but I need to solve this before I sleep, I got your point (pun intended). let me think about the other things you said for a second...I'm trying to solve this again and see where I'm getting stuck

Btw what I said at the end of post#12 is used in the solution that avoids finding the CoR. If we do it your way, then we don’t need to worry about that equation.

Also it’s good to sleep on things

 

[Krushnaraj Pandya]

Yes... but you can be more precise. You should know the exact CoR from that reasoning.

yes (just don't want to apply the trigonometry right now, would probably make a silly mistake since I'm half awake) but I can find x coordinate of pt of contact and of course the height of CoM, combining these gives me the coordinates of CoR

 

[Nathanael]

In this frame dθ/dt is w- the vertical direction is not affected, so the vertical speed is still v, I am visualizing this as a pure rotation about the contact point, I could write dθ/dt(L/2) as the velocity of COM

Right. And this frame is moving horizontally relative the original frame, so the vertical component of velocity is unchanged, and so the v =what? (dθ/dt is unchanged as well if that wasn’t clear)

If you find v correctly in terms of w, then you can find w from the energy equation and then you get L from 2E/w which concludes my method that avoids finding the CoR.

 

[Krushnaraj Pandya]

Btw what I said at the end of post#12 is used in the solution that avoids finding the CoR. If we do it your way, then we don’t need to worry about that equation.

I'm going to do it your way too...as soon as I understand one method I'll try the next

 

[Nathanael]

I am not sure but upon visualizing it seems to be w_CoR=w_CoM + dθ/dt

No. I’m not sure how to hint at it, so I will just say it. They are all the same. If you imagine an arbitrary body purely rotating about some point, you can see it rotates about its CoM at the same rate as it goes about the CoR.

I'm going to do it your way too...as soon as I understand one method I'll try the next

Well now you’re almost done, just a few steps left outlined in post#21.

 

[Krushnaraj Pandya]

Right. And this frame is moving horizontally relative the original frame, so the vertical component of velocity is unchanged, and so the v =what? (dθ/dt is unchanged as well if that wasn’t clear)

v is just the vector addition of the velocity of point of contact (in the opposite direction i.e leftwards) and v downwards, I understood that dθ/dt is unchanged.

 

[Nathanael]

v is just the vector addition of the velocity of point of contact (in the opposite direction i.e leftwards) and v downwards, I understood that dθ/dt is unchanged.

You can write a specific equation for v in terms of w based on what has been said.

I do think you would get more out of this problem if you retried in the morning.

 

[Krushnaraj Pandya]

No. I’m not sure how to hint at it, so I will just say it. They are all the same. If you imagine an arbitrary body purely rotating about some point, you can see it rotates about its CoM at the same rate as it goes about the CoR.Well now you’re almost done, just a few steps left outlined in post#21.

Let's continue this tomorrow(today?) or I'll end up making you do all the work and learn nothing myself
But thanks for the help, you've a lot of patience- I imagine you'd make a really good professor. I'll let you know my progress after I wake up and attempt this. Good night(/day)