Angular momentum about ICOR of a rod

  • #1
Krushnaraj Pandya
Gold Member
697
73

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
 

Answers and Replies

  • #2
Nathanael
Homework Helper
1,650
239

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 LCoR = ICoRwCoR
We know that Etotal = 0.5ICoR(wCoR)2
Thus we can write L in terms of E instead of I.
 
  • #4
Krushnaraj Pandya
Gold Member
697
73
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
 
  • #5
Krushnaraj Pandya
Gold Member
697
73
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)
 
  • #6
Krushnaraj Pandya
Gold Member
697
73
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
 
  • #7
Nathanael
Homework Helper
1,650
239
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?
 
  • #8
Krushnaraj Pandya
Gold Member
697
73
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.
 
  • #9
Krushnaraj Pandya
Gold Member
697
73
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
 
  • #10
Krushnaraj Pandya
Gold Member
697
73
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?)
 
  • #11
Krushnaraj Pandya
Gold Member
697
73
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)
 
  • #12
Nathanael
Homework Helper
1,650
239
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?
 
  • #13
Krushnaraj Pandya
Gold Member
697
73
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
 
  • #14
Nathanael
Homework Helper
1,650
239
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...?
 
  • #15
Krushnaraj Pandya
Gold Member
697
73
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
 
  • #16
Krushnaraj Pandya
Gold Member
697
73
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
 
  • #17
Krushnaraj Pandya
Gold Member
697
73
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
 
  • #18
Krushnaraj Pandya
Gold Member
697
73
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
 
  • #19
Nathanael
Homework Helper
1,650
239
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 :wink:
 
  • #20
Krushnaraj Pandya
Gold Member
697
73
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
 
  • #21
Nathanael
Homework Helper
1,650
239
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.
 
  • #22
Krushnaraj Pandya
Gold Member
697
73
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
 
  • #23
Nathanael
Homework Helper
1,650
239
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.
 
  • #24
Krushnaraj Pandya
Gold Member
697
73
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.
 
  • #25
Nathanael
Homework Helper
1,650
239
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.
 

Related Threads on Angular momentum about ICOR of a rod

Replies
3
Views
458
Replies
5
Views
2K
  • Last Post
2
Replies
43
Views
4K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
9
Views
6K
Replies
15
Views
6K
Top