Angular momentum about ICOR of a rod

In summary: I'll handle it laterIn summary, the conversation involves finding the angular momentum of a rod placed vertically on a smooth horizontal surface and released. The velocity of the center of mass of the rod is v downwards when the rod makes a 60 degree angle with the horizontal. The equations of rotational mechanics are to be used to find the angular momentum about the instantaneous center of rotation. The attempt at a solution involved using energy conservation and intuition to determine the center of rotation, but further progress was hindered by not knowing the value of w or the location of the center of rotation. The conversation also discusses alternative methods for finding the center of rotation, such as choosing another point with known motion and finding the intersection of lines normal to their motion.
  • #1
Krushnaraj Pandya
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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
 
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  • #2
Krushnaraj Pandya said:

The Attempt at a Solution


from energy conservation I wrote (Mg√3L)/4= ...
Mg√3L/4 is not the energy released by gravity.

Krushnaraj Pandya said:
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.
 
  • #3
Nathanael said:
Mg√3L/4 is not the energy released by gravity.
yes, I forgot to subtract it from L/2
 
  • #4
Nathanael said:
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
Nathanael said:
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
Nathanael said:
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
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
Nathanael said:
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
Nathanael said:
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
Nathanael said:
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
Nathanael said:
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
Krushnaraj Pandya said:
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?

Krushnaraj Pandya said:
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

Krushnaraj Pandya said:
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

Krushnaraj Pandya said:
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
Nathanael said:
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
Krushnaraj Pandya said:
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
Nathanael said:
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
Nathanael said:
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
Nathanael said:
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
Nathanael said:
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
Krushnaraj Pandya said:
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.
Krushnaraj Pandya said:
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
Nathanael said:
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
Krushnaraj Pandya said:
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
Nathanael said:
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
Krushnaraj Pandya said:
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.

Krushnaraj Pandya said:
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
Nathanael said:
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
Krushnaraj Pandya said:
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.
 
  • #26
Nathanael said:
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)
 
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1. What is angular momentum about ICOR of a rod?

Angular momentum about ICOR (Instantaneous Center of Rotation) of a rod refers to the measure of the rod's rotational motion around a fixed point at a given moment in time. It takes into account the rod's mass, velocity, and distance from the fixed point.

2. How do you calculate angular momentum about ICOR of a rod?

To calculate the angular momentum about ICOR of a rod, you can use the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. The moment of inertia of a rod can be calculated using the formula I = (1/12)ml^2, where m is the mass of the rod and l is the length of the rod.

3. What factors affect the angular momentum about ICOR of a rod?

The angular momentum about ICOR of a rod is affected by the mass of the rod, its velocity, and its distance from the fixed point. Other factors such as external forces acting on the rod and the shape and orientation of the rod can also affect its angular momentum.

4. Why is it important to consider the angular momentum about ICOR of a rod?

Considering the angular momentum about ICOR of a rod is important because it helps us understand the rotational motion of the rod and predict its behavior. It also has practical applications in fields such as physics, engineering, and sports.

5. Can the angular momentum about ICOR of a rod change?

Yes, the angular momentum about ICOR of a rod can change if there is a change in any of the factors that affect it, such as the mass, velocity, or distance from the fixed point. External forces acting on the rod can also cause a change in its angular momentum.

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