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Rotational dyanamics

  • #1
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Homework Statement



Two rings of masses 'm' (smaller one) and '4m' (bigger one) are placed in the set up as shown in the diagram. The smaller ring is hung through a horizontal frictionless thread. The bigger ring is joined to the smaller ring in such a way that it can rotate freely about the point of suspension. The planes of both the rings are perpendicular to the plane of the paper. The bigger ring has a radius 'R'. After the bigger ring is released from the initial position as shown in the figure, find the angular velocity of the bigger ring when its plane becomes vertical i.e the planes of both the rings become vertical.


2. Homework Equations



3. The Attempt at a Solution

First of all, when the rings reach the final position, we can calculate the relation between the velocities of the center of masses of both the rings (their velocity vectors will be horizontal and since there is no ext. force in the horizontal direction we can conserve the linear momentum). Now when the bigger ring reaches the final position the decrease in the potential energy of the system is, (4m)gR, which will be converted into kinetic energies of the bodies. But i have a doubt; the K.Es of the smaller ring will be only its linear K.E, but for the bigger ring, what will be the K.E? Will it just the linear K.E of the COM of the ring at that instant? or it will be the rotational energy about the point of suspension? or both?

Please help!
 

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Answers and Replies

  • #2
supratim1
Gold Member
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i think we need to consider both the K.E of upper and lower rings, as well as the rotational energy of the lower ring. Using absolute velocity of the center of mass of each ring.
 
  • #3
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Do you think the angular momentum of the two rings + earth about the point of suspension is conserved?
 
  • #4
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I am sorry Quadeer, I couldn't get your question!
As far as i see the question, there is no basic reason for me to conserve angular momentum, coz at any instant the gravitational torque would be acting at the center of mass of the bigger ring, which would make it difficult for me to choose a point about which i get a net zero torque on the ring (which is the essential for applying conservation of angular momentum), i however can conserve linear momentum in the horizontal direction owing to the absence of an external force.

Now, the only way you can match answer is if you, equate the net change in gravitational potential energy of the bigger ring with the translational kinetic energies of both the rings and the rotational KE off the bigger ring w.r.t the point of suspension. But here's my basic doubt:

IN GENERAL WHEN WE TALK ABOUT THE BODY IN TRANSLATIONAL AS WELL AS ROTATIONAL MOTION, WE TAKE ITS TOTAL KINETIC ENERGY AS THE TRANSLATIONAL K.E OF THE C.O.M AND THE ROTATIONAL K.E OF THE BODY ABOUT THE C.O.M, BUT IN THIS CASE THE ANSWER MATCHES WHEN WE TAKE THE TRANSLATIONAL K.E AS STATED BUT THE ROTATIONAL ENERGY ABOUT THE POINT OF SUSPENSION, WHICH IS NOT CORRECT!

WHERE IS THE LOOPHOLE?

I REQUEST ANY MENTOR TO PLEASE GUIDE ME THROUGH THIS QUESTION!!
 
  • #5
1,384
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IN GENERAL WHEN WE TALK ABOUT THE BODY IN TRANSLATIONAL AS WELL AS ROTATIONAL MOTION, WE TAKE ITS TOTAL KINETIC ENERGY AS THE TRANSLATIONAL K.E OF THE C.O.M AND THE ROTATIONAL K.E OF THE BODY ABOUT THE C.O.M, BUT IN THIS CASE THE ANSWER MATCHES WHEN WE TAKE THE TRANSLATIONAL K.E AS STATED BUT THE ROTATIONAL ENERGY ABOUT THE POINT OF SUSPENSION, WHICH IS NOT CORRECT!
Your general case is in the rotational + translational motion about the COM (i.e. when a body rotates about its COM)
Here in the question, the point of suspension is NOT the COM of the bigger ring.

The kinetic energy of a body in a frame R is given by

Kinetic energy of the COM in 'R' frame + Kinetic energy (translational) of the COM in COM frame.

Here the frame R is at the point of suspension of the bigger ring.
 
  • #6
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But Abdul, the Translational kinetic energy of the COM in the COM reference frame will be zero as the COM will have 0 velocity in this frame.
 
  • #7
tiny-tim
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Hi Mandeep! :smile:
But i have a doubt … for the bigger ring, what will be the K.E? Will it just the linear K.E of the COM of the ring at that instant? or it will be the rotational energy about the point of suspension? or both?
IN GENERAL WHEN WE TALK ABOUT THE BODY IN TRANSLATIONAL AS WELL AS ROTATIONAL MOTION, WE TAKE ITS TOTAL KINETIC ENERGY AS THE TRANSLATIONAL K.E OF THE C.O.M AND THE ROTATIONAL K.E OF THE BODY ABOUT THE C.O.M, BUT IN THIS CASE THE ANSWER MATCHES WHEN WE TAKE THE TRANSLATIONAL K.E AS STATED BUT THE ROTATIONAL ENERGY ABOUT THE POINT OF SUSPENSION, WHICH IS NOT CORRECT!
(I assume the upper ring is constrained to stay vertical?)

The KE is always the linear KE of the centre of mass plus the rotational KE about the centre of mass …

perhaps you've used the wrong value for the angular velocity?

Show us your calculations. :smile:
 
  • #8
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But Abdul, the Translational kinetic energy of the COM in the COM reference frame will be zero as the COM will have 0 velocity in this frame.
Yes, you are right. I wrote that in a hurry.
It is Kinetic energy (translational) of the system in COM frame.
 
  • #9
supratim1
Gold Member
279
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(I assume the upper ring is constrained to stay vertical?)
Sir, yes the upper ring is constrained to stay vertical.
 
  • #10
1,384
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(I assume the upper ring is constrained to stay vertical?)
No it is not. The question clearly mentions 'the smaller ring is hung through a horizontal frictionless thread'.
 
  • #11
104
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No, quadeer the ring is constrained to remain vertical, i.e it can have only linear motion.
 
  • #12
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Let the velocities of the COMs of the rings at the instant they are vertical be v1 (bigger ring) and v2 (smaller ring),

Applying conservation of linear momentum,
4v1=v2................................(I)

Now the point about which the bigger ring is rotating, i.e the point at which the bigger ring is attached to the smaller ring, has a velocity same as that of the COM of the smaller ring. So the angular velocity of the bigger ring about its center has value w=(v1+v2)/R [as both v1 and v2 have opposite directions]

Substituting the value of either v1 or v2 in the expression from (I), we can express the velocities v1 and v2, in terms of w.


Applying conservation of energy,
4mgR = (1/2)mv22 + (1/2)4mv12 + (1/2)ICOMw2

Again substituting the values of v1 and v2 in terms of w, we find w to be = 10(r/11R)1/2

This is the angular velocity of the body about the center of mass. Now this value should be equal to the angular velocity of the ring about the point of suspension, isnt it! and hence should be the required answer!!

Please point out the mistake in the solution!
 
  • #13
1,384
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Applying conservation of energy,
4mgR = (1/2)mv22 + (1/2)4mv12 + (1/2)ICOMw2
May be you made a mistake here. 'I' should be about the point of rotation of the bigger ring, not about its COM i.e. I=1.5(4M)R2
 
  • #14
tiny-tim
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Again substituting the values of v1 and v2 in terms of w, we find w to be = 10(r/11R)1/2
How do you get that?
 
  • #15
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w= (v1+v2)/R
And 4v1=v2,

Therefore, v1=(Rw)/5
v2=(4Rw)/5

Putting these values in the energy equation, we get the value of w
(Since, the axis of rotation is on the plane of the ring, its ICOM will be (1/2)4MR2)
 
Last edited:
  • #16
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May be you made a mistake here. 'I' should be about the point of rotation of the bigger ring, not about its COM i.e. I=1.5(4M)R2

Why will 'I' be about the point of suspension?
The net K.E of any body is the translational K.E of the COM of the body and the rotational K.E of the body about its C.O.M!!
 
  • #17
tiny-tim
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Let the velocities of the COMs of the rings at the instant they are vertical be v1 (bigger ring) and v2 (smaller ring),

Applying conservation of energy,
4mgR = (1/2)mv22 + (1/2)4mv12 + (1/2)ICOMw2

Again substituting the values of v1 and v2 in terms of w, we find w to be = 10(r/11R)1/2

This is the angular velocity of the body about the center of mass. Now this value should be equal to the angular velocity of the ring about the point of suspension, isnt it! and hence should be the required answer!!

Please point out the mistake in the solution!
w= (v1+v2)/R
And 4v1=v2,

Therefore, v1=(Rw)/5
v2=(4Rw)/5

Putting these values in the energy equation, we get the value of w
(Since, the axis of rotation is on the plane of the ring, its ICOM will be (1/2)4MR2)
let's see …

4mgR = (1/2)mv22 + (1/2)4mv12 + (1/2)ICOMw2

= (1/2)m((4Rw)/5)2 + (1/2)4m((Rw)/5)2 + (1/2)2MR2w2

= (mR2w2/50) {16 + 4 + 50}

= (7/5)mR2w2,

so w = √((20/7)g/R)

how did you get (100/11)? did you use {16 + 4 + 2} ?
 
  • #18
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Why will 'I' be about the point of suspension?
The net K.E of any body is the translational K.E of the COM of the body and the rotational K.E of the body about its C.O.M!!
ONLY if it rotates about the C.O.M.!!

4mgR = (1/2)mv22 + (1/2)4mv12 + (1/2)(1.5*4m)w2

On simplifying,
4Rg = 10v12 + 3w2R2

Now w is actually the angular velocity of the bigger ring w.r.t to the smaller ring.
V(bigger ring w.r.t smaller ring) = Rw = v1-v2=5v1

So v1=Rw/5

Subsitutute this in the simplified equation above.
I got w = (20g/17R)(1/2)
 
  • #19
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To Tiny Tim:

Apologies!!!!
It was a calculation mistake, but still the answer doesn't match!!
Is this the correct solution?
One more place where i am doubtful, is whether this 'w' is the 'w' about the point of suspension of the bigger ring?? I mean i have proved it but i am really unsure about it!!
 
  • #20
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To Quadeer:

I know the answer matches if you take 'I' about the point of suspension, but it will be incorrect to take the rotational energy about that point.

You tell me, if i have a solid rod hinged at one end and released in a vertical plane to rotate due to gravitational torque, how will you consider its K.E??
You will either take the whole energy as rotational energy about the point of suspension, or you will take the translational K.E of the COM + rotational about its C.O.M. You cannot take the translational KE about the COM and rotational about the point of suspension. It will be incorrect!!!!!
 
  • #21
tiny-tim
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… still the answer doesn't match!!
Is this the correct solution?
Correct method, so it should be. :confused:
 
  • #22
1,384
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You will either take the whole energy as rotational energy about the point of suspension, or you will take the translational K.E of the COM + rotational about its C.O.M. You cannot take the translational KE about the COM and rotational about the point of suspension. It will be incorrect!!!!!
As long as the point about which the rod is hinged does not move w.r.t ground, I would consider only its Rotational Kinetic Energy about that point. The COM of the rod does not translate, it only rotates!. I can also consider the translational K.E. of the COM + rotational K.E. about its COM because the point is fixed ( in other words, it is at rest w.r.t an inertial frame/earth)

In your problem, the point of suspension itself moves w.r.t ground (which is an inertial frame). And there is a different formula for finding its kinetic energy.

I am posting it again for your reference:

The kinetic energy of a body rotating about a fixed point in a frame R is given by

KE (rotational) of COM in reference frame R about fixed point + KE (translational) of the system in reference frame of COM
 
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