Angular velocity of a solid disk

AI Thread Summary
A solid disk with an initial angular velocity of 35.0 rad/s and a mass of 3.6 kg is analyzed after a thin ring of the same radius and mass is added. The conservation of angular momentum is applied, leading to the equation I_iω_i = I_fω_f, where I represents the moment of inertia. The initial moment of inertia of the disk and the ring are calculated, and the final angular velocity is determined by solving the equation with the combined moment of inertia. The final angular velocity is found to remain at 35 rad/s, confirming that angular momentum is conserved despite the addition of the ring.
Nininguyen6
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Homework Statement



A solid disk is spinning around it center with an angular velocity of 35.0 rad/s. The disk has a mass of 3.6kg and a radius of .50m. You drop a thin ring with the same radius and a mass on the disk along the same center of rotation. What is the angular velocity of the disk and ring?

Homework Equations


Id= 1/2mR^2
Ir=mR^2
w=vf/R


The Attempt at a Solution



Id=1/2(3.6kg)(.50)^2 =.45 kg m/s
Ir= (3.6)(.50)^2= .90

W= (I'd+Ir)/ R
= (.45+.90)/(.50)
=2.7 rad/s

Is that the correct answer?
 
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Nininguyen6 said:

Homework Equations



w=vf/R

w stands for angular velocity right?

That's the relationship between linear velocity (at a distance R) and angular velocity.
For this problem you'll want to use conservation of angular momentum.
 
Would it be Li=Lf= i(wf)
 
Yes, it would be L_i=L_f (initial angular momentum = final angular momentum)

So what is the initial angular momentum? What then changes? And what does this change do to the angular velocity?
 
The initial is 35 rad/s aNd nothing changes because the second object is has the same mass and radius. Therefore would the finial be 35 as well?
 
35 rad/s is the initial angular velocity, but it is not the same as the initial angular momentum.

Angular momentum is conserved, not angular velocity.
 
So it would be L= I times w

Since I solved for IDisk and Iring, would I add those two up and mutiple by w?
 
What would be your reasoning for doing that?

You know I_i and ω_i and I_f so how can you solve for ω_f?


(Check posts #3 and #4)
 
((Id)(wd)+(Ir)(wr)) / R
= answer

Wf=w initial - answer
And that would give me the final angular vel?
 
  • #10
Nininguyen6 said:
((Id)(wd)+(Ir)(wr)) / R
= answer

Wf=w initial - answer
And that would give me the final angular vel?

I'm not sure where you're getting these equations.

If:
L_i=L_f

Then:
I_iω_i=I_fω_fSo what is the final rotational inertia?
 
  • #11
I'm a little confused because I solved for the iRing and the iDisk but no I don't know where to plug it in
 
  • #12
iDisk is the initial rotational inertia right?

And after the ring is added, it still has that initial rotational inertia, but now it also has an additional rotational inertia from the ring (iRing)

How would you write that in math?
 
  • #13
So would it be (.45)(2.7rad/s)=(.45+.90)(Wf)?
 
  • #14
Nininguyen6 said:
So would it be (.45)(2.7rad/s)=(.45+.90)(Wf)?

Very close, but where did you get 2.7 rad/s?
 
  • #15
I got 2.7 from the first post so would it be (.45)(35m/s)=(.45+.90)(Wf)?
 
  • #16
Yes, it would be that.

Now just solve for the final angular velocity and you're done!
 
  • #17
THANK YOU! Very much appreciated Nathanael
 
  • #18
No problem, and welcome to the physics forums!
 
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