What Happens to Angular Velocity When a Disc is Dropped onto a Rotating Hoop?

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Homework Help Overview

The discussion revolves around a problem involving a rotating hoop and a disc dropped onto it, focusing on the conservation of angular momentum and the resulting angular velocity after the disc is added. The subject area includes concepts from rotational dynamics and conservation laws.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss which conservation laws apply, particularly angular momentum versus linear momentum. There are questions about the assumptions necessary for applying these laws, such as the role of friction and external forces. Some participants explore the mathematical relationships involved in calculating the final angular velocity.

Discussion Status

The discussion is active, with participants sharing their reasoning and calculations. There is a mix of interpretations regarding the conservation laws and the assumptions needed. Some guidance has been offered on how to approach the problem, particularly in terms of considering the system before and after the disc is dropped.

Contextual Notes

Participants are navigating assumptions about external forces and the nature of the interaction between the hoop and the disc. There is a mention of potential misinterpretations in labeling equations, which could affect the clarity of the discussion.

boognish
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1. A hoop with mass 2.0 kg and radius 1.5 m is rotating with initial angular velocity of 5.4 s-1. A disc with the same mass and radius is dropped on top of the hoop so that the centers coincide. What conservation law applies here? What assumptions must you make to apply it? What is the final angular velocity?



2. Kinetic energy 1 = kinetic energy 2

Ke hoop = m x v^2
ke disc = 1/2 m x v^2




3. Conservation of momentum
There must be no additional external forces like air resistance or friction

ke hoop = 2.0 kg x 5.4 s-
Ke hoop initial = 10.8 kg m s-

Now i don't know if this works like a simple collision where I would factor in the new mass of the disc placed on it... but I am sure that I would have to factor in the moment of inertia of said disc so 1/2 m x v^2 to determine the change in angular velocity.
 
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What conservation law applies here?
-- conservation of angular momentum.

What assumptions must you make to apply it?
-- no friction? What is holding the two objects together then?

What is the final angular velocity?
-- your stated conservation law was "momentum" so you should be using the momentum equations here and not the KE ones.

The way to tackle conservation problems is to think in terms of "before" and "after" stuff happens - ignore the "during". The math works much the same as the linear case you've done before but with the rotational equations - so mass becomes "moment of inertia" and velocity becomes "angular velocity" so ##p=mv## (linear momentum) becomes ##L=I\omega## (angular momentum)
 
Simon Bridge said:
-- no friction? What is holding the two objects together then?
Boognish did specify external forces.
 
so understanding what you have told me i think that it resolves to:

Li= m r^2 w
Li = 2.0 x 1.5^2 x 5.4-s
Li = 24.3

Lf = Li

Lf = (Ii + If) x wf

Lf = ((m R^2 + (1/2 m R^2)) x Wf

24.3 = ((2.0 x 1.5^2) + (.5 x 2.0 x 1.5^2)) x wf
wf = 3.65s-

Any thoughts?
 
haruspex said:
Boognish did specify external forces.
Oh yes, my bad. "Friction was an example of the kind of thing. Question still stands though.
Objects held together by internal forces?

@boognish: that's what I described all right.
I'd watch the labels myself though ... one may expect that Lf = If.wf where If = I(hoop)+I(disk)
You wouldn't want to do all that work and get marked down because the marker misread you.
 

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