Discs/Hoops Rotational Physics Question

In summary, a rotating hoop with mass 2.0 kg and radius 1.5 m has a kinetic energy of 10.8 kg m s-1. When a disc with the same mass and radius is dropped on top of the hoop, the conservation of angular momentum applies, assuming there are no external forces like air resistance or friction. The final angular velocity is 3.65 s-1, calculated using the equation Lf = ((Ii + If) x wf) where Ii is the moment of inertia of the hoop and If is the moment of inertia of the disc.
  • #1
boognish
15
0
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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  • #2
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)
 
  • #3
Simon Bridge said:
-- no friction? What is holding the two objects together then?
Boognish did specify external forces.
 
  • #4
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?
 
  • #5
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.
 

1. What is rotational inertia and how does it relate to discs/hoops?

Rotational inertia is a measure of an object's resistance to changes in its rotational motion. It is directly related to an object's mass and distribution of that mass from the axis of rotation. In discs/hoops, the rotational inertia is determined by the mass and radius of the disc/hoop, as well as the axis of rotation.

2. How does the shape of a disc/hoop affect its rotational motion?

The shape of a disc/hoop can greatly impact its rotational motion. For example, a disc with a larger radius will have a greater rotational inertia and therefore require more torque to change its rotational speed. Additionally, the distribution of mass in a hoop can affect its stability and ability to maintain its rotational motion.

3. What factors affect the stability of a spinning disc/hoop?

The stability of a spinning disc/hoop is affected by several factors, including its rotational speed, mass distribution, and the presence of external forces such as friction or air resistance. A higher rotational speed can make a disc/hoop more unstable, while a more evenly distributed mass can increase its stability.

4. How does torque affect the rotational motion of a disc/hoop?

Torque is a force that causes an object to rotate. In discs/hoops, torque can be applied through the use of a force, such as a hand or a motor, that is perpendicular to the axis of rotation. The greater the torque applied, the greater the change in rotational motion of the disc/hoop.

5. Can the laws of rotational motion be applied to discs/hoops in real-world situations?

Yes, the laws of rotational motion, such as Newton's Second Law for rotation and the Law of Conservation of Angular Momentum, can be applied to real-world situations involving discs/hoops. These laws can help predict how a disc/hoop will behave when subjected to various forces and torques.

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