- #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.
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.