A 20 cm diameter, 2kg solid disk is rotating at 200rpm.?
A 20 cm diameter, 2kg solid disk is rotating at 200rpm. A 20 cm diameter, 1 kg circular loop is dropped straight down onto the rotating disk. Friction causes the loop to accelerate until it is "riding" on the disk. What is the final angular velocity of the combined system?
- oubaasLv 73 weeks agoFavorite Answer
Jd = 2/2*0.1^2 = 0.01 kg*m^2
ω = 2PI/60*n = 0.10472*200 = 20.94 rad/sec
L = Jd*ω = 20.94*0.01 = 0.2094 kg*m^2/sec
after (L is conserved)
Jℓ = 1*0.10^2 = 0.01 kg*m^2
J = Jd+Jℓ = 0.01+0.01 = 0.02 kg*m^2
since J = 2Jd , then ω' = ω/2 = 10.47 rad/sec
- Andrew SmithLv 73 weeks ago
100 rpm. It is obvious but let us go through the steps. For a solid disk the moment of inertia is 1/2 m x^2. = 1 x^2 For the hollow hoop it is m2 x^2 = 1 x^2
Angular momentum is conserved so initially we had an angular momentum of 1 x^2 * 200. After the collision we have ( 1x^2 + 1 x^2 ) * w2 -> w2 = 200/ 2 = 100 rpm
Now if you want angular velocity in conventional units ( radians per second) multiply by 2 pi() radians per turn and divide by 60 seconds in a minute to get 200 pi() / 60 = 20pi() / 6 rad/s