contestada

The small particle of mass m = 3.8 kg and its restraining cord of length r = 0.46 m are spinning with an angular velocity ? = 1.6 rad/s on the horizontal surface of a smooth disk, shown in section. as the force f is slowly relaxed, r increases and ? changes. when r = 0.63 m, what is the angular velocity

Respuesta :

egenriether
Use conservation of angular momentum.  The initial momentum is L=Iω, where I is the moment of inertia and ω=1.6rad/s.  The moment of inertia of a particle of mass m at distance r from the axis of rotation is I = mr².  So in the first case it's
I=3.8(0.46²)=0.804 kgm²
L=0.804 * 1.6 = 1.2865 kgm²/s
In the second case the angular momentum will be the same, but I will increase to I' and ω will decrease to ω'.  We seek to find ω'.
L=1.2865 kgm²/s = I'ω'
I'=3.8(0.63²)=1.508 kgm²
ω' = L/I' = 1.2865/1.508 = 0.853rad/s

Otras preguntas