rossjesse216
contestada

A 40.0-kg child running at 8.00 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 600 kg ∙ m2 and very little friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it?

Respuesta :

onyebuchinnaji

The angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.

What is principle of conservation of angular momentum?

The principle of conservation of angular momentum states that the sum of initial angular momentum is equal to final angular momentum.

Li = Lf

Li = Ii ωi  + Ic ωc

Li = Iiωi  +  MR²(V/R)

Li =  Iiωi + MRV ----- (1)

Angular speed of the merry go round after the child jumps on it

Lf = If ωf

ωf = Lf/If

If = Im  + MR²

ωf = Lf / ( Im  + MR²)

Recall, Lf = Li

ωf = (Iiωi + MRV) / ( Im  + MR²)

where;

  • Iiωi is the initial angular momentum of the merry - go round = 0
  • M is mass of the child
  • R radius of rotation
  • V is tangential speed of the child
  • Im is the moment of inertia of the merry go round

ωf = (Iiωi + MRV) / ( Im  + MR²)

ωf = (0 + 40 x 1.5 x 8) / (600 + 40(1.5)²)

ωf = (480) / (690)

ωf = 0.696 rad/s

Thus, the angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.

Learn more about merry go round here: https://brainly.com/question/6345886

#SPJ1

Otras preguntas