Question

A girl 60kg stands at centre of a playground mery-g-round rotating about frictionles axle at 1 rad/s.treat m-g-round as a uniform disc(m=100kg radius=3m). If the girls jumps to a position 1m from the centre, what will be their angular v after she landed?

Answer 1

The new angular velocity is `0.88 "rad"/s`.

Explanation:

We need the initial angular momentum, `L`. For that we need the rotational inertia, `I`. The formula for rotational inertia of a disk like our m-g-r is

`I = M*R^2/2`.

The girl will not count in the rotational inertia calculation because for her, radius is zero. So our m-g-r has initial rotational inertia
`I_1 = M*R^2/2 = (100 kg*(3 m)^2)/2 = 450 kg*m^2`

The formula for angular momentum is `L = I*omega`. So our m-g-r has angular momentum
`L = 450 kg*m^2*1 "rad"/s = 450 (kg*m^2)/s`

When the girl jumps 1 m out from the center, she becomes an additional part of the rotational inertia.
`I_2 = 450 (kg*m^2) + 60 kg*(1 m)^2 = 510 (kg*m^2)`

The principle of conservation of momentum requires that the momentum after the jump is still `450 (kg*m^2)/s`. To calculate the new angular velocity,

momentum before = momentum after

`450 (kg*m^2)/s = I_2*omega_2 = 510 (kg*m^2)*omega_2`

Solving for `omega_2`
`omega_2 = (450 (kg*m^2)/s)/(510 (kg*m^2)) = 0.88 "rad"/s`

I hope this helps,
Steve