# 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?

Mar 22, 2018

The new angular velocity is $0.88 \frac{\text{rad}}{s}$.

#### 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

$I = M \cdot {R}^{2} / 2$.

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
${I}_{1} = M \cdot {R}^{2} / 2 = \frac{100 k g \cdot {\left(3 m\right)}^{2}}{2} = 450 k g \cdot {m}^{2}$

The formula for angular momentum is $L = I \cdot \omega$. So our m-g-r has angular momentum
$L = 450 k g \cdot {m}^{2} \cdot 1 \frac{\text{rad}}{s} = 450 \frac{k g \cdot {m}^{2}}{s}$

When the girl jumps 1 m out from the center, she becomes an additional part of the rotational inertia.
${I}_{2} = 450 \left(k g \cdot {m}^{2}\right) + 60 k g \cdot {\left(1 m\right)}^{2} = 510 \left(k g \cdot {m}^{2}\right)$

The principle of conservation of momentum requires that the momentum after the jump is still $450 \frac{k g \cdot {m}^{2}}{s}$. To calculate the new angular velocity,

momentum before = momentum after

$450 \frac{k g \cdot {m}^{2}}{s} = {I}_{2} \cdot {\omega}_{2} = 510 \left(k g \cdot {m}^{2}\right) \cdot {\omega}_{2}$

Solving for ${\omega}_{2}$
${\omega}_{2} = \frac{450 \frac{k g \cdot {m}^{2}}{s}}{510 \left(k g \cdot {m}^{2}\right)} = 0.88 \frac{\text{rad}}{s}$

I hope this helps,
Steve