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?

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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?

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Answer 1 · 2018-03-22T21:41:42

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

Explanation:

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

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

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
$I_{1} = M \cdot \frac{R^{2}}{2} = \frac{100 k g \cdot {(3 m)}^{2}}{2} = 450 k g \cdot m^{2}$ $I_1 = M*R^2/2 = (100 kg*(3 m)^2)/2 = 450 kg*m^2$

The formula for angular momentum is $L = I \cdot ω$ $L = I*omega$ . So our m-g-r has angular momentum
$L = 450 k g \cdot m^{2} \cdot 1 \frac{rad}{s} = 450 \frac{k g \cdot m^{2}}{s}$ $L = 450 kg*m^2*1 "rad"/s = 450 (kg*m^2)/s$

When the girl jumps 1 m out from the center, she becomes an additional part of the rotational inertia.
$I_{2} = 450 (k g \cdot m^{2}) + 60 k g \cdot {(1 m)}^{2} = 510 (k g \cdot m^{2})$ $I_2 = 450 (kg*m^2) + 60 kg*(1 m)^2 = 510 (kg*m^2)$

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

momentum before = momentum after

$450 \frac{k g \cdot m^{2}}{s} = I_{2} \cdot ω_{2} = 510 (k g \cdot m^{2}) \cdot ω_{2}$ $450 (kg*m^2)/s = I_2*omega_2 = 510 (kg*m^2)*omega_2$

Solving for $ω_{2}$ $omega_2$
$ω_{2} = \frac{450 \frac{k g \cdot m^{2}}{s}}{510 (k g \cdot m^{2})} = 0.88 \frac{rad}{s}$ $omega_2 = (450 (kg*m^2)/s)/(510 (kg*m^2)) = 0.88 "rad"/s$

I hope this helps,
Steve

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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?

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Explanation:

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