A playground $$merry-go-round$$ of radius R $$=$$ $$2.00$$ m has a moment of inertia I $$=$$ $$250$$ kg.$$m2$$ and is rotating at $$10.0$$ $$rev/min$$ about a frictionless, vertical axle. Facing the axle, a $$25.0-kg$$ child hops onto the $$merry-go-round$$ and manages to sit down on the edge. The new angular speed of the $$merry-go-round$$ is

Question

A playground $$merry-go-round$$ of radius R $$=$$ $$2.00$$ m has a moment of inertia I $$=$$ $$250$$ kg.$$m2$$ and is rotating at $$10.0$$ $$rev/min$$ about a frictionless, vertical axle. Facing the axle, a $$25.0-kg$$ child hops onto the $$merry-go-round$$ and manages to sit down on the edge. The new angular speed of the $$merry-go-round$$ is

Infinity Learn · Accepted Answer

From conservation of angular momentum,

$I_{1} ω_{1} =$ $I_{f} ω_{f}$ .

(250 $k g . m^{2}$ )(10.0 rev/min) = [250 $k g . m^{2}$ +(25.0 kg)(2.0 m)²] $ω_{2}$

$ω_{2} = 7.14 r e v / m i n$

Law of conservation of angular momentum

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From conservation of angular momentum,
$I_{1} ω_{1} =$ $I_{f} ω_{f}$ .
(250 $k g . m^{2}$ )(10.0 rev/min) = [250 $k g . m^{2}$ +(25.0 kg)(2.0 m)²] $ω_{2}$
$ω_{2} = 7.14 r e v / m i n$

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Law of conservation of angular momentum

Remember concepts with our Masterclasses.

From conservation of angular momentum, I1ω1 = Ifωf.(250 kg.m2)(10.0 rev/min) = [250 kg.m2+(25.0 kg)(2.0 m)2]ω2 ω2 = 7.14 rev /min

Ready to Test Your Skills?

free study material

From conservation of angular momentum,
$I_{1} ω_{1} =$ $I_{f} ω_{f}$ .
(250 $k g . m^{2}$ )(10.0 rev/min) = [250 $k g . m^{2}$ +(25.0 kg)(2.0 m)²] $ω_{2}$
$ω_{2} = 7.14 r e v / m i n$