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A playground merry-go-round of radius r = 2.00 m has a moment of inertia = 250 ⋅ , and is rotating at 10.0 rev/min about a frictionless vertical axle. Facing the axle, a 20.0-kg child hops onto the merry-go-round, and manages to sit down on the edge. What is the new moment of inertia of the merry-go-round?

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lublana

Answer:

[tex]330kgm^2[/tex]

Explanation:

We are given that

Radius,r=2 m

Moment of inertia,I=250[tex]kgm^2[/tex]

Angular velocity,[tex]\omega_1=10 rev/min[/tex]

Mass of child,m=20 kg

We have to find the new moment of inertia of the merry go round.

New moment of inertia ,[tex]I'=mr^2+I[/tex]

Using the formula

[tex]I'=20(2)^2+250[/tex]

[tex]I'=80+250=330kgm^2[/tex]

Hence, the new moment of inertia of the merry -go-round=[tex]330kgm^2[/tex]

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