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A 34-kg child runs with a speed of 2.8 m/s tangential to the rim of a stationary merrygo-round. The merry-go-round has a momentum of inertia of 510 kg*m2 and a radius of 2.31 m. When the child jumps onto the merry-go-round, the entire system begins to rotate. What is the angular speed of the system

Sagot :

Answer: [tex]0.43\ rad/s[/tex]

Explanation:

Given

Mass of child [tex]m=34\ kg[/tex]

speed of child is [tex]v=2.8\ m/s[/tex]

Moment of inertia of merry go round is [tex]I=510\ kg.m^2[/tex]

radius [tex]r=2.31\ m[/tex]

Conserving the angular momentum

[tex]\Rightarrow mvr=I\omega \\\Rightarrow 34\times 2.8\times 2.31=510\times \omega\\\\\Rightarrow \omega=\dfrac{219.912}{510}\\\Rightarrow \omega=0.43\ rad/s[/tex]

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