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

Question

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

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Hưng Khoa 3 years 2021-08-16T03:48:25+00:00 1 Answers 48 views 0

Answers ( )

  1. Cherry
    0
    2021-08-16T03:50:16+00:00
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    Answer: 0.43\ rad/s

    Explanation:

    Given

    Mass of child m=34\ kg

    speed of child is v=2.8\ m/s

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

    radius r=2.31\ m

    Conserving the angular momentum

    \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

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