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A 20kg boy stands on a small stationary (at rest) merry-go-round near the edge of the merry-go-round. The total moment of inertia of the system of merry-go-round with the boy on it about the center is 120kg m2 . The boy at the edge of the merry-go-round (radius of 2m) jumps off the merry-go-round in a tangential direction with a liner speed of 1.5m/s. What is the angular speed of the merry-go-round after the boy leaves it?

Answer :

Answer:

The angular speed of the merry-go-round after the boy leaves it is 2.25 rad/s.

Explanation:

Given that,

Mass of boy = 20 kg

Total moment of inertia = 120 kgm²

Radius = 2 m

Speed = 1.5 m/s

We need to calculate the moment of inertia of boy

[tex]I_{total}=I_{marry}+I_{boy}[/tex]

Put the value into the formula

[tex]120=I_{marry}+mr^2[/tex]

[tex]I_{marry}=120-20\times2^2[/tex]

[tex]I_{marry}=40\ kgm^2[/tex]

We need to calculate the

Using conservation of angular momentum

[tex]L_{i}=L_{f}[/tex]

[tex]I_{i}\omega_{i}=I_{f}\omega_{f}[/tex]

[tex]I_{total}\times\dfrac{v}{r}=I_{marry}\omega_{marry}[/tex]

Put the value into the formula

[tex]120\times\dfrac{1.5}{2}=40\times\omega_{marry}[/tex]

[tex]\omega_{marry}=\dfrac{120\times\dfrac{1.5}{2}}{40}[/tex]

[tex]\omega_{marry}=2.25\ rad/s[/tex]

Hence, The angular speed of the merry-go-round after the boy leaves it is 2.25 rad/s.

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