Physics: A20kg 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?

To find the wavelength of a wave you have to divide the speed by the frequency..it will equal 2 m...460/230=2V=F Ωwave speed=frequency*wavelength...

A20kg boy stands on a small stationary (at rest) merry-go-round

lonelynomad00

16.03.2022

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

$I_{total}=I_{marry}+I_{boy}$

Put the value into the formula

$120=I_{marry}+mr^2$

$I_{marry}=120-20\times2^2$

$I_{marry}=40\ kgm^2$

We need to calculate the

Using conservation of angular momentum

$L_{i}=L_{f}$

$I_{i}\omega_{i}=I_{f}\omega_{f}$

$I_{total}\times\dfrac{v}{r}=I_{marry}\omega_{marry}$

Put the value into the formula

$120\times\dfrac{1.5}{2}=40\times\omega_{marry}$

$\omega_{marry}=\dfrac{120\times\dfrac{1.5}{2}}{40}$

$\omega_{marry}=2.25\ rad/s$

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

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