Answer to Question #136732 in Mechanics | Relativity for Subhasmita Dash

Question

A merry-go-round has moment of inertia 4500 kg m2. It is mounted on a frictionless vertical axle. It is initially rotating at an angular speed of 1 rpm. A girl jumps onto the merry-go-round in the radial direction. If the moment of inertia of the girl is 500 kg m 2, find the angular speed of rotation of the merry-go-round.

Answer 1

solution

given data

moment of inertia of marry go round(I₁)=4500 kg m2

angular speed($\omega$₁ )=1 rpm

moment of inertia of girl(I₂)=500 kg m²

there are no any external force if girl and marry go round is considered as system then angular momentum will remain conserved

$I_1\omega_1=(I_2+I_1)\omega_2\\\omega_2=\frac{(I_1\omega_1)}{(I_1+I_2)}$

$\omega_2=\frac{4500\times1}{5000}$ $=0.9 \space rpm$

so angular velocity is 0.9 rpm