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

Answer 1

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

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