Question

A 5.0-m radius playground merry-go-round with a moment of inertia of 1,630 kg m2 is rotating freely with an angular speed of 1.6 rad/s. Two people, each having a mass of 69.5 kg, are standing right outside the edge of the merry-go-round and step on it with negligible speed. What is the angular speed of the merry-go-round right after the two people have stepped on

Answer 1

Answer:

The right solution is "0.511".

Explanation:

Given:

Initial moment of inertia,

= 1630 kg.m²

Radius,

= 5 m

Angular speed,

= 1.6 rad/s

Now,

The moment of inertia after stepping on will be:

= $1630+2\times (69.5\times (5)^2)$

= $1630+2\times (69.5\times 25)$

= $5105 \ Kg.m^2$

hence,

As per the question, the angular speed is conserved, then

⇒ $1630\times 1.6=5105\times \omega'$

            $2608=5105\times \omega'$

                $\omega'=\frac{2608}{5105}$

                     $=0.511$