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

The right solution is "0.511".

Step-by-step 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* (69.5* (5)^2)`

=
`1630+2* (69.5* 25)`

=
`5105 \ Kg.m^2`

hence,

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

⇒
`1630* 1.6=5105* \omega'`

`2608=5105* \omega'`

`\omega'=(2608)/(5105)`

`=0.511`