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

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

in progress 0
Diễm Thu 2 months 2021-07-30T08:20:00+00:00 1 Answers 8 views 0

Answers ( )

    0
    2021-07-30T08:21:15+00:00

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )