1 answer

A 4.5-m-diameter merry-go-round is rotating freely with an angular velocity of 0.84 rad/s . Its total...

Question:

A 4.5-m-diameter merry-go-round is rotating freely with an angular velocity of 0.84 rad/s . Its total moment of inertia is 1700 kg⋅m2 . Four people standing on the ground, each of mass 68 kg , suddenly step onto the edge of the merry-go-round.

What is the angular velocity of the merry-go-round now?

What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?


Answers

here,

diameter , d = 4.5 m

radius , r = d/2 = 2.25 m

initial angular velocity , w0 = 0.84 rad/s

moment of inertia , I = 1700 kg.m^2

mass of each person , m = 68 kg

a)

let the angular velocity of the merry-go-round be w

using conservation of angular momentum

I * w0 = ( I + 4 * m * r^2) * w

1700 * 0.84 = ( 1700 + 4 * 68 * 2.25^2) * w

solving for w

w = 0.46 rad/s

the new angular velocity is 0.46 rad/s

b)

let the angular velocity of the merry-go-round be w

using conservation of angular momentum

(I + 4 * m * r^2) * w0 = I * w

(1700 + 4 * 68 * 2.25^2) * 0.84 = 1700 * w

solving for w

w = 1.52 rad/s

the new angular velocity is 1.52 rad/s

.

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