A merry-go-round with a radius of 2.0 m and a mass of 150 kg can be modeled as a

Question

A merry-go-round with a radius of 2.0 m and a mass of 150 kg can be modeled as a uniform solid disk on a frictionless bearing at its center. The merry-go-round is rotating at 1.50 rad/s with a 60 kg person standing at the rim. The person then walks toward the center of the disk. (a) How fast is the merry-go-round rotating when the person is at a radial distance of 0.50 m from the center? (b) How much does the kinetic energy of the disk-person system change as this occurs? Where does the additional kinetic energy come from?

Explanation / Answer

here,

radius , r = 2 m

mass , m = 150 kg

initial speed , w0 = 1.5 rad/s

mass of the person , M = 60 kg

a)

when person is at x = 0.5 m ,

let the final angular speed be w

using conservation of angular momentum

( 0.5 * m* r^2 + M * r^2) * w0 = ( 0.5 * m * r^2 + M * x^2) * w

(0.5 * 150 * 2^2 + 60 * 2^2) * 1.5 = ( 0.5 * 150 * 2^2 + 60 * 0.5^2) * w

w = 2.57 rad/s

b)

the chnage in kinetic energy , KE = 0.5 * ( 0.5 * m * r^2 + M * x^2) * w^2 - 0.5 * ( 0.5 * m* r^2 + M * r^2) * w0^2

KE = 0.5 * (0.5 * 150 * 2^2 + 60 * 0.5^2) * 2.57^2 - 0.5 * (0.5 * 150 * 2^2 + 60 * 2^2) * 1.5^2

KE = 432.77 J

the kinetic energy gained is 432.77 J

the additional kinetic energy comes from the work done by the person

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