A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round, which can be modeled as a disk with a mass of 290 kg , is spinning at 25 rpm. John runs tangent to the merry-go-round at 5.6 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg.

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wagonbelleville wagonbelleville · Answer 1 · 2019-11-04T07:14:30+01:00

Answer

given,

diameter of merry - go - round = 3 m

mass of the disk = 290 kg

speed of the merry- go-round = 25 rpm

speed = 5.6 m/s

mass of John = 30 kg

[tex]I_{disk} = \dfrac{1}{2}MR^2[/tex]

[tex]I_{disk} = \dfrac{1}{2}\times 290 \times 1.5^2[/tex]

[tex]I_{disk} = 326.25 kg.m^2[/tex]

initial angular momentum of the system

[tex]L_i = I\omega_i + mvR[/tex]

[tex]L_i =326.25 \times 25 \times \dfrac{2\pi}{60} + 30 \times 5.6 \times 1.5[/tex]

[tex]L_i =1106.12\ kg.m^2/s[/tex]

final angular momentum of the system

[tex]L_f = (I_{disk}+mR^2)\omega_{f}[/tex]

[tex]L_f = (326.25 + 30\times 1.5^2)\omega_{f}[/tex]

[tex]L_f= (393.75)\omega_{f}[/tex]

from conservation of angular momentum

[tex]L_i = L_f[/tex]

[tex]1106.12 = (393.75)\omega_{f}[/tex]

[tex]\omega_{f}=2.809 \times \dfrac{60}{2\pi}[/tex]

[tex]\omega_{f}=26.82\ rpm[/tex]