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Four children running alongside push tangentially along the platform's circumference until, starting from rest, the merry-go-round reaches a steady speed of one complete revolution every 2.8 s.

a. If each child exerts a force of 26 N, how far does each child run?
b. What is the angular acceleration of the merry-go-round?
c. How much work does each child do?
d. What is the kinetic energy of the merry-go-round?

Respuesta :

Answer

Assuming mass of the merry-go-round,M = 220 Kg

           diameter of the merry-go-round = 4.10 m

                             radius,r = 2.05 m

time, t = 2.8 s

a) Force,F = 26 N

  Moment of inertia of merry-go-round =

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

                    [tex]I =\dfrac{1}{2}\times 220\times 2.05^2[/tex]

                           I = 462.275 kg.m²

  we know,  

                   [tex]\tau = I\alpha[/tex]

                   [tex]F.r = I\alpha[/tex]

                   [tex]4\times 26\times 2.05 =462.275\times \alpha[/tex]

                   [tex]\alpha = 0.461\ rad/s^2[/tex]

initial angular speed = 0 rad/s

final angular speed[tex]\omega_f=\dfrac{2\pi}{T}[/tex]

                               [tex]\omega_f=\dfrac{2\pi}{2.8}[/tex]

                               [tex]\omega_f=2.24\ rad/s[/tex]

using rotational equation of motion

[tex]\omega_f^2 = \omega_i^2 +2\alpha \theta[/tex]

[tex]2.24^2 = 0^2 +2\times 0.461\times \theta[/tex]

      θ = 5.44 rad

  S = r θ

  S = 5.44 x 2.05

  S = 11.15 m

child are running at 11.15 m.

b) angular acceleration, α = 0.461 rad/s²

c) Work done = τ θ

                       = 26 x 4 x 2.05 x 5.44

                       = 2498.048 J

    Work done by each child = 2498.048/4 = 624.51 J

d) Kinetic energy

        [tex]KE  =\dfrac{1}{2}I\omega^2[/tex]

        [tex]KE  =\dfrac{1}{2}\times 462.275\times 2.241^2[/tex]

                KE = 1160.79 J