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
