Respuesta :
Answer:
F = 176.175 N
Explanation:
given,
radius of merry - go - round = 2.7 m
mass of the disk = 800 kg
speed of the merry- go-round = 18 rpm
= 18 [tex]\dfrac{2\pi }{60}[/tex]
= 1.88 rad/s
time = 13 s
mass of two children = 25 kg
ω = ω₀ + α t
1.88 = 0 + α(13)
α = 0.145 rad/s²
we know
τ = I α
so,
[tex]I = \dfrac{1}{2}MR^2+ 2 MR^2[/tex]
[tex]I = \dfrac{1}{2}\times 800 \times 2.7^2+ 2\times 25\times 2.7^2[/tex]
[tex]I = 3280.5 kg.m^2[/tex]
τ = I α
τ = 3280.5 x 0.145
τ = 475.67 N m
τ = F x r
475.67 = F x 2.7
F = 176.175 N
The torque and force required are respectively; τ = 442.395 N.m and F = 163.85 N
What is the required Torque?
The formula for mass moment of inertia of the system is;
I = ¹/₂Mr² + 2mr
We are given;
M = 800 kg
r = 2.7 m
m = 25 kg
Thus;
I = (¹/₂ * 800 * 2.7²) + (2 * 25 * 2.7)
I = 2916 + 135
I = 3051 kg.m²
Angular velocity achieved in 13 s is;
ω = 18 * 2π/60
ω = 1.885 rad/s
angular acceleration; α = ω/t
α = 1.885/13
α = 0.145 rad/s²
Torque required is gotten from;
τ = Iα
τ = 3051 * 0.145
τ = 442.395 N.m
Force required at edge is;
F = τ/r
F = 442.395/2.7
F = 163.85 N
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