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A dad pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 18 rpm, in 13.0 s. Assume the merry-go-round is a uniform disk of radius 2.7 m; and has a mass of 800 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque. What force is required at the edge?

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

Ï„ = 3280.5 x 0.145

Ï„ = 475.67 N m

Ï„ = F x r

475.67 = F x 2.7

F = 176.175 N

The torque and force required are respectively; Ï„ = 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

Ï„ = 3051 * 0.145

Ï„ = 442.395 N.m

Force required at edge is;

F = Ï„/r

F = 442.395/2.7

F = 163.85 N

Read more about required Torque at; https://brainly.com/question/20691242

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