Physics High School

A day-care center has a merry-go-round that consists of a uniform 240-kg circular wooden platform 4.00m in diameter. Four children run alongside the merry-go-round and push tangentially along the platform’s circumference until, starting from rest, the merry-go-round is spinning at a rate of 2.14 rev/minute. During the spin-up:

a) If each child exerts a sustained force of 26N, how far does each child run?
b) What is the angular acceleration of the merry-go-round?
SHOW ALL WORK AND STEPS PLEASE.

Answers

Answer 1

Answer:

a) The distance each child ran is approximately 0.232 m

b) The angular acceleration is 0.21[tex]\overline 6[/tex] rad/s²

Explanation:

The given parameters of the merry-go-round are;

The mass of the merry-go-round, m = 240-kg

The diameter of the merry-go-round, d = 4.00 m

The number of children pushing tangentially on the merry-go round = Four children

The spinning rate of the merry-go-round, n = 2.14 rev/minute

a) Given that the force exerted by each child = 26 N, we have;

The total force applied by the four children, F = 26 N × 4 = 104 N

The tangential acceleration, [tex]a_t[/tex] = F/m = 104 N/240-kg = 0.4[tex]\overline 3[/tex] m/s²

The angular acceleration, α = [tex]a_t[/tex]/r

Where, the radius of the merry-go-round, r = d/2

∴ r = 4.00 m/2 = 2.00 m

α = 0.4[tex]\overline 3[/tex] m/s²/(2.00 m) = 0.21[tex]\overline 6[/tex] rad/s²

We have;

ω² = ω₀² + 2·α·Δθ

The merry-go-round starts from rest, therefore; ω₀ = 0 rad/s

ω² = 2·α·Δθ

Δθ = ω²/(2·α)

n = 2.14 rev/minute

∴ ω = 2·π×2.14/60 rad/s ≈ 0.22410 rad/s

∴ Δθ = (0.22410 rad/s)²/(2 × 0.21[tex]\overline 6[/tex] rad/s²) ≈ 0.11589 rad

Therefore, the angle each child ran, θ = 0.11589 rad

The distance each child ran = r·θ

∴ The distance each child ran = 2.00 m × 0.11589 rad ≈ 0.232 m

b) From part 'a' above, the tangential acceleration, [tex]a_t[/tex] = 0.4[tex]\overline 3[/tex] m/s²

Angular acceleration, α = [tex]a_t[/tex]/r

∴ α = 0.4[tex]\overline 3[/tex] m/s²/(2.00 m) = 0.21[tex]\overline 6[/tex] rad/s²

The angular acceleration = 0.21[tex]\overline 6[/tex] rad/s²

Related Questions

A 60 kg skateboarder accelerates themselves at 1.5m/s2. How much force was required to do this?

Answers

Answer:

90 N

Explanation:

m = 60 kg

a = 1.5 m/s^2

F = m × a = 1.5 × 60 = 90 N

What is the net force acting on a 52 kg object that has a velocity of 8.0 m/s and is moving in a circle of radius 1.6 m? a. 4000N b. 20880N c. 2500N d. 3500N

Answers

Σf = m a
Σf = m v^2 / r
Σf = 52 8^2 / 1.6
Σf = 2080 N
The net force acting of the object will be "2080 N".

Given values are:

Mass, m = 52 kgVelocity, v = 8.0 m/sRadius, r = 1.6 m

As we know the formula,

→ [tex]\Sigma f = ma[/tex]

or,

→ [tex]\Sigma f = \frac{mv^2}{r}[/tex]

By putting the values, we get

         [tex]= \frac{52\times 8^2}{1.6}[/tex]

         [tex]= \frac{52\times 64}{1.6}[/tex]

         [tex]= 2080 \ N[/tex]

Thus the response above is appropriate.

Learn more about force here:

https://brainly.com/question/14109330

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