Question

A child pushes a merry go-round from rest to a final angular speed of 0.6 rev/s with constant angular acceleration. In doing so, the child pushes the merry-go-round 1.8 revolutions. What is the angular acceleration of the merry go-round?

rad/s^2

This is how i did this, but its wrong:

(1.8/0.6)=3
then, (0.6-0)/3=0.2 , so the answer
should be 0.2 rad/s^2, but its wrong I don't understand why!

Answer 1

v final = v initial + a t

.6 revs/s = a t

average speed during acceleration = .6/2 = .3 revs/s

1.8 revs = .3 revs/s * t
t = 6 seconds

a t = .6 revs/s
a = .6/6 = .1 revs/s^2
but 1 rev = 2 pi radians
a = .1 revs/s^2 * 2 pi radians/rev =
.2 pi radians/s^2
= .628 radians/s^2

Answer 2

Thanks :) Damon!

Answer 3

[AI-generated answer — Explain Bot]

To find the angular acceleration of the merry-go-round, we can use the equation:

ω² = ω₀² + 2αθ

where ω is the final angular velocity, ω₀ is the initial angular velocity (which is 0 in this case), α is the angular acceleration, and θ is the angle through which the merry-go-round has rotated.

Given that the final angular velocity is 0.6 rev/s and the number of revolutions is 1.8, we need to convert these values into radian measures since the equation is in terms of radians.

To convert revolutions to radians, we know that 1 revolution is equivalent to 2π radians. Therefore, 1.8 revolutions is equal to 1.8 * 2π radians.

θ = 1.8 * 2π = 3.6π radians

Now we can substitute the values into the equation:

(0.6 rev/s)² = 0 + 2α * (3.6π radians)

Simplifying:

0.36 rev²/s² = 7.2πα rad

To isolate α, we can divide both sides of the equation by 7.2π:

0.36 rev²/s² ÷ (7.2π) = α

Using a calculator to evaluate the expression on the left side gives us:

0.36 / (7.2π) ≈ 0.015915 rad/s²

So, the angular acceleration of the merry-go-round is approximately 0.0159 rad/s².