A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that is perpendicular to the platform at its center. The angular speed of the merry-go-round is constant, and a child at a distance of 1.4 m from the axis has a tangential speed of 2.2 m/s. What is the tangential speed of another child, who is located at a distance of 2.1 m from the axis?
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1.4
To find the tangential speed of the second child, we can use the concept of angular speed.
The angular speed (ω) is defined as the angle swept out per unit of time. It is constant for the merry-go-round in this case.
The formula to calculate tangential speed (v) is given by:
v = r * ω
where:
v = tangential speed
r = distance from the axis
Given that the first child has a tangential speed (v1) of 2.2 m/s and is located at a distance (r1) of 1.4 m from the axis, we have:
v1 = r1 * ω
We can rearrange this equation to solve for ω:
ω = v1 / r1
Now let's substitute the values given:
ω = 2.2 m/s / 1.4 m
ω ≈ 1.57 rad/s (rounded to two decimal places)
To find the tangential speed (v2) of the second child, who is located at a distance (r2) of 2.1 m from the axis, we can use the same formula:
v2 = r2 * ω
Now let's substitute the values into this equation:
v2 = 2.1 m * 1.57 rad/s
v2 ≈ 3.297 m/s (rounded to three decimal places)
Therefore, the tangential speed of the second child is approximately 3.297 m/s.