Answer:
5 .07 s .
Explanation:
The child will move on a circle of radius r
r = 1.5 m
Let the velocity of rotation = v
radial acceleration = v² / r
v² / r = 2.3
v² = 2.3 r = 2.3 x 1.5
= 3.45
v = 1.857 m /s
Time of revolution = 2π r / v
= 2 x 3.14 x 1.5 / 1.857
= 5 .07 s .
The merry-go round will take 5.07 s to complete one complete revolution.
Given data:
The distance of child from the center is, r = 1.5 m.
The magnitude of radial acceleration is, [tex]a = 2.3 \;\rm m/s^{2}[/tex].
Since, the child is on merry-go round, which is undergoing a rotational motion. And radial acceleration means that it is under the acceleration, whose value is given as,
[tex]a=\dfrac{v^{2}}{r}\\\\v=\sqrt{a \times r}[/tex]
Here, v is the linear velocity.
Solving as,
[tex]v=\sqrt{2.3 \times 1.5} \\\\v=1.857 \;\rm m/s[/tex]
Now, we to obtain the time taken by merry-go round to complete one revolution. Then the expression for the time taken to complete one revolution is,
[tex]T=\dfrac{2 \pi r}{v}\\\\T=\dfrac{2 \pi \times 1.5}{1.857}\\\\T = 5.07 \;\rm s[/tex]
Thus, we can conclude that the merry-go round will take 5.07 s to complete one revolution.
Learn more about the centripetal acceleration here:
https://brainly.com/question/14465119
