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A child is riding a merry-go-round that is turning at 7.18 rpm. if the child is standing 4.65 m from the center of the merry-go-round, how fast is the child moving?

Respuesta :

skyluke89
First we need to convert the angular speed from rpm to rad/s. Keeping in mind that 
[tex]1 rev= 2 \pi rad[/tex]
[tex]1 min = 60 s[/tex]
the angular speed is
[tex]\omega = 7.18 \frac{rev}{min} \cdot \frac{2 \pi}{60} = 0.75 rad/s [/tex]

And so now we can calculate the tangential speed of the child, which is the angular speed times the distance of the child from the center of the motion:
[tex]v= \omega r = (0.75 rad/s)(4.65 m)=3.50 m/s[/tex]

The rate of change of angular displacement is defined as angular speed, Its unit is rad/sec. The child is moving at a pace of 3.48 m/sec.

What is the definition of Angular Speed?

The rate of change of angular displacement is defined as angular speed, and it is stated as follows:

ω = θ t

The given data in the problem is;

n is the rpm of the child=7.18

r is the distance from the center= 4.65 m

v is the linear speed=?

The angular velocity of the child is found by the formula;

[tex]\rm \omega = \frac{2\pi N}{60}\\\\\rm \omega = \frac{2\times 3.14 \times 7.18 }{60}\\\\ \rm \omega=0.75 \; rad/sec[/tex]

The linear speed is given by the product of angular speed and radius,

[tex]\rm v= \omega r \\\\ \rm v= 0.75 \times 4.65 \\\\\ \rm v= 3.48\ m/sec[/tex]

Hence the child is moving at a pace of 3.48 m/sec.

To learn more about the angular speed refer to the link;

https://brainly.com/question/9684874

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