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Circular motion refers to the movement of an object along a circular path. This is a common motion in both everyday and specialized contexts, such as the rotation of a car around a track or an astronaut in a training simulator. In this problem, the astronaut is moving in a circular path with a radius of 11 meters. As she moves around in this circle, she experiences a force directed towards the center of the circle, known as the centripetal force. This force is necessary to keep her moving in a circular path and is caused by the rotation of the device she is in.
In uniform circular motion, even though the speed of the object remains constant, its velocity is not constant. This is because velocity is a vector quantity, which means it has both a magnitude (speed) and a direction. Since the object is constantly changing direction as it moves around the circle, its velocity changes as well. The need for a centripetal force is what provides this change in direction.

The goal in this exercise is to calculate the velocity of the trainee as she rotates in a circular motion. Velocity calculation in circular motion can be intriguing because it involves the balance between gravitational forces and the forces acting due to motion.
To find velocity, we utilize the formula for centripetal force:

• The centripetal force (\( F_c \) ) required to keep the trainee moving in a circle is expressed as \( F_c = \frac{mv^2}{r} \).
• Given that this force is 7.45 times greater than the gravitational force (\( mg \) ), we equate: \( \frac{mv^2}{r} = 7.45mg \).

The mass (\( m \) ) cancels out in this equation, simplifying our task. Now, we can isolate \( v^2 \) ,\( v^2 = 7.45gr \)..
Substituting the known values of \( g = 9.8 \, \text{m/s}^2 \) and \( r = 11 \, \text{m} \), we calculate:\[ v^2 = 7.45 \times 9.8 \times 11 \approx 806.91 \, \text{m}^2/\text{s}^2 \] Thus, \( v \approx \sqrt{806.91} \) which results in a speed of approximately \( 28.41 \, \text{m/s} \).

Once we find the velocity in meters per second, we can determine the rotational speed in terms of revolutions per second, which gives a better idea of how often the trainee completes one full circle.
To convert to revolutions per second, we need to find the circumference of the path, which determines how much distance is covered in one complete revolution. The circumference is calculated by \( C = 2\pi r = 2\pi \times 11 \approx 69.11 \, \text{m} \). Given that the speed \( v \) is \( 28.41 \, \text{m/s} \), we divide this speed by the circumference \( C \) :

• The number of revolutions per second is \( \frac{v}{C} = \frac{28.41}{69.11} \approx 0.411 \text{ rev/s} \).

This means that the trainee experiences just under half a revolution for every second she is in motion. This insight helps connect the linear speed to the rotational speed essential for understanding circular motion in practical applications.