Let's address each part of the question step-by-step.
### Part 1: Finding the Angular Speed
To find the angular speed ([tex]\(\omega\)[/tex]) of an object moving in a circular path, we can use the following formula:
[tex]\[ \omega = \frac{v}{r} \][/tex]
where:
- [tex]\( v \)[/tex] is the linear speed,
- [tex]\( r \)[/tex] is the radius of the circular track.
Given:
- The radius ([tex]\( r \)[/tex]) is 202 meters,
- The linear speed ([tex]\( v \)[/tex]) is 50 meters per second.
Plugging in these values, we get:
[tex]\[ \omega = \frac{50 \, \text{m/s}}{202 \, \text{m}} \][/tex]
[tex]\[ \omega = 0.24752 \, \text{rad/s} \][/tex]
Thus, the angular speed is approximately [tex]\( 0.24752 \, \text{rad/s} \)[/tex].
### Part 2: Finding the Magnitude of the Acceleration
To find the magnitude of the centripetal (radial) acceleration ([tex]\( a \)[/tex]), we use the formula:
[tex]\[ a = \frac{v^2}{r} \][/tex]
where:
- [tex]\( v \)[/tex] is the linear speed,
- [tex]\( r \)[/tex] is the radius of the circular track.
Given:
- The radius ([tex]\( r \)[/tex]) is 202 meters,
- The linear speed ([tex]\( v \)[/tex]) is 50 meters per second.
Plugging in these values, we get:
[tex]\[ a = \frac{(50 \, \text{m/s})^2}{202 \, \text{m}} \][/tex]
[tex]\[ a = \frac{2500 \, \text{m}^2/\text{s}^2}{202 \, \text{m}} \][/tex]
[tex]\[ a = 12.3762 \, \text{m/s}^2 \][/tex]
Thus, the magnitude of the acceleration is approximately [tex]\( 12.3762 \, \text{m/s}^2 \)[/tex].
### Summary
- Angular speed: [tex]\( 0.24752 \, \text{rad/s} \)[/tex]
- Magnitude of acceleration: [tex]\( 12.3762 \, \text{m/s}^2 \)[/tex]
