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