Sure, let's solve the problem step-by-step:
1. Understand the Given Data:
- The radial distance [tex]\( r \)[/tex] from the axis of rotation is given as 34.0 cm.
- This distance needs to be converted to meters for SI units, hence:
[tex]\[ r = 34.0 \, \text{cm} = 0.34 \, \text{m} \][/tex]
- The gravitational acceleration [tex]\( g \)[/tex] is given as 9.8 [tex]\( \text{m/s}^2 \)[/tex].
- The radial acceleration is not to exceed 10.0 times the gravitational acceleration [tex]\( g \)[/tex]:
[tex]\[ \text{Radial acceleration} = 10 \times g = 10 \times 9.8 \, \text{m/s}^2 = 98 \, \text{m/s}^2 \][/tex]
2. Formula for Radial Acceleration:
The radial acceleration in circular motion is given by:
[tex]\[ a_r = \omega^2 \times r \][/tex]
Where:
- [tex]\( a_r \)[/tex] is the radial acceleration.
- [tex]\( \omega \)[/tex] is the angular velocity in radians per second (rad/s).
- [tex]\( r \)[/tex] is the radius in meters.
3. Rearrange the Formula to Solve for [tex]\( \omega \)[/tex]:
Given [tex]\( a_r \)[/tex] and [tex]\( r \)[/tex], we solve for [tex]\( \omega \)[/tex]:
[tex]\[ \omega = \sqrt{\frac{a_r}{r}} \][/tex]
4. Substitute the Known Values:
- [tex]\( a_r = 98 \, \text{m/s}^2 \)[/tex]
- [tex]\( r = 0.34 \, \text{m} \)[/tex]
Thus:
[tex]\[ \omega = \sqrt{\frac{98}{0.34}} \][/tex]
5. Calculate the Angular Velocity [tex]\( \omega \)[/tex]:
[tex]\[ \omega = \sqrt{\frac{98}{0.34}} \approx 16.977 \, \text{rad/s} \][/tex]
Therefore, the angular velocity [tex]\( \omega \)[/tex] of the ice skater, such that the radial acceleration does not exceed 10 times the gravitational acceleration, is approximately [tex]\( 16.98 \, \text{rad/s} \)[/tex].
Here's the summarized information:
- Radius [tex]\( r \)[/tex]: 0.34 meters
- Radial Acceleration [tex]\( a_r \)[/tex]: 98.0 [tex]\( \text{m/s}^2 \)[/tex]
- Angular Velocity [tex]\( \omega \)[/tex]: 16.98 [tex]\( \text{rad/s} \)[/tex]
