To solve this problem, we need to find the angular velocity (ω) of the ice skater given that the radial acceleration at a distance of 24.0 cm from her axis of rotation must not exceed 6 times the gravitational acceleration (g).
### Step-by-Step Solution
1. Convert the radius to meters:
Given the radial distance [tex]\( r \)[/tex] is 24.0 cm, we first convert this to meters:
[tex]\[
r = \frac{24.0 \, \text{cm}}{100} = 0.24 \, \text{m}
\][/tex]
2. Identify the given gravitational acceleration:
The gravitational acceleration [tex]\( g \)[/tex] is:
[tex]\[
g = 9.81 \, \text{m/s}^2
\][/tex]
3. Calculate the maximum radial (centripetal) acceleration:
The problem states that the radial acceleration should not exceed 6 times the gravitational acceleration. Hence:
[tex]\[
a_r = 6.00 \times g = 6.00 \times 9.81 \, \text{m/s}^2 = 58.86 \, \text{m/s}^2
\][/tex]
4. Relate radial acceleration to angular velocity:
The formula for radial (centripetal) acceleration [tex]\( a_r \)[/tex] is given by:
[tex]\[
a_r = \omega^2 \times r
\][/tex]
Where [tex]\( \omega \)[/tex] is the angular velocity and [tex]\( r \)[/tex] is the radius.
5. Solve for the angular velocity [tex]\( \omega \)[/tex]:
Rearrange the equation to solve for [tex]\( \omega \)[/tex]:
[tex]\[
\omega = \sqrt{\frac{a_r}{r}}
\][/tex]
Substitute the values of [tex]\( a_r \)[/tex] and [tex]\( r \)[/tex] into the equation:
[tex]\[
\omega = \sqrt{\frac{58.86 \, \text{m/s}^2}{0.24 \, \text{m}}}
\][/tex]
6. Calculate the angular velocity:
Carry out the division and square root operations:
[tex]\[
\omega = \sqrt{245.25} \approx 15.660 \, \text{rad/s}
\][/tex]
### Final Answer
The angular velocity of the ice skater, given the constraints, is approximately:
[tex]\[
\boxed{15.66 \, \text{rad/s}}
\][/tex]
