To find the angular velocity [tex]\(\omega\)[/tex] of an ice skater given that her radial acceleration at a point [tex]\( r = 26.0 \, \text{cm} \)[/tex] from the axis of rotation should not exceed 12.0 times the gravitational acceleration [tex]\( g \)[/tex], we follow these steps:
1. Convert the radius to meters:
[tex]\[
r = 26.0 \, \text{cm} = 0.26 \, \text{m}
\][/tex]
2. Understand the gravitational acceleration [tex]\( g \)[/tex]:
[tex]\[
g = 9.8 \, \text{m/s}^2
\][/tex]
3. Calculate the maximum radial acceleration:
[tex]\[
\text{Maximum radial acceleration} = 12.0 \times g = 12.0 \times 9.8 \, \text{m/s}^2 = 117.6 \, \text{m/s}^2
\][/tex]
4. Use the formula for radial acceleration:
The radial acceleration [tex]\( a \)[/tex] is given by:
[tex]\[
a = \omega^2 \cdot r
\][/tex]
Where, [tex]\( \omega \)[/tex] is the angular velocity.
5. Solve for angular velocity [tex]\( \omega \)[/tex]:
[tex]\[
\omega^2 = \frac{a}{r}
\][/tex]
[tex]\[
\omega = \sqrt{\frac{a}{r}}
\][/tex]
6. Substitute the given values:
[tex]\[
\omega = \sqrt{\frac{117.6 \, \text{m/s}^2}{0.26 \, \text{m}}}
\][/tex]
7. Calculate [tex]\(\omega\)[/tex]:
[tex]\[
\omega = \sqrt{452.3076923076922}
\][/tex]
[tex]\[
\omega \approx 21.27 \, \text{rad/s}
\][/tex]
Thus, the angular velocity [tex]\(\omega\)[/tex] of the ice skater is approximately [tex]\( 21.27 \, \text{rad/s} \)[/tex].
