Answer:
The final angular speed of the skater is 12 radians per second.
Explanation:
Let consider the skater as a rotating system, given the absence of external forces, the Principle of Angular Momentum Conservation is applied:
[tex]I_{o}\cdot \omega_{o} = I_{f}\cdot \omega_{f}[/tex]
Where:
[tex]I_{o}[/tex], [tex]I_{f}[/tex] - Initial and final moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].
[tex]\omega_{o}[/tex], [tex]\omega_{f}[/tex] - Angular speed, measured in radians per second.
The final angular speed is cleared afterwards:
[tex]\omega_{f} = \frac{I_{o}}{I_{f}} \cdot \omega_{o}[/tex]
Given that [tex]I_{f} = \frac{1}{2}\cdot I_{o}[/tex] and [tex]\omega_{o} = 6\,\frac{rad}{s}[/tex], the final angular speed is:
[tex]\omega_{f} = \frac{I_{o}}{\frac{1}{2}\cdot I_{o} } \cdot \omega_{o}[/tex]
[tex]\omega_{f} = 2 \cdot \omega_{o}[/tex]
[tex]\omega_{f} = 2 \cdot \left(6\,\frac{rad}{s} \right)[/tex]
[tex]\omega_{f} = 12\,\frac{rad}{s}[/tex]
The final angular speed of the skater is 12 radians per second.
The skater's final angular speed is equal to 12 rad/s.
Given the following data:
To determine the skater's final angular speed, we would use the law of conservation of angular momentum:
By applying the law of conservation of angular momentum, we have:
[tex]I_i\omega_i = I_f\omega_f[/tex] ...equation 1.
Making the [tex]\omega_f[/tex] subject of formula, we have:
[tex]\omega_f = \frac{I_i \omega_i}{I_f}[/tex] ...equation 2
Note: The skater decreases her moment of inertia by two times.
Final moment of inertia, [tex]I_f[/tex] = [tex]\frac{I_i}{2}[/tex] ...equation 3.
Substituting eqn. 3 into eqn. 2, we have:
[tex]\omega_f = \frac{I_i \omega_i}{\frac{I_i}{2} }\\\\\omega_f = I_i \omega_i \times \frac{2}{I_i }\\\\\omega_f = 2\omega_i \\\\\omega_f = 2 \times 6\\\\\omega_f = 12\;rad/s[/tex]
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