Respuesta :
Answer:
0.4176 m
Explanation:
L: angular momentum
Just as [tex]p= mv[/tex], [tex]L = I\omega[/tex], I = sum of [tex]mr^{2}[/tex]
[tex]L_o=2mr_o^{2}\omega_o[/tex]
[tex]Lf = 2mrf^{2}\omega_f[/tex]
we know [tex]r_o = 0.705 m[/tex]
[tex]\omega_o = 10 * 2\pi[/tex] rad per minute, and [tex]\omega_f = 28.5 * 2\pi[/tex] rad per minute
[tex]rf = r_o\sqrt{\frac {\omega_o}{\omega_f}} = 0.705 * \frac {10.0}{ 28.5}[/tex]
rf = 0.4176 m
The weights are as far as 0.418 m from the rotational axis after she pulls her arms in
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Further explanation
Let's recall Angular Momentum formula as follows:
[tex]\boxed {L = I \omega}[/tex]
where:
L = angular momentum ( kg.m²/s )
I = moment of inertia ( kg.m² )
ω = angular frequency ( rad/s )
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Given:
initial angular frequency = ω₁ = 10.0 rpm
initial radius of rotation = R₁ = 0.705 m
final angular frequency = ω₂ = 28.5 rpm
Asked:
final radius of rotation = R₂ = ?
Solution:
We will use Conservation of Angular Momentum to solve this problem:
[tex]L_1 = L_2[/tex]
[tex]I_1 \omega_1 = I_2 \omega_2[/tex]
[tex](2m (R_1)^2) \omega_1 = (2m (R_2)^2) \omega_2[/tex]
[tex](R_1)^2 \omega_1 = (R_2)^2 \omega_2[/tex]
[tex](0.705)^2 \times 10.0 = (R_2)^2 \times 28.5[/tex]
[tex]R_2 = \sqrt{ \frac{10.0}{28.5} \times (0.705)^2 }[/tex]
[tex]R_2 \approx 0.418 \texttt{ m}[/tex]
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Conclusion:
The weights are as far as 0.418 m from the rotational axis after she pulls her arms in
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Learn more
- Impacts of Gravity : https://brainly.com/question/5330244
- Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
- The Acceleration Due To Gravity : https://brainly.com/question/4189441
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Answer details
Grade: High School
Subject: Physics
Chapter: Rotational Dynamics
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