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andreyandreev [35.5K]
3 years ago
13

Una mujer de masa m está parada en el borde de una mesa giratoria horizontal de momento de inercia I y radio R. La mesa al princ

ipio está en reposo y tiene libertad de moverse alrededor de un eje vertical sin fricción que pasa por su centro. La mujer empieza a caminar alrededor de la orilla en direción de las manecillas del reloj (vista desde arriba) a una velocidad constante v en relación a la Tierra.
a) En qué dirección y con qué velocidad angular gira la mesa?

b) Cuál es el trabajo efectuado por la mujer para poner en movimiento la mesa?
Physics
1 answer:
Dovator [93]3 years ago
3 0

r

- \frac{mR^2 }{I  } \ vAnswer:

a)      w = - \frac{m r }{I} v  ,  b)   W = - ½ m_woman R² (1 + m_woman R / I²) v²

Explanation:

a) To solve this exercise, let's use the conservation of angular momentum.

We define a system formed by the table and the woman, therefore the torques are internal and the moment is conserved

initial instant. Before starting to move the woman

         L₀ = 0

final instant. After starting to move

         L_f = I w + m v r

the moment is preserved

        L₀ = L_f

         0 = Iw + m v r

         w = - \frac{m r }{I} v                    (1)

the direction of the angular velocity is opposite to the direction of the linear velocity, that is, counterclockwise

b) for this part we use the relationship between work and kinetic energy

        W = ΔK

in this case the initial speed is zero and the final speed of the table, using the relationship between linear and angular variables

         v = w r

we substitute

          W = 0 - ½ I_total w²

          I_total = I + m_{woman} R²

          W = - ½ (I + m_woman R²)  ( \frac{m_{woman} R}{I} \ v) ²

          W = - ½ (m_woman² R² + m_woman³ R³ / I²) v²

          W = - ½ m_woman R² (1 + m_woman R / I²) v²

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