A horizontal applied force of magnitude 25 N acts on a block sliding on a horizontal surface. The force of friction between the
1 answer:
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Answer:N=322.53 rpm
Explanation:
Given
Linear velocity (v)=1.25 m/s
Position from center is 3.7 cm
we know
and
N=322.53 rpm
r
Answer:
a) w = - , 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 = - (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) ( ) ²
W = - ½ (m_woman² R² + m_woman³ R³ / I²) v²
W = - ½ m_woman R² (1 + m_woman R / I²) v²