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Sagot :
Answer:
The speed of the student just before she lands, v₂ is approximately 8.225 m/s
Explanation:
The given parameters are;
The mass of the physic student, m = 43.0 kg
The height at which the student is standing, h = 12.0 m
The radius of the wheel, r = 0.300 m
The moment of inertia of the wheel, I = 9.60 kg·m²
The initial potential energy of the female student, P.E.₁ = m·g·h₁
Where;
m = 43.0 kg
g = The acceleration due to gravity ≈ 9.81 m/s²
h = 12.0 h
∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J
The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;
[tex]K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2[/tex]
The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0
∴ K₁ = 0 + 0 = 0
The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0
Where;
h₂ = 0 m
The final kinetic energy, K₂, of the wheel and student is give as follows;
[tex]K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2[/tex]
Where;
v₂ = The speed of the student just before she lands
ω₂ = The angular velocity of the wheel just before she lands
By the conservation of energy, we have;
P.E.₁ + K₁ = P.E.₂ + K₂
∴ m·g·h₁ + [tex]\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2[/tex] = m·g·h₂ + [tex]\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2[/tex]
Where;
ω₂ = v₂/r
∴ 5061.96 J + 0 = 0 + [tex]\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2[/tex]
5,061.96 J = 21.5 kg × v₂² + 53.[tex]\overline 3[/tex] kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²
v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²
v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s
The speed of the student just before she lands, v₂ ≈ 8.225 m/s.
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