Respuesta :
Answer:
1.57 g
Explanation:
Angular momentum is conserved, so:
I₁ ω₁ = I₂ ω₂
Solving for ω₂:
ω₂ = ω₁ (I₁ / I₂)
Calculating each moment of inertia
I₁ = 5.10×10⁸ kg m² + 150 × 65.0 kg × (151 m)² = 7.32×10⁸ kg m²
I₂ = 5.10×10⁸ kg m² + 50 × 65.0 kg × (151 m)² = 5.84×10⁸ kg m²
Therefore:
ω₂ = ω₁ (7.32×10⁸ / 5.84×10⁸)
ω₂ = 1.25ω₁
Relating angular velocity to centripetal acceleration:
a = v² / r
a = ω² r
r is constant, so acceleration is proportional to the square of angular velocity.
a₂ / a₁ = ω₂² / ω₁²
a₂ / a₁ = (ω₂ / ω₁)²
Plugging in:
a₂ / 1g = (1.25)²
a₂ = 1.57g
The new apparent acceleration experienced at the rim is 1.57 g.
Following are the responses to the given question:
The wheel's and people's moments of inertia
[tex]\to I_1 = I +MR^2\\\\[/tex]
[tex]= 5.10\times 10^{8}\ kg \ m^2 +[ 150 \times 65.0 \ kg \times (151\ m)^2]\\\\= 5.10\times 10^{8}\ kg \ m^2 +[ 9750\ kg \times 22801 \ m^2]\\\\= 5.10\times 10^{8}\ kg \ m^2 +[ 222309750 \ kg \ m^2]\\\\=732309750 \ kg \ m^2\\\\[/tex]
Centripetal acceleration:
[tex]\to g=\omega^2 R\\\[/tex]
Therefore, the angular velocity:
[tex]\omega =\sqrt{\frac{g}{R}}\\\\ =\sqrt{\frac{9.8\ \frac{m}{s^2}}{151\ m}} =0.064\ \frac{rad}{s}\\\\[/tex]
The wheel's and people's moments of inertia:
[tex]\to I_2 =I+ \Sigma MR^2 \\\\[/tex]
[tex]=5.10\times 10^{8}\ kg m^2 + 50 \times 65.0\ kg \times (151\ m)^2 +100 \times 65.0 \ kg \times 0\\\\=5.10\times 10^{8}\ kg m^2 + 50 \times 65.0\ kg \times 22801 +100 \times 65.0 \ kg \times 0\\\\=5.10\times 10^{8}\ kg m^2 +74103250 + 0\\\\=584103250\\\\[/tex]
Centripetal acceleration:
[tex]a= \omega'^2 R[/tex]
Calculating the angular velocity:
[tex]\to \omega'^2 =\sqrt{\frac{a}{R}}=\sqrt{\frac{a}{151\ m}} \\\\[/tex]
As per the angular momentum:
[tex]\to I_1 \ \omega_1 = I_2 \omega_2[/tex]
[tex]\to 732309750 \times 0.064 = 584103250 \times \sqrt{\frac{a}{151\ m}}[/tex]
Calculating the acceleration
[tex]\to \sqrt{\frac{a}{151\ m}} =0.080 \\\\ \to a= 0.9664\ \frac{m}{s^2}[/tex]
Learn more:
brainly.com/question/13720857
