Answer:
1) L = 299.88 kg-m²/s
2) L = 613.2 kg-m²/s
3) L = 499.758 kg-m²/s
4) ω₁ = 0.769 rad/s
5) Fc = 70.3686 N
6) v = 1.2535 m/s
7) ω₀ = 1.53 rad/s
Explanation:
Given
R = 1.63 m
I₀ = 196 kg-m²
ω₀ = 1.53 rad/s
m = 73 kg
v = 4.2 m/s
1) What is the magnitude of the initial angular momentum of the merry-go-round?
We use the equation
L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s
2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?
We use the equation
L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s
3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?
We use the equation
L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s
4) What is the angular speed of the merry-go-round after the person jumps on?
We can apply The Principle of Conservation of Angular Momentum
L in = L fin
⇒ I₀*ω₀ = I₁*ω₁
where
I₁ = I₀ + m*R²
⇒ I₀*ω₀ = (I₀ + m*R²)*ω₁
Now, we can get ω₁
⇒ ω₁ = I₀*ω₀ / (I₀ + m*R²)
⇒ ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)
⇒ ω₁ = 0.769 rad/s
5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?
We have to get the centripetal force as follows
Fc = m*ω²*R
⇒ Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N
6) Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.
What is the linear velocity of the person right as they leave the merry-go-round?
we can use the equation
v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s
7) What is the angular speed of the merry-go-round after the person lets go?
ω₀ = 1.53 rad/s
It comes back to its initial angular speed
