Respuesta :
Answer:
The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s
Explanation:
The rotational inertia of the merry-go-round = 600 kg·m²
The radius of the merry-go-round = 3.0 m
The mass of the boy = 20 kg
The speed with which the boy approaches the merry-go-round = 5.0 m/s
[tex]F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha[/tex]
Where;
[tex]F_T[/tex] = The tangential force
I = The rotational inertia
m = The mass
α = The angular acceleration
r = The radius of the merry-go-round
For the merry go round, we have;
[tex]I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}[/tex]
[tex]I_m[/tex] = The rotational inertia of the merry-go-round
[tex]\alpha _m[/tex] = The angular acceleration of the merry-go-round
[tex]v _m[/tex] = The linear velocity of the merry-go-round
t = The time of motion
For the boy, we have;
[tex]I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}[/tex]
Where;
[tex]I_b[/tex] = The rotational inertia of the boy
[tex]\alpha _b[/tex] = The angular acceleration of the boy
[tex]v _b[/tex] = The linear velocity of the boy
t = The time of motion
When the boy jumps on the merry-go-round, we have;
[tex]I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}[/tex]
Which gives;
[tex]v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}[/tex]
From which we have;
[tex]v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5[/tex]
The velocity of the merry-go-round, [tex]v_m[/tex], after the boy hops on the merry-go-round = 1.5 m/s.
