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
![F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha](https://tex.z-dn.net/?f=F_T%20%5Ccdot%20r%20%3D%20I%20%5Ccdot%20%5Calpha%20%20%3D%20m%20%5Ccdot%20r%5E2%20%20%5Ccdot%20%5Calpha)
Where;
= 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;
![I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}](https://tex.z-dn.net/?f=I_m%20%5Ccdot%20%5Calpha_m%20%20%3D%20I_m%20%5Ccdot%20%5Cdfrac%7Bv_m%7D%7Br%20%5Ccdot%20t%7D)
= The rotational inertia of the merry-go-round
= The angular acceleration of the merry-go-round
= The linear velocity of the merry-go-round
t = The time of motion
For the boy, we have;
![I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}](https://tex.z-dn.net/?f=I_b%20%5Ccdot%20%5Calpha_b%20%20%3D%20m_b%20%5Ccdot%20r%5E2%20%20%5Ccdot%20%5Cdfrac%7Bv_b%7D%7Br%20%5Ccdot%20t%7D)
Where;
= The rotational inertia of the boy
= The angular acceleration of the boy
= The linear velocity of the boy
t = The time of motion
When the boy jumps on the merry-go-round, we have;
![I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}](https://tex.z-dn.net/?f=I_m%20%5Ccdot%20%5Cdfrac%7Bv_m%7D%7Br%20%5Ccdot%20t%7D%20%3D%20m_b%20%5Ccdot%20r%5E2%20%20%5Ccdot%20%5Cdfrac%7Bv_b%7D%7Br%20%5Ccdot%20t%7D)
Which gives;
![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}](https://tex.z-dn.net/?f=v_m%20%3D%20%5Cdfrac%7Bm_b%20%5Ccdot%20r%5E2%20%20%5Ccdot%20%5Cdfrac%7Bv_b%7D%7Br%20%5Ccdot%20t%7D%20%5Ccdot%20r%20%5Ccdot%20t%7D%7BI_m%7D%20%3D%20%5Cdfrac%7Bm_b%20%5Ccdot%20r%5E2%20%20%5Ccdot%20v_b%7D%7BI_m%7D)
From which we have;
![v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5](https://tex.z-dn.net/?f=v_m%20%3D%20%20%5Cdfrac%7B20%20%5Ctimes%203%5E2%20%20%5Ctimes%205%7D%7B600%7D%20%3D%20%201.5)
The velocity of the merry-go-round,
, after the boy hops on the merry-go-round = 1.5 m/s.