Let's approach this problem step-by-step to find the new angular speed of the merry-go-round after the child jumps onto it.
1. Identify and List Given Data:
- Radius of the merry-go-round, [tex]\( r = 2.0 \)[/tex] meters.
- Moment of inertia of the merry-go-round without the child, [tex]\( I_{\text{merry-go-round}} = 250 \)[/tex] kgm[tex]\(^2\)[/tex].
- Initial angular velocity, [tex]\( \omega_{\text{initial}} = 10 \)[/tex] rev/min.
- Mass of the child, [tex]\( m_{\text{child}} = 25 \)[/tex] kg.
2. Convert the Initial Angular Velocity from rev/min to rad/s:
[tex]\[\omega_{\text{initial}} = 10 \, \text{rev/min} \][/tex]
To convert rev/min to rad/s:
[tex]\[ 1 \, \text{rev} = 2\pi \, \text{rad} \][/tex]
[tex]\[ 1 \, \text{min} = 60 \, \text{s} \][/tex]
Thus,
[tex]\[ \omega_{\text{initial}} = 10 \left( \frac{2\pi}{60} \right) \, \text{rad/s} = \frac{10 \times 2\pi}{60} \, \text{rad/s} = 1.0472 \, \text{rad/s} \][/tex]
3. Calculate the Moment of Inertia of the Child as a Point Mass at the Edge:
[tex]\[ I_{\text{child}} = m_{\text{child}} \times r^2 = 25 \times (2.0)^2 = 25 \times 4 = 100 \, \text{kgm}^2 \][/tex]
4. Determine the Total Moment of Inertia (Merry-Go-Round + Child):
[tex]\[ I_{\text{total}} = I_{\text{merry-go-round}} + I_{\text{child}} \][/tex]
[tex]\[ I_{\text{total}} = 250 + 100 = 350 \, \text{kg*m}^2 \][/tex]
5. Apply the Law of Conservation of Angular Momentum:
The initial and final angular momentum of the system must be equal, because no external torques are acting on it.
[tex]\[ I_{\text{initial}} \times \omega_{\text{initial}} = I_{\text{final}} \times \omega_{\text{final}} \][/tex]
Where:
[tex]\[ I_{\text{initial}} = I_{\text{merry-go-round}} \][/tex]
[tex]\[ \omega_{\text{final}} = \][/tex]
Thus:
[tex]\[ 250 \times 1.0472 = 350 \times \omega_{\text{final}} \][/tex]
6. Solve for the Final Angular Velocity in rad/s:
[tex]\[ \omega_{\text{final}} = \frac{250 \times 1.0472}{350} = 0.747998 \, \text{rad/s} \][/tex]
7. Convert the Final Angular Velocity from rad/s to rev/min:
To convert rad/s to rev/min:
[tex]\[ \omega_{\text{final}} = 0.747998 \left( \frac{60}{2\pi} \right) \, \text{rev/min} = 0.747998 \times 9.5493 = 7.14285 \, \text{rev/min} \][/tex]
Therefore, the new angular speed of the merry-go-round after the child jumps onto it is approximately 7.1429 rpm.
