Please do the algebra to show how the answer is 4.1 rad/s **Physics Problem: Rotational Motion**---**Problem Statement:**A horizontal circular disk rotates freely about a vertical axis through its center. It has a mass of 100 kg and a radius of 2.0 m. A person of mass 60 kg stands at the edge of the disk while the disk is rotating at 2.0 rad/s. The person slowly walks from the edge toward the center of the disk. What is the angular speed of the system when the person is 0.50 m from the center?---**Given:**- Mass of the disk \( M = 100 \, \text{kg} \)- Mass of the person \( m = 60 \, \text{kg} \)- Radius of the disk \( R = 2.0 \, \text{m} \)- Initial angular velocity \( \omega_0 = 2.0 \, \text{rad/s} \)- Final distance of the person from the center \( r = 0.50 \, \text{m} \)---**Equations:**1. **Conservation of Angular Momentum:** \[ L = I \omega \]2. **Moment of Inertia for Disk and Person:** \[ I = I_{\text{disk}} + I_{\text{person}} \]3. **Initial Angular Momentum:** \[ I_{\text{disk}} = \frac{1}{2} MR^2 \] \[ I_{\text{person}} = mr^2 \]4. **Final Angular Momentum:** \[ I_o \omega_o = I_f \omega_f \]---**Solution Steps:**- **Initial Moment of Inertia:** \[ I_o = \frac{1}{2} MR^2 + mR^2 \]- **Equation Substitution:** \[ \left(\frac{1}{2} \cdot 100 \cdot 2^2 + 60 \cdot 2^2\right) \cdot 2 = \left(\frac{1}{2} \cdot 100 \cdot 2^2 + 60 \cdot (0.5)^2\right) \cdot \omega_f \]- **Inserting Numbers and Solving for Final Angular Speed \( \omega_f \):

Mass of the circular disc (M)=100 kg Radius of the circular disc (R)=2 m Mass of the person (m)=60 kg Initial angular speed of the disc (ω1)=2 rad/s Final position of the person from the center of the disc (r)=0.5 mSince the person walks in towards the center of the disc, the moment of inertia of the system changes. Initial moment of inertia of the system will be given by the equation; I1=12MR2+mR2 Here, we consider the person as a particle on the rim of the disc Similarly, the final moment of inertia of the system (when the person walks in) will be given by the equation; I2=12MR2+mr2According to the law of conservation of angular momentum, the total initial angular momentum will be equal to the total final angular momentum. L1=L2 ∴I1ω1=I2ω2 ∴12MR2+mR2ω1=12MR2+mr2ω2 Thus, we can write for the final angular speed as; ω2=12MR2+mR2ω112MR2+mr2 Substituting the values; ω2=12(100)(2)2+(60)(2)2×212(100)(2)2+(60)(0.5)2 ∴ω2=4.1 rad/sThus, the angular momentum of the disc and person system when the person is near the center of the disc is 4.1 rad/s.