Final answer:
To find the new angular velocity when the student is 1.73m from the center, we apply the conservation of angular momentum in a frictionless system. By calculating the moments of inertia at the initial and final positions and using the formula for conservation of angular momentum, we can solve for the new angular velocity.
Step-by-step explanation:
The question involves the concept of conservation of angular momentum which is relevant in Physics, particularly in mechanics and rotational motion. In a frictionless system, angular momentum is conserved. This applies to our scenario with a horizontal circular platform and a person moving towards the center. As the person moves closer to the axis of rotation, their moment of inertia decreases, which must result in an increase in angular velocity to conserve angular momentum.
We can calculate the new angular velocity, ω, when the student is 1.73m from the center of the platform, assuming the system is closed and there are no external torques acting on it. The conservation of angular momentum is given by:
L_initial = L_final
(I_platform + I_student_initial) ω_initial = (I_platform + I_student_final) ω_final
Use I = m*r^2 for the moments of inertia of the student and I = 1/2*m*r^2 for the platform (which is a solid cylinder).
Solving for the new angular velocity gives us:
ω_final = ω_initial * (I_platform + I_student_initial) / (I_platform + I_student_final)
Plugging in the values:
ω_final = 2.5 rad/s * [(1/2*112.1kg*(3.01m)^2 + 66.3kg*(3.01m)^2) / (1/2*112.1kg*(3.01m)^2 + 66.3kg*(1.73m)^2)]
Through this formula, we can determine the change in angular velocity as the student walks towards the center of the platform.