a person 0f mass m = 60 kg walks along the radius from the periphery to the center of a circular platform of moment of inertia= 200kgm square and radius r = 3m. the disk is a rotating frictionless with angular velocity= 0.3 rad\s.find the angular velocity of the system when the person is 50 cm from the center. find the change of mechanical energy during the displacement. assuming that the walk takes 2.50s, find the average torque exerted by the person on the platform.
To find the angular velocity of the system when the person is 50 cm from the center, we can use principles of conservation of angular momentum.
The angular momentum of the system is given by L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Initially, the person is at the periphery, so the angular momentum of the system is L_initial = (m + M)r_initial^2ω_initial, where M is the moment of inertia of the circular platform and r_initial is the initial radius (periphery). In this case, the person's mass is m = 60 kg, the moment of inertia of the platform is M = 200 kgm^2, the initial radius r_initial = 3 m, and the initial angular velocity ω_initial = 0.3 rad/s.
Finally, when the person is 50 cm (or 0.5 m) from the center, the final radius r_final = 0.5 m. We need to find the final angular velocity ω_final.
To conserve angular momentum, we equate L_initial = L_final:
(m + M)r_initial^2ω_initial = (m + M)r_final^2ω_final
Now, we can plug in the values:
(60 kg + 200 kgm^2)(3 m)^2(0.3 rad/s) = (60 kg + 200 kgm^2)(0.5 m)^2(ω_final)
Simplifying the equation, we can solve for ω_final as follows:
(260 kgm^2)(9 m^2/s)(0.3 rad/s) = (260 kgm^2)(0.25 m^2)(ω_final)
(2340 kgm^4/s) = (65 kgm^4)(ω_final)
ω_final = (2340 kgm^4/s) / (65 kgm^4)
ω_final ≈ 36 rad/s
Therefore, the angular velocity of the system when the person is 50 cm from the center is approximately 36 rad/s.
To find the change in mechanical energy during the displacement, we need to calculate the initial and final mechanical energies.
The initial mechanical energy, E_initial, is given by E_initial = (1/2)I_initialω_initial^2, where I_initial is the moment of inertia of the system initially and ω_initial is the initial angular velocity.
The final mechanical energy, E_final, is given by E_final = (1/2)I_finalω_final^2, where I_final is the moment of inertia of the system finally and ω_final is the final angular velocity.
In this case, the moment of inertia initially, I_initial, is (m + M)r_initial^2, and the moment of inertia finally, I_final, is (m + M)r_final^2.
Plugging in the values, we can calculate the initial and final mechanical energies:
E_initial = (1/2)(60 kg + 200 kgm^2)(3 m)^2(0.3 rad/s)^2
E_final = (1/2)(60 kg + 200 kgm^2)(0.5 m)^2(36 rad/s)^2
Simplifying these equations will give you the initial and final mechanical energies.
To find the average torque exerted by the person on the platform, we can use the formula:
Torque = Change in angular momentum / Change in time
The change in angular momentum is given by ΔL = L_final - L_initial, where L_final and L_initial are the final and initial angular momenta, respectively.
The change in time, Δt, is the time taken for the walk, which is given as 2.50 s.
The average torque can be calculated as:
Average Torque = ΔL / Δt
Substituting the values, we can calculate the average torque exerted by the person on the platform.