Respuesta :
Answer:
Part 1)
[tex]\omega_2 = 1.74 rad/s[/tex]
Part 2)
initial kinetic energy of the system
[tex]K_1 = 1600 J[/tex]
Final kinetic energy is given as
[tex]K_2 = 696.35 J[/tex]
Energy is decreased due to friction between parachutist and the disc
Explanation:
Part a)
As we know that the system of turn table and parachutist is isolated and there is no external force on this system
So here we can use angular momentum conservation for this
[tex]L_i = L_f[/tex]
[tex]I_1\omega_1 = (I_1 + I_2)\omega_2[/tex]
[tex]\frac{1}{2}Mr^2 \omega_1 = (\frac{1}{2}Mr^2 + mr^2)\omega_2[/tex]
[tex]\frac{1}{2}(100)(2^2) 4 = (\frac{1}{2}(100)(2^2) + 65(2^2))\omega_2[/tex]
[tex]800 = 460\times \omega_2[/tex]
[tex]\omega_2 = 1.74 rad/s[/tex]
Part 2)
initial kinetic energy of the system
[tex]K_1 = \frac{1}{2}I\omega_1^2[/tex]
[tex]K_1 = \frac{1}{2}(\frac{1}{2}Mr^2)(4^2)[/tex]
[tex]K_1 = \frac{1}{2}(\frac{1}{2}(100)(2^2))(16)[/tex]
[tex]K_1 = 1600 J[/tex]
Final kinetic energy is given as
[tex]K_2 = \frac{1}{2}(\frac{1}{2}Mr^2 + mr^2)\omega_2^2[/tex]
[tex]K_2 = \frac{1}{2}(\frac{1}{2}(100)(2^2) + 65(2^2))(1.74^2)[/tex]
[tex]K_2 = 696.35 J[/tex]
Energy is decreased due to friction between parachutist and the disc
The final angular speed of the turntable after the parachutist lands is equal to 1.74 rad/s.
Given the following data:
Radius = 2.00 m.
Total mass = 100 kg.
Initial angular velocity = 4.00 rad/s.
Mass of parachutist = 65.0 kg.
How to calculate the final angular speed.
In order to determine the final angular speed of the turntable after the parachutist lands, we would apply the law of conservation of momentum:
[tex]L_i=L_f\\\\I_i \omega_i=I_f \omega_f\\\\I_i \omega_i=(I_i+I_f) \omega_f\\\\(\frac{1}{2} mr^2)\omega_i=(\frac{1}{2} mr^2+mr^2) \omega_f\\\\(\frac{1}{2} \times 100 \times 2^2)4=(\frac{1}{2} \times 100 \times 2^2+ 65.0 \times 2^2) \omega_f\\\\800=460\omega_f\\\\\omega_f=\frac{800}{460} \\\\\omega_f=1.74 \;rad/s[/tex]
How to calculate the kinetic energy.
For the kinetic energy of the system before the parachutist lands, we have:
[tex]K.E =\frac{1}{2} I_i \omega_i^2\\\\K.E =\frac{1}{2} (\frac{1}{2} mr^2) \omega_i^2\\\\K.E = \frac{1}{2} \times 200 \times 4^2 \\\\[/tex]
K.E = 1600 Joules.
For the kinetic energy of the system after the parachutist lands, we have:
[tex]K.E =\frac{1}{2} I_f \omega_f^2\\\\K.E =\frac{1}{2} (\frac{1}{2} mr^2+mr^2) \omega_i^2\\\\K.E = \frac{1}{2} \times (200+(65 \times 2^2)) \times 1.74^2 \\\\[/tex]
K.E = 696.35 Joules.
Read more on moment of inertia here: https://brainly.com/question/3406242
