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A disk with moment of inertia 9.15 × 10−3 kg∙m2 initially rotates about its center at angular velocity 5.32 rad/s. A non-rotating ring with moment of inertia 4.86 × 10−3 kg∙m2 right above the disk’s center is suddenly dropped onto the disk. Finally, the two objects rotate at the same angular velocity ???????? about the same axis. There is no external torque acting on the system during the collision. Please compute the system’s quantities below. 1. Initial angular momentum ???????? 3. Final angular velocity ???????? 2. Initial rotational kinetic energy K???? 4. Final rotational kinetic energy K????

Respuesta :

Answer:

1) Li = 0048678 kg -m^2/s

2) 0.129483 joule

3) 3.4745 Rad/s

4) 0.08456 joule

Explanation:

1) INITIAL ANGULAR MOMENTUM

we know that

[tex]Li =Ii \omega _i[/tex]

 [tex] = 9.15\times 10^{-3} \times 5.32 = 0.048678 kg-m^2/s[/tex]

2) INITIAL ROTATIONAL KINETIC ENERGY

[tex]KE_i = 1/2 I_i \omega_i^2[/tex]

[tex]KE_i = 1/2 9.15\times 10^{-3} \times 5.32^2 = 0.129483 joule[/tex]

3) FINAL ANGULAR VELOCITY

from conservation of angular momentum

Li =Lf

[tex]0.048678 = (I_{disk} + I_{ring}) w_f[/tex]

[tex]w_f = \frac{0.048678}{9.15\times 10^{-3} + 4.86\times 10^{-3}}[/tex]

      = 3.4745 Rad/s

4) FINAL ROTATIONAL KINETIC ENERGY[tex] KE_f[/tex]

[tex]KE_f = 1/2 (I_{disk} + I_{ring}} w_f^2[/tex]

[tex]= 1/2 (9.15\times 10^{-3} + 4.86\times 10^{-3}) 3.4745^2 = 0.08456 J[/tex]

onyebuchinnaji

(1) The initial angular momentum of the disk is 0.0487 kgm²/s.

(2) The final angular veocity of the system is 3.48 rad/s.

(3) The initial rotational kinetic energy is 0.13 J.

(4) The final kinetic energy of the system is 0.085 J.

Initial angular momentum

The initial angular momentum of the disk is calculated as follows;

Li = I₁ω₂

Li = (9.15 x 10⁻³ x 5.32)

Li = 0.0487 kgm²/s

Final angular velocity

The final angular veocity of the system is determined by applying the principle of conservation of angular momentum.

[tex]L_i = L_f\\\\I_i \omega_i = I_f\omega_f\\\\\omega _f = \frac{I_i \omega_i }{I_f} \\\\\omega _f = \frac{0.0487}{(9.15 \times 10^{-3} \ + \ 4.86\times 10^{-3})} \\\\\omega_ f = 3.48 \ rad/s[/tex]

Initial rotational kinetic energy

The initial rotational kinetic energy is calculated as follows;

[tex]K.E = \frac{1}{2} I\omega ^2\\\\K.E = \frac{1}{2} \times (9.15\times 10^{-3})\times (5.32)^2\\\\K.E = 0.13 \ J[/tex]

Final rotational kinetic energy

The final kinetic energy of the system is calculated as follows;

[tex]K.E_f = \frac{1}{2} I_f \omega _f^2\\\\K.E_f = \frac{1}{2} \times (9.15 \times 10^{-3} \ + \ 4.86 \times 10^{-3} ) \times (3.48)^2\\\\K.E_f = 0.085 \ J[/tex]

Learn more about rotational kinetic energy here: https://brainly.com/question/25803184

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