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You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 9.3 kg * m^2. When you pull the weights in toward your body, the moment of inertia decreases to 5.1 kg * m^2.
(a) Whatis the resulting angular speed of the platform?
(b) What is thechange in kinetic energy of the system?
(c) Where did thisincrease in energy come from?

Answer :

boffeemadrid

Answer:

20.62361 rad/s

489.81804 J

Explanation:

[tex]I_i[/tex] = Initial moment of inertia = 9.3 kgm²

[tex]I_f[/tex] = Final moment of inertia = 5.1 kgm²

[tex]\omega_i[/tex] = Initial angular speed = 1.8 rev/s

[tex]\omega_f[/tex] = Final angular speed

As the angular momentum of the system is conserved

[tex]I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s[/tex]

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

[tex]\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J[/tex]

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

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