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Some special vehicles have spinning disks (flywheels) to store energy while they roll downhill. They use that stored energy to lift themselves uphill later on. Their flywheels have relatively small rotational masses but spin at enormous angular speeds. How would a flywheel’s kinetic energy change if its rotational mass were 7 times larger but its angular speed were 7 times smaller?

Respuesta :

Answer:

Smaller by 7 times

Explanation:

Rotational mass is called moment of inertia

So, initial moment of inertia, I1 = I

initial angular velocity, ω1 = ω

Final moment of inertia, I2 = 7I

Final angular velocity, ω2 = ω/7

The kinetic energy in rotational motion is given by

[tex]K = \frac{1}{2}I\omega ^{2}[/tex]

So, initial kinetic energy of rotation

[tex]K_{1} = \frac{1}{2}I_{1}\omega_{1} ^{2}= \frac{1}{2}I\omega ^{2}[/tex]

So, final kinetic energy of rotation

[tex]K_{2} = \frac{1}{2}I_{2}\omega_{2} ^{2}= \frac{1}{2}7I \left (\frac{\omega}{7}  \right ) ^{2}[/tex]

[tex]K_{2} = \frac{K_{1}}{7}[/tex]

Thus, teh kinetic energy becomes smaller by 7 times.

oladayoademosu

The flywheel's kinetic energy would be one-seventh of its initial value.

Rotational kinetic energy

The rotational kinetic energy of the flywheel is given by

K = 1/2Iω² where

  • I = rotational mass and
  • ω = angular speed

Let

  • K₀ = initial rotational kinetic energy,
  • I₀ = initial rotational mass and
  • ω₀ = initial angular speed

When its rotational mass were 7 times larger but its angular speed were 7 times smaller, we have that

  • I₁ = final rotational mass = 7I₀ and
  • ω₁  = final angular speed = ω₀/7

Final rotational kinetic energy

So, the final rotational kinetic energy, K₁ = 1/2I₁ω₁²

= 1/2 × 7I₀(ω₀/7)²

= 1/2 × 7I₀ω₀²/49

= 1/2I₀ω₀²/7

= K₀/7

So, the flywheel's kinetic energy would be one-seventh of its initial value.

Learn more about rotational kinetic energy here:

https://brainly.com/question/15076457

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