annie413
contestada

You are helping to design a new fire escape for your third-floor apartment. You step out the window onto a crossbar
that is attached to a rope, which in turn is wound around a massive solid cylindrical flywheel that can turn on its axis.
As you fall at increasing speed, the rope causes the flywheel to turn at increasing angular velocity. Your task is to
choose a flywheel that will enable you to move at a safe 2.0 m/s when you reach 10 m below your starting position.
Your mass is 80 kg. Determine the rotational inertia and mass of the uniform 0.80 m radius flywheel you need to
buy.

Respuesta :

oladayoademosu

The rotational inertia of the flywheel is 2457.6 kgm² and the mass of the flywheel is 7680 kg

Conservation of Energy

The total mechanical energy of the cylindrical flywheel and man is conserved.

So, mgh + 1/2Iω² + 1/2mv² = mgh' + 1/2Iω'² + 1/2mv'² where

  • m = mass of man = 80 kg,
  • g = acceleration due to gravity = 9.8 m/s²,
  • h = initial height of crossbar = 0 m,
  • I = rotational inertia of cylindrical flywheel,
  • ω = initial angular speed of flywheel = 0 rad/s (since it starts from rest),
  • v = initial velocity of flywheel and man = 0 m/s (since it starts fom rest),
  • h' = final height of crossbar = -10 m(negative since it is below our starting point),
  • ω' = final angular velocity of flywheel = v'/R where
  • v' = final velocity of flywheel = 2.0 m/s and
  • R = radius of flywheel = 0.80 m. So, ω' = v'/R = 2.0 m/s/0.80 m = 2.5 rad/s

So, with h = 0 m, ω = 0 and v = 0. Substituting these into the equation, we have

mgh + 1/2Iω² + 1/2mv² = mgh' + 1/2Iω'² + 1/2mv'²

mg(0) + 1/2I(0)² + 1/2m(0)² = mgh' + 1/2Iω'² + 1/2mv'²

0 + 0 + 0 = mgh' + 1/2Iω'² + 1/2mv'²

0 = mgh' + 1/2Iω'² + 1/2mv'²

Rotational inertia of flywheel

The rotational inertia of the flywheel is 2457.6 kgm²

Making I subject of the formula, we have

I = -m(v'² + 2gh')/ω'²

So, substituting the values of the variables into the equation, we have

I = -m(v'² + 2gh')/ω'²

I = -80 kg[(2.0 m/s)² + 2 × 9.8 m/s² × -10 m)/(2.5 rad/s)²

I = -80 kg[4.0 m²/s² - 196 m²/s²)/6.25 rad²/s²

I = -80 kg[-192 m²/s²)/6.25 rad²/s²

I = 15360 kgm²/s²/6.25 rad²/s²

I = 2457.6 kgm²

The rotational inertia of the flywheel is 2457.6 kgm²

Mass of cylindrical flywheel

We know that the rotational inertia of a cylinder I = 1/2MR² where

  • M = mass of cylinder and
  • R = radius of cylinder

So, making M subject of the formula, we have

M = 2I/R²

Substituting the values of the variables into the equation, we have

M = 2I/R²

M = 2 × 2457.6 kgm²/(0.80 m)²

M = 4915.2 kgm²/0.64 m²

M = 7680 kg

So, the mass of the flywheel is 7680 kg

Learn more about rotational inertia here:

https://brainly.com/question/14585807

Otras preguntas