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You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius 27.0 cm . Starting from rest at t = 0, the flywheel rotates with constant angular acceleration 3.00 rad/s2 about an axis perpendicular to the flywheel at its center If the flywheel has a density (mass per unit volume) of 8600 kg/m3, what thickness must it have to store 800 J of kinetic energy at t = 8.00 s?

Respuesta :

Answer:

[tex]h\approx1.0443\,cm[/tex]

Explanation:

Given:

  • Radius of flywheel, [tex]r= 27 \, cm[/tex]
  • initial time,[tex]t_i=0\,s[/tex]
  • Angular acceleration of flywheel, [tex]\alpha=3\,rad.s^{-1}[/tex]
  • Density of flywheel, [tex]\rho=8600\,kg.m^{-3}[/tex]
  • kinetic energy after final time,[tex]KE=800\, J[/tex]
  • Final time, [tex]t_f=8\,s[/tex]

We know for rotational kinetic energy:

[tex]KE=\frac{1}{2} I.\omega^2[/tex].................................(1)

where:

I = mass moment of inertia for the given mass  geometry

[tex]\omega[/tex]= angular velocity in radians per second,

Here,

[tex]I= \frac{1}{2} M.r^2[/tex]

∵[tex]mass=density\times volume[/tex]

[tex]I= \frac{1}{2} \times (8600\times \pi\times 0.27^2\times h)\times 0.27[/tex]

[tex]I=265.8947\times h[/tex]

Now,

[tex]\omega= \alpha\times t_f[/tex]

[tex]\omega=3\times8[/tex]

[tex]\omega=24 \,rad.s^{-1}[/tex]

∴Using eq. (1)

[tex]800=\frac{1}{2} \times (265.8947\times h)\times 24^2[/tex]

[tex]h\approx1.0443\,cm[/tex]

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