Answer:
230
Explanation:
[tex]\omega[/tex] = Rotational speed = 3600 rad/s
I = Moment of inertia = 6 kgm²
m = Mass of flywheel = 1500 kg
v = Velocity = 15 m/s
The kinetic energy of flywheel is given by
[tex]K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J[/tex]
Energy used in one acceleration
[tex]K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J[/tex]
Number of accelerations would be given by
[tex]n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4[/tex]
So the number of complete accelerations is 230
The rotational kinetic energy of a flywheel is 3.89 × 10⁷ Joules. The number of accelerations the car could make before the flywheel's energy is dissipated is 230.52
The rotational kinetic energy of a flywheel can be determined by using the formula:
[tex]\mathbf{K_i = \dfrac{1}{2} I \omega ^2}[/tex]
[tex]\mathbf{K_i = \dfrac{1}{2} (6.0 \ kg.m^2) \times (3600)^2}[/tex]
[tex]\mathbf{K_i = 3.89 \times 10^7 \ Joules}[/tex]
The kinetic energy of the system is:
[tex]\mathbf{K.E = \dfrac{1}{2}mv^2}[/tex]
[tex]\mathbf{K.E = \dfrac{1}{2}\times 1500 \times (15) ^2}[/tex]
K.E = 168750 Joules
The number of accelerations the car could make before the flywheel's energy is dissipated can be computed as:
[tex]\mathbf{n = \dfrac{3.89 \times 10^7 \ Joules }{168750 \ Joules}}[/tex]
n = 230.52
Learn more about rotational kinetic energy here:
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