Respuesta :
Answer:
The kinetic energy of the flywheel is 5000kJ.
Explanation:
We have that the moment of inertia of a solid and uniform cylinder is equal to:
[tex]I=\frac{1}{2} mR^{2}[/tex]
Where m is the mass of the cylinder and R its radius.
Next, the rotational kinetic energy of an object rotating about an axis is given by:
[tex]K_R=\frac{1}{2} I\omega^{2}[/tex]
Where ω is the angular velocity. From these two equations, we can derive that:
[tex]K_R=\frac{1}{4} mR^{2} \omega^{2}[/tex]
Plugging the given values in this equation, we obtain:
[tex]K_R=\frac{1}{4}(500kg)(1.0m)^{2}(200rad/s)^{2}=5*10^{6}J=5000kJ[/tex]
In words, the kinetic energy of the flywheel is 5000kJ.
Answer:
Explanation:
Radius, r = 1 m
Mass, m = 500 kg
w = 200 rad/s
Moment of inertia for a dolid symmetric cylinder, I = 1/2 × mr^2
KE = 1/2 × I × w^2
Therefore,
KE = 1/4 × m × w^2 × r ^2
= 1/4 × 500 × 200^2 × 1^2
= 5.0 × 10^6 J.
