Calculate the rotational inertia of a wheel that has a kinetic energy of 24,400 J when rotating at 677 rev/min.
I know that I = mr^2 and that
T = Ia; also I calculated 677rpm to be 70.9 radians/sec. But I don't understand how to integrate KE into this.
Iw^2=KE
w=677*2PI/60 rad/sec\
solve for I.
rotational inertia then= I w
To calculate the rotational inertia (I) of a wheel, we need to consider its kinetic energy (KE) and angular velocity (ω). The formula for rotational kinetic energy is KE = 1/2Iω^2, where I is the rotational inertia and ω is the angular velocity.
In this case, we are given the kinetic energy (KE) of the wheel, which is 24,400 J, and the angular velocity (ω), which is 70.9 radians/sec (obtained by converting 677 revolutions/minute).
To integrate the kinetic energy into the equation, we can rearrange the equation for KE and solve for I:
KE = 1/2Iω^2
Solving for I:
I = 2KE/ω^2
Now, we can substitute the given values into the equation:
I = 2 * 24,400 J / (70.9 rad/sec)^2
Calculating the value:
I ≈ 113 kg·m²
Therefore, the rotational inertia of the wheel is approximately 113 kg·m².