Answer to Question #117072 in Physics for MICHAEL

Question #117072
A wheel of mass 6.0kg and radius of gyration 40cm is rotating at 300prm. Find it's moment of inertia and it's rotational KE
1
Expert's answer
2020-05-25T11:20:06-0400

By definition, the moment of inertia of the body with the radius of gyration Rg=40cmR_g = 40 cm and mass m=6kgm = 6kg will be:


I=mRg2=60.42=0.96kgm2I = mR_g^2 = 6\cdot 0.4^2 = 0.96 kg\cdot m^2


The rotational KE:


Erot=12Iω2E_{rot} = \dfrac12I\omega^2

where ω\omega is the angular velocity. If the frequency of rotation is ν=300rpm=5Hz\nu = 300 rpm = 5Hz, then the angular velocity will be:


ω=2πν=2π531.4rad/s\omega = 2\pi\nu = 2\pi\cdot 5 \approx 31.4 rad/s

Thus, the rotational KE will be:


Erot=120.9631.42=473.7JE_{rot} = \dfrac12\cdot 0.96\cdot 31.4^2 = 473.7 J

Answer. I = 0.96 kg*m^2, Erot = 473.7 J.


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