Answer:
See explanation
Explanation:
We have a mass [tex]m[/tex] revolving around an axis with an angular speed [tex]\omega[/tex], the distance from the axis is [tex]r[/tex]. We are given:
[tex]\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg][/tex]
and also the formula which states that the kinetic rotational energy of a body is:
[tex]K =\frac{1}{2}I\omega^2[/tex].
Now we use the kinetic energy formula
[tex]K =\frac{1}{2}mv^2[/tex]
where [tex]v[/tex] is the tangential velocity of the particle. Tangential velocity is related to angular velocity by:
[tex]v=\omega r[/tex]
After replacing in the previous equation we get:
[tex]K =\frac{1}{2}m(\omega r)^2[/tex]
now we have the following:
[tex]K =\frac{1}{2}m(\omega r)^2 =\frac{1}{2}Iw^2[/tex]
therefore:
[tex]mr^2=I[/tex]
then the moment of inertia will be:
[tex]I = 13*(0.5)^2=3.25 [Kg*m^2][/tex]
