The rotational or kinetic energy of the thin hoop with the given radius, mass and angular speed is 0.092J
Given the data in the question;
Rotational or kinetic energy; [tex]E_{rotational} = \ ?[/tex]
Rotational energy or angular kinetic energy is simply kinetic energy due to the rotation of a rigid body.
It is expressed as;
[tex]E_{rotational} = \frac{1}{2}Iw^2[/tex]
Where [tex]I[/tex] is the moment of inertia around the axis of rotation and [tex]w[/tex] is the angular speed or velocity.
For the moment of inertia around the axis of rotation.
[tex]I = \frac{1}{2}mr^2[/tex]
Hence
[tex]E_{rotational} = \frac{1}{2}(\frac{1}{2}mr^2)w^2 \\\\E_{rotational} = (\frac{1}{4}mr^2)w^2[/tex]
Now, we substitute our given values into the above equation to find the rotational or kinetic energy.
[tex]E_{rotational} = (\frac{1}{4}*3.0kg * (0.1m)^2) * (3.5rad/s)^2 \\\\E_{rotational} = 0.0075kgm^2 * 12.25rad/s^2\\\\E_{rotational} = 0.092kg.m^2/s^2\\\\E_{rotational} = 0.092J[/tex]
Therefore, the rotational or kinetic energy of the thin hoop with the given radius, mass and angular speed is 0.092J
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