Answer:
The moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²
Explanation:
Given;
radius of the turnable, r = 60 cm = 0.6 m
rotational kinetic energy, E = 0.25 J
angular speed of the turnable, ω = 45 rpm
The rotational kinetic energy is given as;
[tex]E_{rot} = \frac{1}{2} I \omega ^2[/tex]
where;
I is the moment of inertia about the axis of rotation
ω is the angular speed in rad/s
[tex]\omega = 45 \frac{rev}{\min} \times \frac{2 \pi \ rad}{1 \ rev} \times \frac{1 \ \min}{60 \ s} \\\\\omega = 4.712 \ rad/s[/tex]
[tex]E = \frac{1}{2} I \omega ^2\\\\I = \frac{2E}{\omega ^2} \\\\I = \frac{2 \ \times \ 0.25}{(4.712)^2} \\\\I = 0.0225 \ kg.m^2[/tex]
Therefore, the moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²
