Question

The design of a 60.0 cm industrial turntable requires that it has a kinetic energy of 0.250 j when turning at 45.0 rpm. What must be the moment of inertia of the turntable about the rotation axis

Answer 1

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;

$E_{rot} = \frac{1}{2} I \omega ^2$

where;

I is the moment of inertia about the axis of rotation

ω is the angular speed in rad/s

$\omega = 45 \frac{rev}{\min} \times \frac{2 \pi \ rad}{1 \ rev} \times \frac{1 \ \min}{60 \ s} \\\\\omega = 4.712 \ rad/s$

$E = \frac{1}{2} I \omega ^2\\\\I = \frac{2E}{\omega ^2} \\\\I = \frac{2 \ \times \ 0.25}{(4.712)^2} \\\\I = 0.0225 \ kg.m^2$

Therefore, the moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²