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A 2.00 kg mass is located at (4.00 m, 0.00 m, 0.00 m) and a 4.00 kg mass is located at (0.00 m, 3.00 m, 0.00 m). The rotational inertia of this system of masses about the X-axis, perpendicular to the Z-Y plane, is

Question

A 2.00 kg mass is located at (4.00 m, 0.00 m, 0.00 m) and a 4.00 kg mass is located at (0.00 m, 3.00 m, 0.00 m). The rotational inertia of this system of masses about the X-axis, perpendicular to the Z-Y plane, is

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Solution

The rotational inertia, also known as the moment of inertia, is given by the equation I = m*r^2, where m is the mass and r is the distance from the axis of rotation.

For the 2.00 kg mass, it is located on the X-axis, so its distance from the X-axis is 0. Therefore, its contribution to the rotational inertia is 0.

For the 4.00 kg mass, it is located 3.00 m away from the X-axis. Therefore, its contribution to the rotational inertia is I = mr^2 = 4.00 kg * (3.00 m)^2 = 36.00 kgm^2.

So, the total rotational inertia of the system about the X-axis is 36.00 kg*m^2.

This problem has been solved

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