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). If this system of masses rotated about the Z-axis perpendicular to the X-Y plane, then the moment of inertia of this system is:_______

Answer 1

Answer:

The moment of inertia of this system is 68 kilogram-square meters.

Explanation:

We have two particles rotating about the z-axis, which is orthogonal to xy plane, the moment of inertia of the system ($I_{z}$), measured in kilogram-square meters, is determined by the following formula:

$I_{z} = \Sigma \limits_{i=1}^{2}m_{i}\cdot r_{i}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$r_{i}$ - Distance of the i-th particle from axis of rotation, measured in meters.

By Pythagorean Theorem we calculate each distance:

$(x_{1}, y_{1}, z_{1}) = (4\,m, 0\,m,0\,m)$

$r_{1}=\sqrt{(4\,m-0\,m)^{2}+(0\,m-0\,m)^{2}+(0\,m-0\,m)^{2}}$

$r_{1} = 4\,m$

$(x_{2},y_{2}, z_{2}) = (0\,m, 3\,m, 0\,m)$

$r_{2} = \sqrt{(0\,m-0\,m)^{2}+(3\,m-0\,m)^{2}+(0\,m-0\,m)^{2}}$

$r_{2} = 3\,m$

If we know that $m_{1} = 2\,kg$, $r_{1} = 4\,m$, $m_{2} = 4\,kg$ and $r_{2} = 3\,m$, then the moment of inertia of the system is:

$I_{z} = m_{1}\cdot r_{1}^{2}+m_{2}\cdot r_{2}^{2}$ (1b)

$I_{z} = (2\,kg)\cdot (4\,m)^{2}+(4\,kg)\cdot (3\,m)^{2}$

$I_{z} = 68\,kg\cdot m^{2}$

The moment of inertia of this system is 68 kilogram-square meters.