Answer
(a) The rotational inertia about the x-axis is $1.3\times 10^{-3}~kg~m^2$
(b) The rotational inertia about the y-axis is $2.51\times 10^{-3}~kg~m^2$
(c) The rotational inertia about the z-axis is $3.81\times 10^{-3}~kg~m^2$
(d) The center of mass is located at the point $(-1.3~cm, -1.0~cm, 0~cm)$
Work Step by Step
(a) In general, $I = \sum M_i~R_i^2$, where $M_i$ is the mass of each particle and $R_i$ is the distance from each particle to the axis of rotation.
We can find the rotational inertia about the x-axis:
$I = (0.20~kg)(0.050~m)^2+(0.30~kg)(0)^2+(0.50~kg)(0.040~m)^2$
$I = 1.3\times 10^{-3}~kg~m^2$
The rotational inertia about the x-axis is $1.3\times 10^{-3}~kg~m^2$
(b) We can find the rotational inertia about the y-axis:
$I = (0.20~kg)(0.030~m)^2+(0.30~kg)(0.060~m)^2+(0.50~kg)(0.050~m)^2$
$I = 2.51\times 10^{-3}~kg~m^2$
The rotational inertia about the y-axis is $2.51\times 10^{-3}~kg~m^2$
(c) We can find the rotational inertia about the z-axis:
$I = (0.20~kg)(\sqrt{0.0034}~m)^2+(0.30~kg)(0.060~m)^2+(0.50~kg)(\sqrt{0.0041}~m)^2$
$I = 3.81\times 10^{-3}~kg~m^2$
The rotational inertia about the z-axis is $3.81\times 10^{-3}~kg~m^2$
(d) We can find the coordinates for the center of mass:
$x_{com} = \frac{(0.20~kg)(-3.0~cm)+(0.30~kg)(6.0~cm)+(0.50~kg)(-5.0~cm)}{0.20~kg+0.30~kg+0.50~kg}$
$x_{com} = -1.3~cm$
$y_{com} = \frac{(0.20~kg)(5.0~cm)+(0.30~kg)(0)+(0.50~kg)(-4.0~cm)}{0.20~kg+0.30~kg+0.50~kg}$
$y_{com} = -1.0~cm$
$z_{com} = \frac{(0.20~kg)(0)+(0.30~kg)(0)+(0.50~kg)(0)}{0.20~kg+0.30~kg+0.50~kg}$
$z_{com} = 0~cm$
The center of mass is located at the point $(-1.3~cm, -1.0~cm, 0~cm)$
