Question

A mass of \[1\,{\text{kg}}\] is placed at \[(1m,{\text{ }}2m,{\text{ }}0)\] another mass of \[2\,{\text{kg}}\] is placed at \[(3m,{\text{ 4}}m,{\text{ }}0)\]. Find the moment of inertia of the system about the z-axis.
A.\[50\,{\text{kg}}{{\text{m}}^{\text{2}}}\]
B.\[55\,{\text{kg}}{{\text{m}}^{\text{2}}}\]
C.\[60\,{\text{kg}}{{\text{m}}^{\text{2}}}\]
D.\[65\,{\text{kg}}{{\text{m}}^{\text{2}}}\]

Answer 1

Hint: To solve the above question first find the distance of the masses from the z-axis after finding that calculate the moment of inertia for individual masses with their distance from the z-axis.

Complete Step by step answer: Given, mass of \[1\,{\text{kg}}\] is placed at \[(1m,{\text{ }}2m,{\text{ }}0)\]
Mass of \[2\,{\text{kg}}\] is placed at \[(3m,{\text{ 4}}m,{\text{ }}0)\]

Moment of inertia is the sum of the product of the masses and their perpendicular distance from the axis, which is written as
\[I = \sum {{m_i}{r_ \bot }^2} \],
where \[{m_i}\] are the masses and \[{r_ \bot }\]is their perpendicular distance from the axis.
We observe from the coordinates for both masses that their z-coordinate is zero, that is they are in the x-y plane. So, the perpendicular distance of the masses from z-axis is
\[{r_ \bot } = \sqrt {{x^2} + {y^2}} \],
where \[x\] is the x-coordinate and \[y\] is the y-coordinate.
Now, for mass \[{m_1} = 1{\text{kg}}\], the perpendicular distance from z-axis is
\[{r_1} = \sqrt {{1^2} + {2^2}} \\
   \Rightarrow {r_1} = \sqrt 5 \,{\text{m}} \]
For mass \[{m_2} = 2{\text{kg}}\], the perpendicular distance from z-axis is
\[{r_2} = \sqrt {{3^2} + {4^2}} \\
   \Rightarrow {r_2} = \sqrt {25} \,{\text{m}} \]
Now, the moment of inertia of the system about the z-axis is
\[ I = \sum {{m_i}{r_ \bot }^2} \\
   \Rightarrow I = {m_1}{r_1}^2 + {m_2}{r_2}^2 \]
Substituting the values of \[{m_1}\], \[{r_1}\] and \[{m_2}\],\[{r_2}\] we get
\[ I = 1 \times {\left( {\sqrt 5 } \right)^2} + 2 \times {\left( {\sqrt {25} } \right)^2} \\
   \Rightarrow I = 5 + 50 \\
   \Rightarrow I = 55\,\,{\text{kg}}{{\text{m}}^{{\text{ - 2}}}} \]
Therefore, the moment of inertia of the system about z-axis is \[55\,\,{\text{kg}}{{\text{m}}^{{\text{ - 2}}}}\]

Hence, the correct answer is option (B) \[55\,\,{\text{kg}}{{\text{m}}^{{\text{ - 2}}}}\]

Note: In problems where we have to find moment of inertia about an axis for system of many bodies or masses, it will be an easy approach if we find the moment of inertia for individual bodies and then add them to find the total moment of inertia of the whole system.