Question

Consider a solid cube of mass \[m\] and side \[L\] . What will be the value of \[\sum {{m_i}} {r_i}^2\] for this body, when the point of intersection of axes is the centre of the cube ?
A. \[\dfrac{{M{L^2}}}{2}\]
B. \[\dfrac{{M{L^2}}}{4}\]
C. \[\dfrac{{M{L^2}}}{3}\]
D. \[\dfrac{{M{L^2}}}{6}\]

Answer 1

Hint: In the above given question, we are given a cube which has a mass \[m\] and the length of each edge of the cube is \[L\] . Also it is given that the point of intersection of the three axes is located at the centre of the cube. We have to find the value of \[\sum {{m_i}} {r_i}^2\] for the given body i.e. cube. The quantity \[\sum {{m_i}} {r_i}^2\] is called the moment of inertia of the body around its axis of rotation.

Complete step by step answer:
Given that, a solid cube of mass \[m\] and side \[L\]. The intersection of the three axes is given as the centre of the solid cube. We have to find the value of \[\sum {{m_i}} {r_i}^2\] for the given cube. In other words, we have to find the moment of inertia of the cube around its axis of rotation. Now, let \[{I_x},{I_y},{I_z}\] be the moments of inertia about the three mutually perpendicular axes, which intersect at the centre of the cube.Taking the three axes as coordinate axes, we can write the sum of \[{I_x},{I_y},{I_z}\] as,
\[ \Rightarrow {I_x} + {I_y} + {I_z}\]
That gives us,
\[ \Rightarrow {I_x} + {I_y} + {I_z} = \sum {{m_i}\left( {{y_i}^2 + {z_i}^2} \right)} + \sum {{m_i}\left( {{x_i}^2 + {z_i}^2} \right)} + \sum {{m_i}\left( {{x_i}^2 + {y_i}^2} \right)} \]

Adding these,
\[ \Rightarrow {I_x} + {I_y} + {I_z} = 2\sum {{m_i}\left( {{y_i}^2 + {z_i}^2 + {x_i}^2} \right)} \]
We can write it as,
\[ \Rightarrow {I_x} + {I_y} + {I_z} = 2\sum {{m_i}{r_i}^2} \]
Now, since we know that, for a cube we have
\[ \Rightarrow {I_x} = {I_y} = {I_z} = \dfrac{{M{L^2}}}{6}\]
That gives us,
\[ \Rightarrow {I_x} + {I_y} + {I_z} = 3{I_x} = 2\sum {{m_i}{r_i}^2} \]
\[ \Rightarrow 3\dfrac{{M{L^2}}}{6} = 2\sum {{m_i}{r_i}^2} \]
Therefore, we can write
\[ \Rightarrow \sum {{m_i}{r_i}^2} = \dfrac{{3M{L^2}}}{{2 \times 6}}\]
Hence, we get
\[ \therefore \sum {{m_i}{r_i}^2} = \dfrac{{M{L^2}}}{4}\]
That is the required value of \[\sum {{m_i}} {r_i}^2\].

Therefore the correct answer is B.

Note: The moment of inertia of a body is defined as the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a turning force called torque.In other words, the moment of inertia of a body is the resistance offered by the body in acquiring motion while being is the state of rest or in acquiring the state of rest while being in a motion. It is denoted by \[I\].