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

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 $I$.