Question

The moment of inertia of a thin uniform rectangular plate relative to the axis passing perpendicular to the plane of the plate through one of its vertices, if the sides of the plate are equal to $a$ and $b$ , and the mass $m$ is $I={\displaystyle \frac{m}{x}}(a^2+b^2)$. Find the value of $$x$$.

Answer 1

Hint: Find the moment of inertia of a rectangle plate with respect to the axis that is towards the center of mass and is perpendicular to the plane of the plate.
• Apply the parallel axis theorem to find the moment of inertia of the plate about the axis that is perpendicular to the plane of the plate and passing through one of the vertices.
• Find the perpendicular distance by using the given sides of the rectangular plate.
• Compare the given equation with the relation you get and find the value of the required term.

Formula used:
The moment of inertia of a rectangle plate with respect to the axis that is towards the center of mass and is perpendicular to the plane of the plate $I_c={\displaystyle \frac{m}{12}}(a^2+b^2)$
According to the Parallel axes theorem, $I_v=I_c+m{l}^2$
$I_v$ = the moment of inertia of the plate about the axis that is perpendicular to the plane of the plate and passing through one of the vertices.
$m=$ The mass of the plate
$l=$ The perpendicular distance $=\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^2+{\left({\displaystyle \frac{b}{2}}\right)}^2}$.

Complete step by step answer:
Let The moment of inertia of a rectangle plate with respect to the axis that is towards the center of mass and is perpendicular to the plane of the plate is $I_c$ and the moment of inertia of the plate about the axis that is perpendicular to the plane of the plate and passing through one of the vertices is $I_v$.
Since the two sides of the rectangle are given $a$ and $b$ the mass of the plate is $m$,
$I_c={\displaystyle \frac{m}{12}}(a^2+b^2)$
Now according to the Parallel axis theorem, $I_v=I_c+m{l}^2$
Where, $m=$ The mass of the plate
$l=$ The perpendicular distance

The perpendicular distance can be written as, $$l=\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^2+{\left({\displaystyle \frac{b}{2}}\right)}^2}$$
So, $I_v={\displaystyle \frac{m}{12}}(a^2+b^2)+m\left[{\left({\displaystyle \frac{a}{2}}\right)}^2+{\left({\displaystyle \frac{b}{2}}\right)}^2\right]$
$\Rightarrow I_v={\displaystyle \frac{m}{12}}(a^2+b^2)+{\displaystyle \frac{m}{4}}(a^2+b^2)$
$\Rightarrow I_v=(a^2+b^2)\left[{\displaystyle \frac{m}{12}}+{\displaystyle \frac{m}{4}}\right]$
$\Rightarrow I_v={\displaystyle \frac{4m}{12}}(a^2+b^2)$
$\Rightarrow I_v={\displaystyle \frac{m}{3}}(a^2+b^2)$
The relation given in the problem is, $\Rightarrow I={\displaystyle \frac{m}{x}}(a^2+b^2)$
By comparing the two equations we get $x=3$
Hence, the value of $x=3$.

Note: The role of the moment of inertia in a circular motion is the same as the mass in the linear motion. We know that the mass of an object is called translational inertia because to change the motion of an object the applied constraint by the object is its mass. For the rotational motion, the torque ( it is the circular term of a force) that is applied on the object, changes the motion of that object. So the moment of inertia that resists the changes in a circular motion or tries to hold the object rigid is actually the rotational inertia.