The moment of inertia of 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 mass $m$ is $I = \dfrac{m}{x}\left( {{a^2} + {b^2}} \right)$. Find the value of $x$.
Answer
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Hint: We will use Parallel axis theorem to find the moment of inertia of the plate with respect to axis which is perpendicular and passing through the vertices of the plate as we know the moment of inertia of the plate with respect to axis which is perpendicular to the plane and passing through the center of the plate.
Complete step by step answer:
Let the moment M.I of the plate about an axis which is perpendicular to the plane and passing through the center is ${I_c}$ and the M.I of the plate about an axis which is perpendicular to the plane and passing through one of the vertices of the rectangular plate is ${I_v}$.
Respective axes are shown in the figure below
According to parallel Axis Theorem,
${I_v} = {I_c} + m{l^2}$
Where,
$l = $perpendicular distance between the parallel axes.
$l = \sqrt {{{\left( {\dfrac{a}{2}} \right)}^2} + {{\left( {\dfrac{b}{2}} \right)}^2}} $
${I_c} = \dfrac{m}{{12}}\left( {{a^2} + {b^2}} \right)$
So,
${I_v} = \dfrac{m}{{12}}\left( {{a^2} + {b^2}} \right) + m\left( {\dfrac{{{a^2}}}{4} + \dfrac{{{b^2}}}{4}} \right)$
${I_v} = m\left( {{a^2} + {b^2}} \right)\left[ {\dfrac{1}{{12}} + \dfrac{1}{4}} \right]$
${I_v} = \dfrac{m}{3}\left( {{a^2} + {b^2}} \right)$
On comparing,
$x = 3$
Additional information:
> Moment of inertia plays the role in rotational mechanics is same as mass plays in linear kinetics. They both are characterized as the resistance of the body to change in its state of rest or state of motion.
> Moment of inertia depends upon the distribution of mass in space and the axis of rotation. For a point-like mass the moment of inertia is directly given as $m{r^2}$but it will change as we change the distribution of mass around the axis of rotation.
Note: Parallel axis theorem is also valid in case of 3D objects like cone, hemisphere. Perpendicular axis theorem is always valid for 2D shapes only.
Complete step by step answer:
Let the moment M.I of the plate about an axis which is perpendicular to the plane and passing through the center is ${I_c}$ and the M.I of the plate about an axis which is perpendicular to the plane and passing through one of the vertices of the rectangular plate is ${I_v}$.
Respective axes are shown in the figure below
According to parallel Axis Theorem,
${I_v} = {I_c} + m{l^2}$
Where,
$l = $perpendicular distance between the parallel axes.
$l = \sqrt {{{\left( {\dfrac{a}{2}} \right)}^2} + {{\left( {\dfrac{b}{2}} \right)}^2}} $
${I_c} = \dfrac{m}{{12}}\left( {{a^2} + {b^2}} \right)$
So,
${I_v} = \dfrac{m}{{12}}\left( {{a^2} + {b^2}} \right) + m\left( {\dfrac{{{a^2}}}{4} + \dfrac{{{b^2}}}{4}} \right)$
${I_v} = m\left( {{a^2} + {b^2}} \right)\left[ {\dfrac{1}{{12}} + \dfrac{1}{4}} \right]$
${I_v} = \dfrac{m}{3}\left( {{a^2} + {b^2}} \right)$
On comparing,
$x = 3$
Additional information:
> Moment of inertia plays the role in rotational mechanics is same as mass plays in linear kinetics. They both are characterized as the resistance of the body to change in its state of rest or state of motion.
> Moment of inertia depends upon the distribution of mass in space and the axis of rotation. For a point-like mass the moment of inertia is directly given as $m{r^2}$but it will change as we change the distribution of mass around the axis of rotation.
Note: Parallel axis theorem is also valid in case of 3D objects like cone, hemisphere. Perpendicular axis theorem is always valid for 2D shapes only.
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