Question

The moment of inertia of a solid cylinder of mass M, length L and radius R about the diameter of one of its faces will be:
A.$M\left( \dfrac{{{L}^{2}}}{12}+\dfrac{{{R}^{2}}}{4} \right)$
B.$M\left( \dfrac{{{L}^{2}}}{4}+\dfrac{{{R}^{2}}}{4} \right)$
C.Zero
D.$\dfrac{M{{R}^{2}}}{4}$

Answer 1

To solve this problem we will be using perpendicular axis theorem and parallel axis theorem in which moment of inertia along two parallel axis or moment of inertia along an axis perpendicular to two axis can be calculated.

Formula used:
$\Rightarrow I=m{{r}^{2}}$
Perpendicular axis theorem
$\Rightarrow I={{I}_{X}}+{{I}_{Y}}$
Parallel axis theorem
$\Rightarrow I={{I}_{X}}+M{{(r)}^{2}}$

Complete answer:
The property of a body to resist angular acceleration is known as moment of inertia of the body, it’s value can be calculated as the summation of product of mass of each particle and square of its distance from the axis of rotation.
Now, the moment of inertia of the solid cylinder along its rotational axis is given by,
$\Rightarrow I=\dfrac{M{{R}^{2}}}{2}$
Now, the moment of inertia along the two perpendicular axis passing through centroid will be,
$\Rightarrow I={{I}_{X}}+{{I}_{Y}}$
Since the cylinder is uniform so,
$\Rightarrow {{I}_{X}}={{I}_{Y}}$
$\Rightarrow I=2{{I}_{X}}$
$\Rightarrow {{I}_{X}}=\dfrac{M{{R}^{2}}}{4}$
Now the distance of face diameter from the centroid axis whose inertia is calculate in the above equations is $\dfrac{L}{2}$,
So by applying parallel axis theorem, we can calculate moment of inertia along one of the face diameter as,
$\Rightarrow I={{I}_{X}}+M{{(r)}^{2}}$
Here,${{I}_{X}}$ (moment of inertia along one of the centroid axes calculated above), r (distance between the two parallel axes).
Now, by putting the values in above equation we have,
$\Rightarrow I=\dfrac{M{{R}^{2}}}{4}+M{{\left( \dfrac{L}{2} \right)}^{2}}$
$\Rightarrow I=M\left( \dfrac{{{R}^{2}}}{4}+{{\left( \dfrac{L}{2} \right)}^{2}} \right)$
$\therefore$ The moment of inertia of solid cylinder about the diameter of one of its faces will be,
$\Rightarrow I=M\left( \dfrac{{{R}^{2}}}{4}+{{\left( \dfrac{L}{2} \right)}^{2}} \right)$

Note:
Parallel axis theorem can be applied to any object irrespective of shape or size but parallel axis theorem can be applied only on planar objects as it needs perpendicular axis in planes to be applied. The moment of inertia of system of particles of masses,
${{m}_{1}},{{m}_{2}},{{m}_{3}}....$
And, their distance from the axis of rotation is,
${{r}_{1}},{{r}_{2}},{{r}_{3}}....$
Then, moment of inertia is given by,
$\Rightarrow I={{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}{{r}_{2}}^{2}+{{m}_{3}}{{r}_{3}}^{2}....$
If we considered mass to be concentrated at one point, inertia will be
$\Rightarrow I=\sum\limits_{{}}^{{}}{m{{r}^{2}}}$