Question

Match the following moment of inertia of a thin circular plate about different axes of rotations.

a- about any diameter	e- \[\dfrac{3M{{r}^{2}}}{2}\]
b- about any tangent in its plane	f- \[\dfrac{M{{r}^{2}}}{4}\]
c- about natural axis	g- \[\dfrac{5M{{r}^{2}}}{4}\]
d- about any tangent perpendicular to its plane	h- \[\dfrac{M{{r}^{2}}}{2}\]

1- a-g, b-f, c-h, d-e
2- a-f, b-g, c-h, d-e
3- a-g, b-f, c-e, d-h
4- a-f, b-g, c-e, d-h

Answer 1

Hint: The moment of inertia of a thin circular disk is the same as that for a solid cylinder of any length, it is often used as an element for building up the moment of inertia expression for other geometries, such as the sphere or the cylinder about an end diameter.

Complete step by step answer:
The moment of inertia about an axis passing through the centre of mass is \[\dfrac{M{{r}^{2}}}{2}\], this is also known as its natural axis. So, c-h.
Now to find the moment of inertia about tangent perpendicular to the plane we use theorem of parallel axis, thus it can be found out as,
\[\dfrac{M{{r}^{2}}}{2}+M{{r}^{2}}=\dfrac{3M{{r}^{2}}}{2}\]
So, d-e.
Now to find moment of inertia about any diameter we use theorem of perpendicular axis, thus,
\[\Rightarrow {{I}_{xx}}+{{I}_{yy}}=\dfrac{M{{r}^{2}}}{2}\]
Since, both the diameters are same,
\[\Rightarrow 2{{I}_{xx}}=\dfrac{M{{r}^{2}}}{2}\]
=\[\dfrac{M{{r}^{2}}}{4}\]
So, a-f. and the last option left will be b-g.

So, the correct answer is “Option 2”.

Note:
Moment of inertia is the name given to rotational inertia, and it plays the same role in rotation as is played by mass in translation. Moment of inertia is defined concerning a specific rotation axis. Here we have four rotational axes and corresponding to them we have found out the same.