Question

Calculate the moment of inertia of a wheel about its axis which has a rim of mass $24\;M$ and twenty four spokes each of mass $M$ and length $l$.

Answer 1

Hint: You might find it easier to deconstruct the wheel to its rim and spokes and determine the moment of inertia in each case separately. Since the moment of inertia is the rotational analogue of mass, we can call it an additive quantity. In other words, the net moment of inertia for the wheel can be found by adding the moments of inertia of the rims and spokes together. In better analogy, treat the rim as a hoop and the spokes as rods.

Formula used:
Moment of inertia of thin hoop with axis through the centre of the hoop $I=mr^2$, where m is the hoop mass and r is the distance of the hoop from the axis of rotation.
Moment of inertia of a long uniform rod with the axis though one end of the road $I = \dfrac{1}{3}ml^2$ where m is the mass of the rod and l is the length of the rod from the axis of rotation.

Complete answer:
Let us deconstruct the wheel into two parts: rim and spokes.
We are given that the wheel rotates about its axis, which means that the axis of rotation passes through the centre of the wheel.

When this is the case, we can analogously relate the rim to a hoop with its axis of rotation passing through the centre, and each spoke to a rod with the axis of rotation passing through one end.
The mass of the rim is $24M$ and it’s at a distance equal to the radius of the wheel (which is just the length of the spokes) from the axis of rotation. Therefore, its distance from the axis will be $l$
Therefore, the moment of inertia of the rim can be given by:

$I_{rim} = mr^2 = 24Ml^2$

Now, the mass of each spoke is $M$ and is at a distance $l$ from the axis of rotation, and there are 24 spokes in total.
Therefore, the moment of inertia of the spokes can be given by:

$I_{spokes} = 24 \times \dfrac{1}{3}ml^2 = 24 \times \dfrac{1}{3}Ml^2 = 8Ml^2$

Therefore, the moment of inertia of the wheel altogether will be the sum of moments of inertia of the rim and the spokes.

$I_{wheel} = I_{rim} + I_{spokes} = 24Ml^2 + 8Ml^2$
$ \Rightarrow I_{wheel} = 32Ml^2$

Note:
Always remember that the expressions for the moments of inertia do not only depend on the shape of the object but also on the location of the axis of rotation. For example, consider the case of a long uniform rod of length L and mass M.
If the axis of rotation passes through the centre of the rod, moment of inertia
$I = \dfrac{1}{12}ML^2$
However, for the same rod, if the axis of rotation passes through one end of the road, moment of inertia
$I =\dfrac{1}{3}ML^2$
Thus, it is essential to determine where the axis of rotation is located every time we reconstruct a system to its bones.