Derive an expression for the rotational inertia of the disk about the pivot.
Answers
To calculate the moment of inertia of this disk about the z-axis, we sum the moment of inertia of a volume element
dV from the centre (where
r=0) to the outer radius
r.
I_{zz} = \int_{r=0} ^{r=r} r_{\perp} ^{2} dm \text{ (Equ. 1)}
The mass element
dm is related to the volume element
dV via the equation
dm = \rho dV (where
\rho is the density of the volume element). We will assume in this example that the density
\rho(r) of the disk is uniform; but in principle if we know its dependence on
r, \; \rho (r) = f(r), this would not be a problem.
The volume element
dV can be calculated by considering a ring at a radius
r with a width
dr and a thickness
t. The volume of this ring is just this rings circumference multiplied by its width multiplied by its thickness.
dV = (2 \pi r dr) t
so we can write
dm = \rho (2 \pi r dr) t
and hence we can write equation (1) as
I_{zz} = \int_{r=0} ^{r=r} r_{\perp} ^{2} \rho (2 \pi r dr) t = 2 \pi \rho t \int_{r=0} ^{r=r} r_{\perp} ^{3} dr
Integrating between a radius of
r=0 and
r, we get
I_{zz} = 2 \pi \rho t [ \frac{ r^{4} }{ 4 } -0 ] = \frac{1}{2} \pi \rho t r^{4} \text{ (Equ. 2)}
If we now define the total mass of the disk as
M, where
M = \rho V
and
V is the total volume of the disk. The total volume of the disk is just its area multiplied by its thickness,
V = \pi r^{2} t
and so the total mass is
M = \rho \pi r^{2} t
Using this, we can re-write equation (2) as
\boxed{ I_{zz} = \frac{1}{2} \pi \rho t r^{4} = \frac{1}{2} Mr^{2} }