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Posts Tagged ‘Moment of inertia of a disk’

In physics, the rotational equivalent of mass is something called the moment of inertia. The definition of the moment of inertia of a volume element dV which has a mass dm is given by


dI = r_{\perp}^{2} dm


where r_{\perp} is the perpendicular distance from the axis of rotation to the volume element. To find the total moment of inertia of an object, we need to sum the moment of inertia of all the volume elements in the object over all values of distance from the axis of rotation. Normally we consider the moment of inertia about the vertical (z-axis), and we tend to denote this by I_{zz}. We can write


I_{zz} = \int _{r_{1}} ^{r_{2}} r_{\perp}^{2} dm


The moment of inertia about the other two cardinal axes are denoted by I_{xx} and I_{yy}, but we can consider the moment of inertia about any convenient axis.

Derivation of the moment of inertia of a disk

In this blog, I will derive the moment of inertia of a disk. In upcoming blogs I will derive other moments of inertia, e.g. for an annulus, a solid sphere, a spherical shell and a hollow sphere with a very thin shell.

For our purposes, a disk is a solid circle with a small thickness t (t \ll r, small in comparison to the radius of the disk). If it has a thickness which is comparable to its radius, it becomes a cylinder, which we will discuss in a future blog. So, our disk looks something like this.



A disk of negligible thickness, with a radius of r.

A disk of small thickness t, with a radius of r



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} }

What are the moments of inertia about the x and y-axes?

To find the moment of inertia about the x or the y-axis we use the perpendicular axis theorem. This states that, for objects which lie within a plane, the moment of inertia about the axis parallel to this plane is given by


I_{zz} = I_{xx} + I_{yy}


where I_{xx} and I_{yy} are the two moments of inertia in the plane and perpendicular to each other.

We can see from the symmetry of the disk that the moment of inertia about the x and y-axes will be the same, so I_{zz} = 2I_{xx}. Therefore we can write



\boxed{ I_{xx} = I_{yy} = \frac{1}{2}I_{zz} = \frac{1}{4} Mr^{2} }

Flywheels

Flywheels are used to store rotational energy. This is useful when the source of energy is not continuous, as they can help provide a continuous source of energy. They are used in many types of motors including modern cars.

It is because of an disk’s moment of inertia that it can store rotational energy in this way. Just as with mass in the linear case, it requires a force to change the rotational speed (angular velocity) of an object. The larger the moment of inertia, the larger the force required to change its angular velocity. As we can see above from the equation for the moment of inertia of a disk, for two flywheels of the same mass a thinner larger one will store more energy than a thicker smaller one because its moment of inertia increases as the square of the radius of the disk.

Sometimes mass is a critical factor, and next time I will consider the case of an annulus, where the inner part of the disk is removed.

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