Moment Of Intertia Of Disk

The Moment of Inertia of a Rod and a Disk

Last edited: 2021-06-04 13:28:28

If yous are merely searching for the moment of inertia of a rod or a disk I volition but requite them to you. For a rod $I = \frac{Fifty^2}{12}$ if the rotation axis is located at the middle the rod and $I = M \frac{L^2}{3}$ if the rotational centrality is located at the finish of the rod. For a homogeneous disk/cylinder $I = M \frac{R^2}{2}$ and for a disk/cylinder with just a shell $I = MR^ii$ . Now if y'all are interested in how we become these values feel free to read further.

Nosotros know that the moment of a point of inertia of a point mass is

$$M R^2,$$

where $Thou$ is the mass and $R$ is the altitude betwixt the mass and the rotational axis. We can apply this to calculate the moment of inertia for objects that is non point masses. Nosotros practise this by integrating over all infinitesmal mass elements $dm$ of the object. So our moment of inertia is now

$$I = \int_0^M r^2 dm.$$

The Moment of Inertia of a Rod

The moment of inertia is dependent on the location of the rotational centrality, therefore the moment of inertia volition vary for a rod depending on the centrality of rotation. We take our general formula for the moment of inertia

$$I = \int_0^Grand r^ii dm.$$

Nosotros now have to identify what $dm$ is. Since the rod tin be looked at in 1 dimension

$$dm = \frac{M}{L} dr$$

where $L$ is the length of the rod. $\frac{M}{L}$ tin exist looked at every bit an i dimensional density for the rod. Now lets calculate the moment of inertia if the rod is rotating at one of its ends. The moment of inertia then becomes

$$I = \int_0^L r^2 \frac{M}{L} dr = \frac{Thou}{Fifty} \Bigg[ \frac{r^iii}{3} \Bigg]_0^Fifty = \frac{One thousand}{L} \frac{L^3}{3} = G \frac{L^2}{3}.$$

If the axis of rotation is in the middle of the rod the moment of inertia then becomes

$$I = \int_{-L/2}^{L/2} r^ii \frac{G}{L} dr = \frac{Thou}{L} \Bigg[ \frac{r^3}{3} \Bigg]_{-Fifty/2}^{Fifty/ii} = \frac{M}{3L} \left[ \frac{L^3}{viii} - \left(\frac{-L^3}{8} \right) \right] = Yard \frac{Fifty^two}{12}.$$

Lets create a more full general formula. Lets use the variable $x$ for the point of rotation. Our moment of inertia integral then becomes

$$I = \int_{-x}^{50-10} r^two \frac{M}{Fifty} dr = \frac{M}{L} \Bigg[ \frac{r^iii}{3} \Bigg]_{-x}^{L-x} = \frac{1000}{3L} \left[ (L-ten)^3 + 10^3 \right].$$

The Moment of Inertia of a Disk

Lets motility on to calculate the moment of inertia for a disk (this also applies to a cylinder which is kind of an elongated disk). The axis of rotation will be located in the center of the disk merely we volition go back to calculating the moment of inertia of a displaced rotational axis. Like usual we beginning of with the archetype integral for the moment of inertia

$$I = \int_0^M r^two dm.$$

Defining $dm$ is a little bit more tricky than for the rod. In this problem we accept to go 2 dimensional. Lets start with defining our two dimensional density which is

$$\frac{One thousand}{\pi R^2}.$$

The denominator is only the surface area of the disk. Our surface area chemical element is

$$r dr d\theta$$

I won't get into to much detail here, you'll have to do some repetition on polar coordinates if don't sympathise how we get this area chemical element. So our moment of inertia becomes

$$I = \frac{M}{\pi R^2} \int_0^{2 \pi} \int_0^R r^3 dr d\theta = \frac{2M}{R^2} \int_0^R r^three dr = \frac{2M}{R^2} \Bigg[ \frac{r^4}{4} \Bigg]_0^R = \frac{2M}{R^2} \frac{R^4}{4} = Yard \frac{R^2}{2}.$$

At present, what happens if the axis of rotation was shifted (only still parallel to the centrality through the center of mass)? Well luckily for us we don't take to do a super complicated integral, nosotros tin can only use the parallel centrality thereom which says

$$I = I_\text{CM} + Doctor^2,$$

where $I_\text{CM}$ is the moment of inertia through the center of mass, in this instance $M \frac{R^2}{two}$ . $d$ is the distance between the axis of rotation and the rotational axis throught the center of mass. You can test this formula past calculating the moment of inertia at an end of the rod. In that case we accept $I_\text{CM} = M \frac{L^2}{12}$ and $d=\frac{L}{2}$ .

And so what if the disk/cylinder is hollow? And so nosotros only alter the limits of integration and the density. If the deejay is hollow upwards to $r=a$ and and then has mass upward to $r=b$ the 2D density changes to

$$\frac{Yard}{\pi (b^2-a^2)}.$$

So we get the post-obit moment of inertia

$$I = \frac{2M}{b^ii-a^2} \int_a^b r^3 dr = \frac{2M}{b^two-a^2} \Bigg[ \frac{r^4}{4} \Bigg]_a^b = \frac{2M}{b^ii-a^2} \frac{b^4-a^4}{4} = M \frac{b^two+a^2}{2}.$$

Hither nosotros can meet that if the disk/cylinder only consisted of a crush ( $a \rightarrow b$ ) we become

$$I = K b^2.$$

So now y'all know how to calculate the moment of inertia of a rod and a disk.

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Moment Of Intertia Of Disk,

Source: https://www.engineeringguidance.com/the-moment-of-inertia-of-a-rod-and-a-disk

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