The Riemann Sum is a method of numerical analysis to calculate the area under the curve. But this time, we will use Riemann Sum to analyze the moment of inertia.
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00:00This time, we will learn to calculate the moment of inertia of a rigid body using Riemann
00:09sums.
00:10The rigid object is a thin stick.
00:11A thin stick is lying in the x-y plane, rotating about an axis on the z-axis, where one end
00:12of the stick is the axis.
00:13Before going any further, let's get acquainted with the concept of Riemann sums briefly.
00:38In Riemann sums, we will divide the rod into several elements in equal intervals of length
00:42and units.
00:43This element is also known as an increment.
00:44So, the element in calculating the moment of inertia is a mass element.
00:45We can think of them as very small particles.
00:46Based on the discrete particle concept, the moment of inertia value is the sum of the
00:47moments of inertia of each mass element.
00:48That value can be written as I of n discrete object units is sigma m-I or I-squared.
01:14The true value of the moment of inertia will be known when the value of n goes to infinity.
01:23Or I is lim of I-n, when n goes to infinity.
01:24Back to calculating the moment of inertia of the stick.
01:25Let's assume the length of the stick is L.
01:26We will divide the stick into infinitesimal mass elements, the number of which is n.
01:27The width of each mass element is the same, namely L over n.
01:53The maximum point in the coordinate plane is 0, L over n, 2 L over n, 3 L over n, 4
02:01L over n, and so on until it reaches L.
02:02If the mass of the stick is big M, then the mass of each mass element is big M over n.
02:03This is because uniform objects will have the same density.
02:04So mass elements with the same length will have the same mass.
02:05The distance of the first mass element to the shaft is 0, so the moment of inertia of
02:32the first mass element is big M over n times 0 squared.
02:34In the same way, the moment of inertia of the second mass element is big M over n, multiplied
02:55by the square of L over n.
03:01And so on, until the moment of inertia of the last mass element is big M over n, multiplied by the square of n minus one L over n.
03:11To simplify calculations, exclude the big M over n factor.
03:18Apart from that, we will also exclude the square of L over n factor.
03:25The values in brackets indicate regularity as a series. It can be written as sigma of I minus one squared.
03:37Now, decompose the value of I minus one squared into I squared, minus two I, plus one.
03:46Using the properties of sigma notation you will get sigma I squared, minus two sigma I, plus sigma one.
03:55Based on the previous tutorial, sigma I squared is equal to one sixth of n, times n plus one, times two n plus one.
04:04Sigma I is equal to half n, times n plus one.
04:09And sigma one is equal to n.
04:14Simplifying the values in brackets we get one third n to the third power, minus one half n squared, plus one sixth n.
04:24So, the approximate value of the moment of inertia is I n equals big ML squared, times one third, minus one over two n, plus one over six n squared.
04:37Well, the actual value of the moment of inertia is obtained when n goes to infinity.
04:44Well, the factor of one over n, and one over n squared will go to zero.
04:49So, what we're left with is I equals one third of big ML squared.
04:56This is the moment of inertia of a stick rotating horizontally at one end, I equals one third of big ML squared.
05:07To summarize, in the Riemann sum method, a rigid body can be considered as a collection of discrete particles.
05:16Hopefully this tutorial can provide useful benefits.
05:20And, don't forget to watch the next video.