Calculate the moment of inertia of the following rigid...

Calculate the moment of inertia of the following rigid bodies about the given fixed axis n?. In each case, the rigid body has uniform mass density. (a) A uniform circular hoop of radius R and mass m. The axis n? is taken to pass through the centre of the hoop and be orthogonal to the plane containing the hoop. (b) The same setup as part (a), with the hoop replaced by a uniform disk. (c) A uniform rod of length l and mass m, with two point-like masses M attached to its endpoints. The axis n? is taken to pass through the midpoint of the rod and be orthogonal to the rod. Hint: I_n? is additive, so, I_n?^total = I_n?^rod + I_n?^point masses.

Transcript

00:02 The question is about moment of inertia.

00:05 There are multiple parts in the question.

00:08 In first part of the question we have to find moment of inertia of a ring about its central axis.

00:17 So, moment of inertia for ring.

00:31 So to find this, the diagram will be like this.

00:40 In the diagram dm is smallest mass of the ring at a distance of r.

00:47 So its moment of inertia will be d .i, which is differential moment of inertia for differential mass, will be d .m multiplied by distance square.

01:02 And distance is r.

01:05 So, and r is remains same for all the dm masses.

01:10 So to get full moment of inertia integrate this equation.

01:19 So it will become d .m .r.

01:24 Integration on both the side now it will become total moment of inertia and r square out of integration and integration of dm now our final answer will be i equal to m r square because integration of dm is m this is our first answer in part b of the question there is a thin ring of thickness d r at a distance of r from the center so in part b we have to given that in ring as shown in figure thin ring of radius r in the previous ring of radius capital r there is thin ring of radius small r mass per unit length will be same mass per unit area that is mass density for ring will be same.

02:56 Let us consider sigma is mass density so it will be mass offering divided by pi r square.

03:08 Now from this equation we can simplify we can obtain dm equal to 2 pi r d r d r multiplied by sigma so it will be equal to m upon m divided by pi r squared multiplied by 2 pi r d r now integrating we will get integration d i equal to integration d m r square this is a new equation now, integration will be from 0 to capital i, which is total moment of inertia.

04:27 Here, dm is 2m divided by r2, integration is from 0 to capital r.

04:40 And r, dr, multiplied by r square as it is.

04:48 Now, from this equation after getting integration, we will get total moment of inertia for a given ring i equal to m r square by this is our second answer.

05:13 Now, in part c there is one rod for which first we have to find moment of inertia without point man and another.

05:29 With point man.

05:31 So first of all we will find moment of inertia for rod adds in at its own axis...

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