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Find the moment of inertia of a uniform circular disc about an axis passing through its geometrical centre and perpendicular to its plane and radius of gyration.

Answer» <html><body><p></p>Solution :As shown in fig. consider a uniform circular disc with mass M and radius <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>. For the moment of inertia of this disc about the axis ZZ. passing through its geometrical centre, area of the disc `A=pir^(2)` and mass per unit area of the disc. <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P1_C07_E02_071_S01.png" width="80%"/> <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> <a href="https://interviewquestions.tuteehub.com/tag/sigma-1207107" style="font-weight:bold;" target="_blank" title="Click to know more about SIGMA">SIGMA</a>=("Mass of the disc")/("Area of the disc")` <br/> `(M)/(piR^(2))` <br/> Let imagine this disc consisting of serveral concentric rings with difference radii and their centre is O. <br/> Consider one of such rings with radius x and width dx as shown in the fig. <br/> Area of this ring `S=2pix.dx` and <br/> `therefore` mass `m=lambdaS` <br/> `=(M)/(piR^(2))xx2pixdx` <br/> `therefore m=(2Mxdx)/(R^(2))""...(1)` <br/> If dI is the moment of inertia of this ring about the axis ZZ. then <br/> `dI=mr^(2)=(2Mx^(3))/(R^(2))dx""....(2)` <br/> The total of the moment of inertia of such concentric rings with different radius <a href="https://interviewquestions.tuteehub.com/tag/gives-1007647" style="font-weight:bold;" target="_blank" title="Click to know more about GIVES">GIVES</a> the moment of inertia of the disc as a whole about the ZZ. axis. <br/> For this integrating <a href="https://interviewquestions.tuteehub.com/tag/eqn-973463" style="font-weight:bold;" target="_blank" title="Click to know more about EQN">EQN</a>. (1) in the interval `x=0` to `x=R`, <br/> `I=intdl=underset(0)overset(R)int(2Mx^(3))/(R^(2))dx` <br/> `=(2M)/(R^(2))underset(0)overset(R)intx^(3)dx=(2M)/(R^(2))[(x^(4))/(4)]_(0)^(R)` <br/> `=(2M)/(R^(2))[(R^(4))/(4)-0]` <br/> `I=(MR^(2))/(2)` <br/> Now comparing the eqn. (3) with `I=MK^(2)` <br/> `k^(2)=(R^(2))/(2)` <br/> `therefore k=(R)/(sqrt(2))` is a radius of gyration</body></html>


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