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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» Solution :As shown in fig. consider a uniform circular disc with mass M and radius R. 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.  `THEREFORE SIGMA=("Mass of the disc")/("Area of the disc")` `(M)/(piR^(2))` Let imagine this disc consisting of serveral concentric rings with difference radii and their centre is O. Consider one of such rings with radius x and width dx as shown in the fig. Area of this ring `S=2pix.dx` and `therefore` mass `m=lambdaS` `=(M)/(piR^(2))xx2pixdx` `therefore m=(2Mxdx)/(R^(2))""...(1)` If dI is the moment of inertia of this ring about the axis ZZ. then `dI=mr^(2)=(2Mx^(3))/(R^(2))dx""....(2)` The total of the moment of inertia of such concentric rings with different radius GIVES the moment of inertia of the disc as a whole about the ZZ. axis. For this integrating EQN. (1) in the interval `x=0` to `x=R`, `I=intdl=underset(0)overset(R)int(2Mx^(3))/(R^(2))dx` `=(2M)/(R^(2))underset(0)overset(R)intx^(3)dx=(2M)/(R^(2))[(x^(4))/(4)]_(0)^(R)` `=(2M)/(R^(2))[(R^(4))/(4)-0]` `I=(MR^(2))/(2)` Now comparing the eqn. (3) with `I=MK^(2)` `k^(2)=(R^(2))/(2)` `therefore k=(R)/(sqrt(2))` is a radius of gyration |
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