MI of a Disc and a Cylinder PDF

MI of a Disc and a cylinder What is a disc?  A Disc is an object that is generally thin and has cylindrical geometry A disc is supposed to be composed of many circular rings of increasing radius and it has also thickness. MI of a circular disc about an axis passing through its centre an...

MI of a Disc and a cylinder What is a disc?  A Disc is an object that is generally thin and has cylindrical geometry A disc is supposed to be composed of many circular rings of increasing radius and it has also thickness. MI of a circular disc about an axis passing through its centre and perpendicular to its plane  Derivation M=Mass, Radius R, and t = thickness of the disc Mass M = Density Volume of the disc = R2 t dI = dmx 2 (For a small element of mass dm).......(i) dm = V = (2xdx)t dI = (2xdx)tx 2  dI = (2x3 dx)t Integrating above equation within the limits of x = 0 to x = +R We can get I = 1/2MR2  Poll The moment of inertia of a disc is given by (A) 1/2MR2 (B) 1/4MR2 (C) 2/5MR2 (D) None of the above Cylinder Axes of a cylinder Moment of inertia of a solid cylinder about geometrical axis  Moment of inertia of a solid cylinder Mass per unit volume, = M/R2 l MI of the disc about XX’ = 1/2mR2 MI of the cylinder = ½mR2 m = M (Mass of the cylinder) MI of the cylinder about the geometrical axes = ½MR2 MI of a cylinder about an axis passing through centre and  to geometrical axis  Derivation Area of the disc = R2 Volume of the disc = R2 dx Mass of the disc = m = (R2 dx) MI of the disc about PQ = ¼ MassRadius2 = ¼ (R2 dx)R2 = ¼ (R4 dx)  By theorem of Parallel axes I = Ic + Mh2 dI2 = ¼ (R4 dx) + (R2 dx)x2 dI2 = R2 dx(¼ R2 + x2 ) To find the MI of whole of the cylinder between the limits x=-l/2 and x = +l/2 I2 = M[R 2 /4 + l2 /12]  Poll The moment of inertia of a solid cylinder about the geometric axes is given by (A) 1/2MR2 (B) 1/4MR2 (C) 2/5MR2 (D) None of the above Relation between length and radius of the cylinder for minimum moment of inertia about an axis passing through centre and  to length We know that I2 = M[R 2 /4 + l2 /12].......(1) M = R2 l , where  = Density of the solid cylinder R2 = M/l Putting in equation 1, we have I2 = M[M/4 l + l2 /12]  The moment of inertia will be minimum if dI2 /dl = 0 Solving using this condition l/R = (3/2)1/2 = 3/2 Relation between length and radius of the cylinder for moment of inertia equal to its geometrical axis and the axis passing through centre of cylinder and  to length  MI of geometrical axes = I1 = ½ MR2 MI of a cylinder about an axis passing through the centre and perpendicular to its length I 2 = M[R 2 /4 + l2 /12] I1=I2 Solving we get, l = R3