Passing through the center of mass is, I= (1/12) M l 2 Its moment of inertia with respect to a perpendicular axis Particles of the body from the axis of rotation.įact, the moment of inertia of any object could be expressed in the form, I = MK 2.Įxample, let us take the moment of inertia of a uniform rod of mass M and Gyration indicates that it is the root mean square (rms) distance of the Nm is the total mass M of the body and K is the radius of gyration. We take all the n number of individual masses to be equal, Moment of inertia of that object can be written as, r n respectively as shown in Figure 5.24. Perpendicular distances (or positions) r 1, r 2, r 3 Rotating rigid body with respect to any axis, is considered to be made up of The radius of gyration is distance, its unit is m. Mass, which would have the same mass as well as the same moment of inertia of Is the perpendicular distance from the axis of rotation to an equivalent point M is the total mass of the object and K is called the radius of gyration,
The general expression for moment of inertia is given as, That it is applicable even for objects of irregular shape and non-uniformĭistribution of mass. But, we need an expression for the moment of inertia whichĬould take care of not only the mass, shape and size of objects, but also its Moment of inertia (I) of the entire disc is,īulk objects of regular shape with uniform mass distribution, the expressionįor moment of inertia about an axis involves their total mass and geometricalįeatures like radius, length, breadth, which take care of the shape and the (2 π r is the length and dr is the thickness). The term ( 2 π r dr ) is the area of this elemental ring Mass of the infinitesimally small ring is, The moment of inertia (dI ) of the small ring This disc is made up of many infinitesimally As the mass is uniformly distributed,Ĭover the entire length of the ring, the limits of integration are taken from 0Ī disc of mass M and radius R. Length of the ring is its circumference ( 2 π R ). Moment of inertia (dI) of this small mass (dm) is, This (dm) is located at a distance R, which is the radius of the ring To the plane, let us take an infinitesimally small mass (dm) of length (dx) of Inertia of the ring about an axis passing through its center and perpendicular Us consider a uniform ring of mass M and radius R. The mass is distributed on either side of the origin, the limits for Moment of inertia (I) of the entire rod can be found by integrating dI, (dm) mass of the infinitesimally small length as, dm = λdx = M/ l dx
The mass is uniformly distributed, the mass per unit length ( λ ) of the rod is, λ = M/ l Moment of inertia of this rod about an axis that passes through the center ofįirst an origin is to be fixed for the coordinate system so that itĬoincides with the center of mass, which is also the geometric center of the Us consider a uniform rod of mass (M) and length ( l ) as shown in Figure 5.21. The common bulk objects of interest like rod, ring, disc, sphere etc. Get the moment of inertia of the entire bulk object by integrating the aboveĬan use the above expression for determining the moment of inertia of some of The moment of inertia of this point mass can now be The way the mass is distributed around the axis of rotation.įind the moment of inertia of a uniformly distributed mass we have to considerĪn infinitesimally small mass (dm) as a point mass and take its position (r) It depends not only on the mass of the body, but also on But, the moment of inertia of a body is not an In general, mass is an invariable quantity of matter (except for motionĬomparable to that of light). Rotational motion, moment of inertia is a measure of rotational inertia. Translational motion, mass is a measure of inertia in the same way, for For point mass m iĪt a distance r i from the fixed axis, the moment of inertia is given This quantity isĬalled moment of inertia (I) of the bulk object. The expressions for torque and angular momentum for rigid bodies (which areĬonsidered as bulk objects), we have come across a term Σ m i r i 2.
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