Moment of Inertia & Theorems Moment of Inertia & Theorems When a body rotates, some axes feel “easy” to spin about and others feel “stubborn”. Moment of inertia (MI) is the rotational analogue of mass: it quantifies resistance to angular acceleration about a specified axis. The farther mass is spread from the axis, the larger the MI, and the harder it is to change the rotational state. You feel this in daily life: opening a door is easiest when you push far from the hinges; a skater spins faster by pulling arms in (reducing MI). In rotational dynamics, torque plays the role of force and is tied to angular acceleration through MI. For point masses, distance from the axis matters quadratically; for extended bodies, every tiny mass element contributes based on how far it is from the axis. That is why the exact axis must be specified for every MI value. This topic is a pillar for rotational kinematics and dynamics: it connects angular acceleration under a given torque, the rotational share of kinetic energy, and the energy partition during rolling. You will learn clean definitions, standard MI values for rods, rings, discs, cylinders, plates, and spheres, and two powerful theorems—the parallel and perpendicular axes theorems—to shift or reorient the axis quickly without re-integrating from scratch. Equally important are the boundaries: which theorem applies when, what “plane lamina” means, and how to combine MI for composite bodies. remember Big picture: "Mass tells how hard it is to change speed; moment of inertia tells how hard it is to change spin—about a particular axis." Pulling mass closer to the axis reduces I and makes spinning easier. We assume a rigid body (relative distances between mass elements stay fixed) and a fixed, clearly specified rotation axis. Always state the axis with your MI answer. tip An ideal body whose size and shape do not change; distances between all pairs of points remain fixed. Rigid body Rotational analogue of mass about a chosen axis. For particles, I = m i r i 2 ; for a continuous body, I = r 2 ,dm . Moment of inertia (I) Axis of rotation The fixed line about which the body rotates; all MI values are defined with respect to a specific axis. A length defined by I = Mk 2 for a given axis; it is the distance from the axis at which the whole mass M could be concentrated to keep the same MI. Radius of gyration (k) Parallel axis theorem For any axis parallel to a centroidal axis: I = I cm + Md 2 , where d is the perpendicular distance between axes. Perpendicular axis theorem For a plane lamina: I z = I x + I y , where x and y axes lie in the plane and z is perpendicular through the same point. The rotational effect of a force about an axis; relates to angular acceleration by = I for a rigid body about a fixed axis. Torque (τ) Rate of change of angular velocity: = d /dt . Angular acceleration (α) Definition of MI for a system of point masses at distances r i from the axis. Discrete MI Continuous MI Definition of MI for a continuous mass distribution. In dynamics, torque changes rotation the way force changes translation. For a rigid body about a fixed axis, = I . Energy also has a rotational part. If a body spins with angular speed , its rotational kinetic energy is K rot = 1 2 I 2 . These relations make MI central in predicting angular acceleration under a given torque and in computing energy partition during rolling. Rotational KE Rotational kinetic energy of a rigid body about the given axis. This formula calculates the energy stored in an object due to its rotation about a fixed axis, assuming the object's mass distribution is kn Definition of the radius of gyration k for a body of mass M about a specified axis. Radius of gyration I = I cm + M d 2 Rigid body with total mass M. Two axes are parallel; one passes through the center of mass. d is the perpendicular distance between the axes. Parallel axis theorem: I = I cm + M d 2 r is position vector of dm from CM axis; r' from the parallel axis at distance d. neet-alert In the parallel axis theorem, d is the perpendicular distance between the axes. Many errors come from using a slanted or in-plane offset instead of the shortest perpendicular distance. I z = I x + I y Perpendicular axis theorem: I z = I x + I y Applies only to a plane lamina (all mass in a plane). x and y axes lie in the plane and intersect at the same point as z which is perpendicular to the plane. tip Perpendicular axis theorem is valid only for thin planar bodies (laminae). It fails for 3D bodies like cylinders or spheres. Radius of gyration k compresses the shape information into a single length: k = I/M . A larger k means mass is effectively farther from the axis. For example, for a ring about its symmetry axis, I = MR 2 so k = R . For a solid disc, I = 1 2 MR 2 , so k = R/ 2 . Standard Moments of Inertia (about common axes) Moments of Inertia Constants Body Shape Axis of Rotation Moment Formula ( I ) Radius of Gyration ( k ) Geometric Constraint Ring is 1, Disc is half, Sphere is two-fifths of the path; Rod is one-twelfth when centered, and one-third when cornered. Thin Ring (Axis perpendicular to plane, through center) I = MR 2 k = R Mass M , Radius R Thin Disc (Axis perpendicular to plane, through center) I = 1 2 MR 2 k = R 2 Mass M , Radius R Solid Sphere (Axis through diameter) I = 2 5 MR 2 k = 2 5 R Mass M , Radius R Hollow Sphere (Axis through diameter) I = 2 3 MR 2 k = 2 3 R Mass M , Radius R Thin Rod (Axis perpendicular to length, through center) I = 1 12 ML 2 k = L 12 Mass M , Length L Thin Rod (Axis perpendicular to length, through one end) I = 1 3 ML 2 k = L 3 Mass M , Length L Solid Cylinder (Axis through its own symmetry axis) I = 1 2 MR 2 k = R 2 Mass M , Radius R Hollow Cylinder (Axis through its own symmetry axis) I = MR 2 k = R Mass M , Radius R Thin Ring (Axis through diameter) I = 1 2 MR 2 k = R 2 Mass M , Radius R Thin Disc (Axis through diameter) I = 1 4 MR 2 k = R 2 Mass M , Radius R Rectangular Plate (Axis perpendicular to plane, through center) I = M(a 2 + b 2) 12 k = a 2 + b 2 12 Sides a, b Annular Disc (Axis perpendicular to plane, through center) I = 1 2 M(R 1 2 + R 2 2) k = R 1 2 + R 2 2 2 Inner R 1 , Outer R 2 Solid Cylinder (Axis perpendicular to length, through center) I = M( R 2 4 + L 2 12 ) k = R 2 4 + L 2 12 Radius R , Length L moments of inertia constants Memorize with meaning: compare shapes pairwise. A ring has all mass at radius R, so I = MR 2 . A disc spreads mass inside, so its I is smaller, 1 2 MR 2 . A rod about its end is four times the rod about center: 1 3 ML 2 vs 1 12 ML 2 , consistent with the parallel axis theorem using d = L/2 . Ring MR 2; Disc 1/2 MR 2; Solid sphere 2/5 MR 2; Hollow sphere 2/3 MR 2; Rod center 1/12 ML 2; Rod end 1/3 ML 2; Plate (a×b) center 1/12 M( a 2 + b 2 ). Quick mental card for NEET last-minute recall. Using the theorems avoids fresh integration: shift the axis with the parallel axis theorem or rotate axes in a lamina with the perpendicular axis theorem. Combine with additivity: the MI of a composite body about an axis equals the sum of MI of the parts about the same axis. easy L = 1.2 m M = 0.60 kg Use standard results for a thin rod and the parallel axis theorem to relate end and center axes. kg m 2 Find the MI of a thin uniform rod of length L = 1.2 m and mass M = 0.60 kg about an axis perpendicular to the rod through (a) its center and (b) one end. (a) I center, (b) I end Check with intuition: the end axis places more mass farther from the axis, so MI must be larger. The factor increase from center to end for a uniform rod is 4. Use I cm = 1 2 MR 2 and shift by d = R using the parallel axis theorem. kg m 2 I tangent (axis perpendicular to plane and tangent to rim) A solid disc of radius R = 0.20 m and mass M = 2.0 kg rotates about a tangent axis perpendicular to its plane. Find its MI. medium R = 0.20 m M = 2.0 kg The tangent axis is parallel to the central symmetry axis, so the parallel axis theorem applies directly with d = R . a = 0.30 m b = 0.20 m M = 1.5 kg hard I about the in-plane axis through center and along b (i.e., x-axis if b is along x) A rectangular plate (a = 0.30 m, b = 0.20 m, M = 1.5 kg) rotates about an axis through its center and parallel to side b (i.e., in-plane axis along b). Find its MI. kg m 2 Use the perpendicular axis theorem to relate in-plane axes to the perpendicular (out-of-plane) axis. Composite bodies and transfer of axes For a composite body, compute each part’s MI about the same target axis and add them. If the part’s standard MI is about its own CM, shift it to the target axis using the parallel axis theorem. If you need an in-plane CM axis from a known perpendicular CM axis (for a lamina), use the perpendicular axis theorem first, then shift. medium m = 0.20 kg each L = 0.50 m Treat as two point masses at distance L/2 from the axis. kg m 2 Two small spheres, each of mass m = 0.20 kg, are attached at the ends of a light rod of length L = 0.50 m. Find the MI about an axis through the rod’s midpoint and perpendicular to the rod. I about the midpoint, perpendicular to the rod In rolling without slipping on an incline of angle , acceleration is a = g 1 + I/(MR 2) . Lower I/(MR 2) gives higher acceleration. That is why a solid sphere (with 2/5 ) beats a disc ( 1/2 ), which beats a ring ( 1 ) on the same slope. control M (mass) control I cm I (shifted) dependent Parallel axis theorem visual: I varies linearly with d 2 with slope equal to mass. dsq 2D PLOT I about CoM Icm Mass I = Icm + M dsq Moment of inertia vs d² (parallel-axis theorem) A straight line with slope M and intercept I cm illustrating I = I cm + M d 2 $. I about shifted axis kg m 2 I cm d = 0 → I = I cm d 2 Linear growth I cm + M d 2 custom m 2 d 2 (square of distance between parallel axes) The linear relation I = I cm + Md 2 means that if you plotted I against d 2 , you would get a straight line. The slope is the mass M and intercept is I cm . This is a useful way to validate data or estimate M or I cm experimentally. Angular position under constant angular acceleration: = 0 + 0 t + 1 2 t 2 (valid for fixed axis, constant , angles in radians). Angular acceleration is constant. Rotation is about a fixed axis. Angles are measured in radians. = 0 + 0 t + 1 2 t 2 = 0 + 0 t + 1 2 t 2 Use radians in rotational kinematics formulae. Plugging degrees makes the numerical value wrong by a factor of 180/π. neet-alert Moment of inertia is just the mass of the body. MI depends on both mass and how that mass is distributed relative to the axis; changing the axis changes I even if mass stays the same. It applies only to plane laminae (thin, planar bodies). It does not hold for 3D volumes. The perpendicular axis theorem can be used for any 3D body like a sphere or cylinder. Parallel axis theorem works between any two axes that intersect. The axes must be parallel. If they intersect or are skewed, the theorem does not apply directly. remember Additivity rule: I total about a given axis equals the sum of I of parts about the same axis. Use symmetry to simplify whenever possible. State the exact axis clearly (through which point, direction, in/out of plane). Pick a known standard MI about a convenient axis (often through CM). Use perpendicular axis (lamina) to switch between in-plane and out-of-plane CM axes if needed. Use parallel axis to shift to the required axis (add Md 2). For composites, add MI of parts after shifting each to the target axis. Double-check dimensions and limits (I ≥ 0, increases with distance from axis). How to compute I efficiently I = 1 2 MR 2 Uniform disc of radius R and mass M. Axis perpendicular to the plane through center. I of a solid disc about its central axis: I = 1 2 MR 2 The disc can be viewed as a stack of concentric rings. Each ring contributes dI = r 2 ,dm , and integration over radius naturally yields the 1 2 MR 2 result. Uniform thin rod of length L and mass M. Axis perpendicular to the rod through its center. I of a thin rod about its center: I = 1 12 ML 2 I = 1 12 ML 2 Using the parallel axis theorem with d = L/2 immediately gives the end-axis value: I end = I center + M(L/2) 2 = 1 3 ML 2 . Use I disc = 1/2 MR 2; then = I ; then rotational kinematics with constant $. rad s -2 , rad A rigid wheel (mass 3.0 kg, radius 0.25 m) is spun from rest by a constant torque of 1.5 N m about its axle. The wheel is a uniform disc. (a) Find its angular acceleration. (b) Find the angle turned in 2.0 s. (a) , (b) in 2.0 s Rotational dynamics + kinematics combo medium M = 3.0 kg R = 0.25 m = 1.5 N , m Initial 0 = 0 Angles in kinematics are dimensionless but must be in radians for formulae like = 0 + 0 t + 1/2 , t 2 . Do not plug degrees. neet-alert Practical check: If you replace the disc with a ring of the same M and R (so I = MR 2 ), the angular acceleration becomes smaller for the same torque: = /I . This matches the intuition that rings are "harder" to spin than discs. Unit sanity: MI has units of kg , m 2 . If your answer ends in kg , m or kg , m 3 , you likely missed a square or an integral factor. tip Sometimes, you are given the radius of gyration k instead of a shape formula. Then I = Mk 2 instantly. This is handy in lab problems where k is measured from oscillation periods. easy M = 4.0 kg k = 0.15 m = 10 rad , s -1 kg m 2 , J Use I = Mk 2 and K = 1/2 I 2. A body has mass 4.0 kg and radius of gyration 0.15 m about an axis. (a) Find its MI. (b) If it spins at 10 rad s -1 , find its rotational KE. (a) I, (b) K rot Edge awareness: MI grows without bound as the axis moves far away (via Md 2 ). In contrast, about the CM axis, MI is minimum among all parallel axes for a given direction. Among all axes parallel to a given direction, the axis through the center of mass gives the minimum I. remember Worked logic for plates: For a square plate (side a, mass M), about a central in-plane axis along a side, I = 1 12 Ma 2 . About the edge along the same direction, shift by d = a/2 to get I edge = 1 12 Ma 2 + M(a/2) 2 = 1 3 Ma 2 . Beware of mixing in-plane and out-of-plane axes. For a lamina, I z (out-of-plane) is usually larger than either I x or I y because it includes both x 2 and y 2 in the integrand. Radius of gyration k is the physical radius of the body. k depends on both shape and axis via k = I/M . For a disc about its central axis, k = R/ 2 , not R . Composite with cavity: For a disc with a circular hole (same center), subtract the MI of the missing part (treated as a disc of negative mass density) from the full disc MI about the same axis. Theorems still apply because integration is linear. R = 0.30 m r = 0.10 m M = 2.4 kg medium Find the MI of an annular disc (outer radius R = 0.30 m, inner radius r = 0.10 m, total mass M = 2.4 kg) about the central symmetry axis perpendicular to the plane. I about central axis kg m 2 Use areal density to split mass: I = (1/2) M out R 2 - (1/2) M in r 2 but with M fixed overall; or use I = (1/2)M( R 2 + r 2 ) for an annulus of uniform density. If axes do not pass through the same point, use the perpendicular axis theorem first (for a lamina) to relate in-plane CM axes to the out-of-plane CM axis. Only then apply the parallel axis theorem to shift to an edge or tangent. Do not apply the perpendicular axis theorem when the x and y axes do not intersect at the same point as the z axis. The theorem requires all three axes to be mutually perpendicular and concurrent. neet-alert Dimension and scaling checks prevent mistakes: If you scale all lengths by factor s at fixed density, mass scales as s 3 (for 3D bodies) and MI scales as s 5 (since I M L 2 ). For laminae (2D), mass scales as s 2 and MI as s 4 . Strategy reminder: Prefer CM axes first. Many standard formulas are tabulated at the CM. From there, the parallel axis theorem takes you to edge/tangent axes in one clean step. Rolling without slipping couples translation and rotation: v = R . The total kinetic energy is K = 1 2 Mv 2 + 1 2 I 2 . For a given v , larger I means more energy in rotation, leaving less for translation—hence different accelerations down the same incline. For symmetric bodies, principal axes through the center diagonalize MI: cross terms vanish and MI takes standard scalar forms. In NEET problems, symmetry is your best friend to avoid tensor complications. Edge case: For a point mass on the axis itself ( r=0 ), its MI contribution is zero. This often simplifies composite systems where some masses lie on the axis. Another edge case: As r , MI diverges like r 2 . This aligns with the parallel axis theorem’s Md 2 term dominating at large d . Common transfer: For a disc, the in-plane diameter axis MI is I diam = 1 4 MR 2 . A tangent in-plane axis parallel to the diameter adds Md 2 with d = R/2 from the diameter through center to the tangent through the edge: I = 1 4 MR 2 + M(R/2) 2 = 1 2 MR 2 . Experimental perspective: Plotting I vs d 2 using a trifilar or torsion pendulum measurement can extract both M and I cm from the line’s slope and intercept. Consistency check for a plate: Using I z = I x + I y and symmetry for a square ( I x = I y ), we get I x = I y = I z/2 . This shortcut frequently reduces algebra. Key terms recap Resistance to angular acceleration about a specified axis: I = m i r i 2 = r 2 ,dm Moment of inertia (I) Distance such that I = Mk 2 for a given axis Radius of gyration (k) I = I cm + Md 2 Parallel axis theorem For a lamina: I z = I x + I y Perpendicular axis theorem Torque Rotational effect of force, = I Rate of change of angular velocity, = d /dt Angular acceleration