Parallel Axis Theorem & Perpendicular Axis Theorem
Rotation
12
JEE Qs
8%
Hard
60
min
Master the correct application conditions for each theorem and accurately identify the center of mass and perpendicular distances to avoid common pitfalls in MOI calculations.
🧮 Key Formulas
✅ Key Points for JEE
- 1Parallel Axis Theorem is applicable for *any* rigid body to find its Moment of Inertia (MOI) about an axis parallel to an axis passing through its center of mass (CM). I_CM is always the minimum MOI for a given set of parallel axes.
- 2Perpendicular Axis Theorem is strictly applicable *only* for planar (2D) bodies or thin laminae. The three axes must be mutually perpendicular and pass through a common point, with two axes (x, y) lying *in the plane* of the body.
- 3These theorems significantly simplify the calculation of MOI for complex axes without resorting to integration from scratch, provided the MOI about the center of mass (I_CM) or MOI about two planar axes (I_x, I_y) are known.
- 4Always ensure that the I_CM used in the Parallel Axis Theorem refers to the moment of inertia about an axis that *actually* passes through the center of mass of the body.
- 5Memorizing the standard MOI formulas for common uniform rigid bodies (e.g., ring, disc, rod, sphere, cylinder) about their symmetry axes and CM is crucial as these form the basis for applying the theorems.
⚠️ Common Mistakes
- ✕Applying the Parallel Axis Theorem incorrectly by not using the MOI about the center of mass (I_CM) as the reference, or by using 'd' as an incorrect distance (it must be the perpendicular distance between the two parallel axes).
- ✕Applying the Perpendicular Axis Theorem to 3D bodies (e.g., solid spheres, cubes) instead of only planar bodies or thin laminae.
- ✕Not ensuring that the two axes (I_x, I_y) for the Perpendicular Axis Theorem lie *in the plane* of the body, or that all three axes (x, y, z) are mutually perpendicular and intersect at a common point.
📝 Practice Questions
See allQ28.A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in 2025 (22 Jan Shift 1) JEE Main Previous Year Paper the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above. (1) 32 7 MR2 (2) 329 MR2 (3) 17 32 MR2 (4) 1332 MR2
Q48.The position vectors of two 1 kg particles, (A) and (B), are given by→rA = (α1t2^i + α2t^j + α3t^k)m , respectively; →rB = (β1t^i + β2t2^j + β3t^k)m (α1 = 1 m/s2, α2 = 3nm/s, α3 = 2 m/s, β1 = 2 m/s, β2 = −1 m/s2, β3 = 4pm/s), where t is time, n and p −→ are constants. At t = 1 s, VA = →VB and velocities →VA and →VB of the particles are orthogonal to each other. At t = 1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is √Lkgm2 s−1 . The value of L is _______ .
Q28.The torque due to the force (2^i + ^j + 2^k) about the origin, acting on a particle whose position vector is (^i + ^j + ^k), would be (1) ^i −^k (2) ^i + ^k (3) ^j + ^k (4) ^i −^j + ^k
Q49.A tube of length 1 m is filled completely with an ideal liquid of mass 2 M , and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is F then angular velocity of the tube is in SI unit. The value of α is __________. √FαM
Q45.A solid sphere of mass ' m ' and radius ' r ' is allowed to roll without slipping from the highest point of an inclined plane of length ' L ' and makes an angle 30∘ with the horizontal. The speed of the particle at the bottom of the plane is v1 . If the angle of inclination is increased to 45∘ while keeping L constant. Then the new speed of the sphere at the bottom of the plane is v2 . The ratio v21 : v22 is (1) 1 : √2 (2) 1 : √3 (3) 1 : 3 (4) 1 : 2 2025 (23 Jan Shift 1) JEE Main Previous Year Paper
Q40.A uniform solid cylinder of mass ' m ' and radius ' r ' rolls along an inclined rough plane of inclination 45∘ . If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be (1) 1 g (2) 1 g √2 3√2 (3) √2 g (4) √2g 3
NCERT Chapters
- Class 11 Physics Ch 7: System of Particles and Rotational Motion