Moment of Inertia + Parallel Axis Theorem
Rotation
52
JEE Qs
22%
Hard
90
min
Master the standard MOI formulas for common shapes and meticulously practice the correct application of the Parallel Axis Theorem, ensuring 'd' is always the distance from the CM axis.
🧮 Key Formulas
✅ Key Points for JEE
- 1Moment of inertia (MOI) quantifies an object's resistance to angular acceleration; it is the rotational analogue of mass.
- 2MOI depends critically on the axis of rotation and how mass is distributed relative to that axis.
- 3The Parallel Axis Theorem (I = I_CM + Md^2) is an indispensable tool for finding MOI about any axis parallel to one passing through the center of mass (CM).
- 4For discrete particles, MOI is a simple sum (sum(m_i * r_i^2)); for continuous bodies, it generally requires integration (integral(r^2 * dm)).
- 5Symmetry often allows direct use of standard formulas or significantly simplifies the integration process for complex shapes.
⚠️ Common Mistakes
- ✕Incorrectly applying the Parallel Axis Theorem, especially using 'd' as the distance between any two parallel axes, rather than specifically the distance between the desired axis and the *parallel axis passing through the Center of Mass*.
- ✕Errors in identifying or calculating the perpendicular distance 'r' from the mass element 'dm' or discrete mass 'm_i' to the axis of rotation.
- ✕Mistaking or misremembering standard MOI formulas for common shapes, or applying the wrong formula for a given axis (e.g., using a rod formula for a disk).
- ✕Incorrectly setting up the limits or the differential mass element 'dm' during integration for continuous bodies.
📝 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