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PhysicsMediumClass 11

Moment of Inertia — Standard bodies (ring, disc, rod, sphere)

Rotation

12

JEE Qs

8%

Hard

60

min

Thoroughly memorize standard MOI formulas and master the conditions for applying Parallel and Perpendicular Axis Theorems to solve complex problems efficiently.

🧮 Key Formulas

I = ∫r^2 dm
I_ring (axis through center, perpendicular to plane) = MR^2
I_ring (axis through diameter) = MR^2/2
I_disc (axis through center, perpendicular to plane) = MR^2/2
I_disc (axis through diameter) = MR^2/4
I_rod (axis through center, perpendicular to length) = ML^2/12
I_rod (axis through end, perpendicular to length) = ML^2/3
I_solid_sphere (axis through diameter) = 2MR^2/5
I_hollow_sphere (axis through diameter) = 2MR^2/3
Parallel Axis Theorem: I = I_CM + Md^2
Perpendicular Axis Theorem (for planar bodies): I_z = I_x + I_y

✅ Key Points for JEE

  • 1Memorize the Moment of Inertia (MOI) values for the standard bodies (ring, disc, rod, solid sphere, hollow sphere) about their principal axes (e.g., through CM).
  • 2Master the application of the Parallel Axis Theorem (I = I_CM + Md^2) to find MOI about any axis parallel to a known axis passing through the center of mass.
  • 3Understand and correctly apply the Perpendicular Axis Theorem (I_z = I_x + I_y) only for planar laminae (e.g., disc, ring, rectangular plate) when the three axes are mutually perpendicular and intersect at a common point, with two axes lying in the plane of the body.
  • 4MOI is a scalar quantity, but its value is critically dependent on the chosen axis of rotation and the distribution of mass relative to that axis. It is analogous to mass in translational motion.
  • 5For composite bodies, the total MOI about a given axis is the scalar sum of the MOIs of individual components about the same axis.

⚠️ Common Mistakes

  • Incorrectly recalling or confusing the standard MOI formulas for different shapes or for different axes of the same shape.
  • Applying the Parallel Axis Theorem using a reference MOI (I_ref) that is NOT about an axis passing through the center of mass (I_CM).
  • Incorrectly using the Perpendicular Axis Theorem for non-planar bodies or when the axes are not appropriately chosen (e.g., two axes not in the plane of the body).
  • Using the wrong 'd' in the Parallel Axis Theorem; 'd' must be the perpendicular distance between the two parallel axes, not from an arbitrary origin.

📝 Practice Questions

See all

Q28.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

2025·MCQMedium

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 _______ .

2025·NumericalHard

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

2025·MCQEasy

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

2025·NumericalHard

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

2025·NumericalMedium

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

2025·MCQMedium

NCERT Chapters

  • Class 11 Physics Ch 7: System of Particles and Rotational Motion