Physics · System Of Particles And Rotational Motion · NEET
The parallel axis theorem says the moment of inertia about any axis equals the moment of inertia about a parallel axis through the centre of mass plus M times the square of the distance between the two axes: I = I_cm + M d^2. It lets you shift a known I_cm to any parallel axis without new integration. Memory hook: "Move away from the centre, you always ADD M d-squared" (I is always larger away from the centre of mass).
The moment of inertia I about the red axis equals I_cm about the parallel green axis through the centre of mass, plus M d^2, where d is the perpendicular distance between the two axes. Since M d^2 is always positive, I is minimum at the centre of mass.
Your doubts, answered
What does each symbol in I = I_cm + M d^2 actually mean?
I is the moment of inertia about the axis you want. I_cm is the moment of inertia about a parallel axis passing through the centre of mass (COM). M is the total mass of the body. d is the perpendicular distance between the two parallel axes. Example: a rod of mass M and length L has I_cm = ML^2/12 about its centre. About one end, d = L/2, so I = ML^2/12 + M(L/2)^2 = ML^2/12 + ML^2/4 = ML^2/3.
Does the parallel axis theorem work for ANY axis, or only special ones?
It works for any axis, but with ONE strict condition: the new axis must be parallel to an axis that passes through the centre of mass. You cannot shift from just any axis to any other axis. You must always start from the COM axis. If the two axes are parallel but neither passes through the COM, you cannot apply the theorem directly in one step.
Why do we always ADD M d^2 and never subtract it?
Because the moment of inertia is smallest about the axis through the centre of mass. Moving the axis away from the COM always spreads the mass farther from the axis, so I only grows. That is why d^2 (always positive) is always added. Trap: if a problem gives you I about an off-centre axis and asks for I_cm, you must SUBTRACT: I_cm = I - M d^2.
When exactly should I reach for the parallel axis theorem in a NEET problem?
Use it whenever the axis does NOT pass through the centre of mass but you know (or can look up) the standard I_cm. Common cases: a rod about its end, a disc/ring about a tangent, a sphere about a tangent, or a body hanging/pivoted at a point. If the axis already passes through the COM, you do not need it.
How is this different from the perpendicular axis theorem?
Parallel axis theorem shifts an axis sideways (to a parallel line) and works for any solid 3D body: I = I_cm + M d^2. Perpendicular axis theorem relates three mutually perpendicular axes and works ONLY for flat (planar) laminae: I_z = I_x + I_y. Different job, different condition.
⚠️ The NEET trap ✗ Shifting the moment of inertia between two off-centre parallel axes directly, e.g. going from the end of a rod to a point L/4 away by adding M(L/4)^2 to the end value. ✓ Always route through the centre of mass. First get I_cm, then apply I = I_cm + M d^2 to the target axis. The d in the formula is measured from the COM axis, not from any arbitrary axis. 🧠 d is measured FROM the centre of mass, never between two random axes.
Real NEET questions
ReNEET 2026
A solid sphere A of radius R and mass M is attached at a point to a smaller solid sphere B of radius r (r < R) and mass m (m < M), their line of centres horizontal. The moment of inertia of the system about a vertical axis through the centre of A is I_A, and about a vertical axis through the centre of B is I_B. The difference I_A - I_B is:
A · (M - m)(R + r)^2
B · (m - M)(R + r)^2 ✓
C · (m - M)(R - r)^2
D · 0
Solution: Step 1: The two spheres touch, so the distance between their centres is d = R + r.
Step 2: About the axis through centre of A, sphere A is on its own COM (I = (2/5)MR^2) and sphere B is shifted by d: I_A = (2/5)MR^2 + (2/5)mr^2 + m(R + r)^2.
Step 3: About the axis through centre of B, sphere B is on its own COM and sphere A is shifted by d: I_B = (2/5)MR^2 + (2/5)mr^2 + M(R + r)^2.
Step 4: Subtract. The two (2/5) self-terms cancel: I_A - I_B = m(R + r)^2 - M(R + r)^2 = (m - M)(R + r)^2. Answer: B.
NEET 2026
A thin wire of length L and linear mass density m (mass per unit length) is bent into a circular ring lying in the x-y plane with centre C. The moment of inertia of the ring about a tangential axis yy' lying in the plane of the ring is:
A · mL^3/(8 pi^2)
B · 3 m L^3/(8 pi^2) ✓
C · m L^3/(8 pi^3)
D · 3 m L^2/(8 pi^2)
Solution: Step 1: Total mass = mL. The circumference equals L, so 2(pi)r = L, giving radius r = L/(2 pi).
Step 2: Moment of inertia of a ring about a diameter (a line through the centre, in the plane) is I_d = (1/2)(mass)r^2 = (1/2)(mL)r^2.
Step 3: A tangent in the plane is parallel to a diameter, shifted out by d = r. Apply the parallel axis theorem: I = I_d + (mass)r^2 = (1/2)(mL)r^2 + (mL)r^2 = (3/2)(mL)r^2.
Step 4: Substitute r = L/(2 pi): I = (3/2)(mL) * L^2/(4 pi^2) = 3 m L^3/(8 pi^2). Answer: B.
Solved System Of Particles And Rotational Motion NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The moment of inertia about any axis equals the moment of inertia about a parallel axis through the centre of mass plus the total mass times the square of the perpendicular distance between the axes: I = I_cm + M d^2.
Can the parallel axis theorem be used for a 3D solid body?
Yes. Unlike the perpendicular axis theorem (planar bodies only), the parallel axis theorem holds for any rigid body, flat or solid, such as spheres, cylinders and cones.
What is the moment of inertia of a ring about a tangent perpendicular to its plane?
I_cm about the central perpendicular axis is MR^2, and the tangent is a distance d = R away, so I = MR^2 + MR^2 = 2MR^2.
What is the moment of inertia of a solid sphere about a tangent line?
I_cm about a diameter is (2/5)MR^2 and the tangent is d = R from the centre, so I = (2/5)MR^2 + MR^2 = (7/5)MR^2.
Why is the moment of inertia smallest about the centre of mass?
Because M d^2 is always positive, the term added when you move away from the COM is always positive, so I_cm is the minimum possible value for any set of parallel axes.
