Moment of inertia of disc about tangent parallel to diameter

Question

Find the moment of inertia of a uniform disc of mass $M$ and radius $R$ about a tangent drawn to the disc in its own plane (i.e., a tangent parallel to a diameter).

Solution — Step by Step

We start with a standard result that must be memorised:

The moment of inertia of a disc about its diameter (an axis through the centre, in the plane of the disc) is:

I_{diameter} = \frac{MR^2}{4}

This is derived from the perpendicular axes theorem: if $I_z = MR^2/2$ (about the axis through the centre, perpendicular to disc), then $I_x = I_y = MR^2/4$ by symmetry, and $I_x + I_y = I_z$ .

The Parallel Axes Theorem states: If $I_{CM}$ is the moment of inertia about an axis through the centre of mass, then the moment of inertia about any parallel axis at distance $d$ from the CM axis is:

I = I_{CM} + Md^2

We want the moment of inertia about a tangent that is parallel to a diameter and lies in the plane of the disc. This tangent axis is at the edge of the disc, so its distance from the diameter through the centre is $d = R$ (the radius).

The axis through the centre parallel to the tangent is the diameter. So:

I_{CM} = I_{diameter} = \frac{MR^2}{4}

d = R

I_{tangent} = I_{CM} + Md^2 = \frac{MR^2}{4} + MR^2 = \frac{MR^2}{4} + \frac{4MR^2}{4} = \frac{5MR^2}{4}

\boxed{I_{tangent} = \frac{5MR^2}{4}}

The moment of inertia of the disc about a tangent in its own plane is $\frac{5MR^2}{4}$ .

Why This Works

The Parallel Axes Theorem works because angular inertia increases when the rotation axis is moved away from the centre of mass. The $Md^2$ term represents the “extra” inertia due to the offset — as if the entire mass were concentrated at the CM, rotating about the new axis at distance $d$ .

The tangent is the farthest possible axis in the plane of the disc (from any point on the disc), which is why $d = R$ (exactly at the edge). This gives the maximum moment of inertia for any axis in the plane of the disc.

Standard disc moment of inertia results to memorise for JEE:

About axis perpendicular to disc through centre: $MR^2/2$
About diameter: $MR^2/4$
About tangent perpendicular to disc: $3MR^2/2$
About tangent in plane of disc: $5MR^2/4$

All of these can be derived using the perpendicular axes theorem and/or parallel axes theorem from just $I_{centre} = MR^2/2$ .

Alternative — Direct Application of Perpendicular Axis Theorem (Not needed here)

The perpendicular axes theorem states: for a planar body, $I_z = I_x + I_y$ where z is perpendicular to the plane and x, y are in the plane. This was used to derive $I_{diameter} = MR^2/4$ from $I_{perpendicular} = MR^2/2$ .

Common Mistake

The most common error is applying the parallel axis theorem with $d = 2R$ (diameter of disc) instead of $d = R$ (radius). The tangent is at a distance equal to the RADIUS from the centre, not the diameter. The distance from the axis of symmetry (at centre) to the tangent (at the edge) is simply $R$ . Using $d = 2R$ gives $I = MR^2/4 + 4MR^2 = 17MR^2/4$ — completely wrong.

Question

Solution — Step by Step

Why This Works

Alternative — Direct Application of Perpendicular Axis Theorem (Not needed here)

Common Mistake

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