Moment of InertiaJEE Main

Rotational mass of common rigid bodies — interactive Physics simulation for IIT-JEE.

Concept

Moment of inertia $I$ is rotational mass — how hard it is to angularly accelerate a body. It depends not just on mass but on how far that mass sits from the axis. The same $M$ and $R$ give very different $I$ for a ring versus a solid sphere.

Key formula

I = \sum m_i r_i^2 = cMR^2, \qquad \tau = I\alpha, \qquad k = \sqrt{I/M}

Derivation

Each mass element contributes $dm\,r^2$ . Integrating over the body gives $I = cMR^2$ with a shape-dependent coefficient $c$ :

- Solid sphere $c = \tfrac25$ , hollow sphere $\tfrac23$ - Disc/cylinder $\tfrac12$ , ring $1$ - Rod about centre $\tfrac1{12}$ (uses $L$ ), about end $\tfrac13$

Newton's law for rotation is $\tau = I\alpha$ , the rotational analogue of $F = ma$ .

Scenarios to explore

Moment of Inertia — Rotational mass of spheres, discs, rings & rods.

Real-world applications

Flywheels store rotational energy ( $\tfrac12 I\omega^2$ ).
Why a hollow cylinder rolls down a slope slower than a solid one.
Figure skaters spinning faster by pulling their arms in (lower $I$ ).

JEE exam tips

Memorise the coefficients; most rotation problems reduce to picking the right $c$ .
Parallel-axis ( $+Md^2$ ) and perpendicular-axis ( $I_z = I_x + I_y$ for laminae) theorems are exam staples.

Common mistakes

Using the rod's length where a radius is expected (or vice-versa).
Forgetting the parallel-axis theorem $I = I_{cm} + Md^2$ for shifted axes.
Assuming two bodies of equal mass have equal $I$ — distribution matters.

Exam traps to avoid

Rod about its end ( $\tfrac13ML^2$ ) is four times the value about its centre ( $\tfrac1{12}ML^2$ ).
Radius of gyration $k$ is not the physical radius — it is $\sqrt{I/M}$ .