Class 11 > Unit # 04: Rotational & Circular Motion > Moment of Inertia

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Unit No. 4 Rotational and Circular Motion - Moment of Inertia & Two-Particle Systems

Define Moment of Inertia, Its formula and Unit.

Moment of Inertia

"Moment of inertia is the property of a body by virtue of which it resists angular acceleration. It is defined as the sum of the products of the mass of each particle in the body and the square of its perpendicular distance from the axis of rotation."

Also known as rotational inertia or angular mass, it quantifies how much torque is required for a specific angular acceleration about a rotational axis. A higher moment of inertia means the body is more resistant to changes in its rotational state.

Formula

Mathematically, the moment of inertia ($I$) for a collection of point masses or an extended body is expressed as:

$I = \sum m_i r_i^2 \quad \text{or} \quad I = m_1 r_1^2 + m_2 r_2^2 + \dots$

Where:
• $I$ = Moment of Inertia
• $m$ = Mass of each particle
• $r$ = Perpendicular distance from the axis of rotation

Unit and Dimension

• SI Unit: Kilogram square meter ($\text{kg}\cdot\text{m}^2$)
• Dimensional Formula: $[\text{ML}^2]$

Factors on Which Moment of Inertia Depends

1. Shape and size of the body
2. Density (mass distribution) of the body
3. Orientation and position of the axis of rotation relative to the mass distribution

Determine the moment of inertia of a two-mass system when (i) Axis of rotation through the center (ii) Axis of rotation is at the end.

Consider a rigid dumbbell-like system consisting of two identical point masses, each of mass $m$, connected by a light massless rod of total length $L$.

(i) When Axis of Rotation is at the Center

When the axis of rotation passes perpendicularly through the exact center of the connecting rod, each mass is situated at a perpendicular distance of $r = \frac{L}{2}$ from the rotational axis.

Applying the moment of inertia formula for both particles ($m_1 = m_2 = m$):

$I = m_1 r_1^2 + m_2 r_2^2$

$I = m\left(\frac{L}{2}\right)^2 + m\left(\frac{L}{2}\right)^2 = m\frac{L^2}{4} + m\frac{L^2}{4}$

$I = \frac{1}{2} m L^2$

(ii) When Axis of Rotation is at the End

When the axis of rotation passes perpendicularly through one of the ends (e.g., the left end) of the rod, the perpendicular distances for the two masses are $r_1 = 0$ (for the mass at the axis) and $r_2 = L$ (for the mass at the opposite end).

Calculating the total moment of inertia:

$I = m_1 r_1^2 + m_2 r_2^2$

$I = m(0)^2 + m(L)^2 = 0 + mL^2$

$I = m L^2$

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