Rotational motion explains how bodies rotate about an axis. The chapter covers moment of inertia, its calculation for point masses and extended bodies, and standard formulas for different shapes.
The Parallel Axis Theorem and Perpendicular Axis Theorem are also important concepts in this topic. Correctly identifying the axis of rotation and the distance from it helps students solve numerical problems for NEET, JEE Main, and JEE Advanced.
Moment of inertia is the rotational analogue of inertia. In translational motion, a body resists changes in its state of motion. Similarly, a rotating body tends to continue rotating unless an external torque changes its motion.
For a point mass:
I = m r^2
Here, m is the mass and r is the perpendicular distance from the axis of rotation.
For a system of discrete particles:
I = Sum of m_i r_i^2
The distance must always be measured perpendicular to the axis. Since the distance is squared, its sign does not affect the result. For example, a distance of -4 is used as 4 because (-4)^2 = 16.
The moment of inertia of a point mass is zero when the mass lies directly on the axis because r = 0.
To calculate the moment of inertia about any specified axis:
Find the perpendicular distance of every mass from the axis.
Calculate m_i r_i^2 for each mass.
Add all contributions.
For the x-axis, distances are measured vertically from the x-axis. For the y-axis, distances are measured horizontally from the y-axis. For the z-axis, the perpendicular distance must be found geometrically.
The central principle remains:
I = Sum of m_i r_i^2
A particle located on the selected axis contributes zero.
Three equal masses m are placed at the vertices of an equilateral triangle of side l.
For the x-axis, two masses may lie on the axis and contribute zero. The third mass is at a perpendicular distance of √3 l / 2. Therefore:
I_x = m (√3 l / 2)^2
For the y-axis, one mass lies on the axis, while the other two are at distances l and l / 2:
I_y = m l^2 + m (l / 2)^2
For the z-axis perpendicular to the triangle’s plane:
I_z = m l^2 + m l^2
For this planar arrangement:
I_x + I_y = I_z
This is a common JEE Main pattern and directly demonstrates the Perpendicular Axis Theorem.
For an extended body, divide it into small elements and integrate their contributions.
For a uniform rod of mass M and length l, a small element of length dx at distance x has mass:
dm = (M / l) dx
Its differential moment of inertia is:
dI = x^2 dm
Thus:
I = Integral of x^2 dm
For an axis through one end and perpendicular to the rod:
I = M l^2 / 3
For a nonuniform rod, the linear density must remain inside the integration. If the density is:
Lambda = Lambda_0 x
then:
dm = Lambda_0 x dx
and:
I = Integral of Lambda_0 x^3 dx = Lambda_0 l^4 / 4
The main rule is that position-dependent density cannot be treated as a constant.
The moment of inertia of a thin rod depends on the position and direction of the axis of rotation. The following table shows the standard results for a rod of mass M:
|
Axis |
Moment Of Inertia |
|---|---|
|
Through one end and perpendicular to the rod |
M l^2 / 3 |
|
Through the centre of mass and perpendicular to the rod |
M l^2 / 12 |
|
Along the rod’s own axis |
0 |
The result for the rod’s own axis is zero because the rod is considered thin and its mass lies on the axis.
If a uniform rod of length l makes an angle theta with the axis, the perpendicular distance of an element at distance x is:
r = x sin theta
Therefore:
I = M l^2 sin^2 theta / 3
A useful shortcut is to use the projected length:
Effective length = l sin theta
Then apply the end-axis rod formula:
I = M (Effective length)^2 / 3
For a system of rods, calculate the moment of inertia of each rod about the same axis and add the results.
The following results should generally be remembered for examinations.
|
Body Or Configuration |
Moment Of Inertia |
|---|---|
|
Point mass or mass concentrated at distance R |
M R^2 |
|
Thin ring about central axis |
M R^2 |
|
Hollow cylinder about central axis |
M R^2 |
|
Disc about central axis |
M R^2 / 2 |
|
Solid cylinder about central axis |
M R^2 / 2 |
|
Solid cone about symmetry axis |
3 M R^2 / 10 |
|
Hollow cone about symmetry axis |
M R^2 / 2 |
|
Solid sphere about diameter |
2 M R^2 / 5 |
|
Hollow sphere about diameter |
2 M R^2 / 3 |
|
Thin rod about one end |
M l^2 / 3 |
|
Thin rod about centre |
M l^2 / 12 |
The Parallel Axis Theorem relates the moment of inertia about an arbitrary axis to the moment of inertia about a parallel axis through the centre of mass.
I = I_C + M D^2
Here, I_C is the moment of inertia about the centre-of-mass axis, M is the total mass, and D is the separation between the parallel axes.
The two axes must be parallel, and the reference axis for I_C must pass through the centre of mass.
A direct consequence is that the moment of inertia is minimum about an axis passing through the centre of mass.
For example, for a ring about a tangent axis:
I = M R^2 + M R^2 = 2 M R^2
For a disc about an axis parallel to its central axis:
I = M R^2 / 2 + M D^2
The Perpendicular Axis Theorem applies only to a two-dimensional planar body such as a disc, ring, plate, sheet, or square lamina.
If x and y are mutually perpendicular axes in the plane and z is perpendicular to that plane through their intersection:
I_z = I_x + I_y
The theorem must not be applied directly to three-dimensional bodies such as solid spheres and hollow spheres.
For a ring:
I_z = M R^2
By symmetry, the moments about two perpendicular diameters are equal:
I_x = I_y = M R^2 / 2
For a disc:
I_z = M R^2 / 2
Therefore, about any diameter through its centre:
I_x = I_y = M R^2 / 4
For multiple bodies:
I_total = I_1 + I_2 + I_3 + …
Calculate the contribution of each part about the required axis. Apply the Parallel Axis Theorem when a component’s central axis does not match the required axis.
For a removed part, treat it as negative mass:
I_remaining = I_complete body - I_removed part
This method is useful for discs with circular portions removed. For identical repeated components, calculate one contribution and multiply it by the number of components.
Students can make mistakes in Rotational Motion by choosing the wrong axis, using incorrect formulas, or applying the axis theorems incorrectly. Keep the following points in mind while solving numerical problems.
Using the wrong distance: Always measure the perpendicular distance of a mass from the axis of rotation when calculating moment of inertia.
Applying the Parallel Axis Theorem incorrectly: The two axes must be parallel, and the reference axis must pass through the centre of mass.
Misusing the Perpendicular Axis Theorem: Apply it only to planar bodies, not directly to three-dimensional bodies such as solid spheres.
Confusing standard formulas: Check the body's shape, mass distribution, and axis of rotation before selecting a formula.
Ignoring the axis position: The moment of inertia changes when the axis shifts. For a rod, the values differ for an axis through its centre and one through its end.
Adding removed parts instead of subtracting them: When calculating the moment of inertia of a body with a portion removed, subtract the removed part's contribution from that of the complete body.
Understanding moment of inertia, standard formulas, and the Parallel and Perpendicular Axis Theorems is important for solving Rotational Motion problems. Revise these concepts and practise numerical questions to strengthen your preparation for NEET.
The formula is I = m r^2, where r is the perpendicular distance from the axis of rotation.
It is zero when the particle lies directly on the axis because its perpendicular distance is zero.