The centre of mass of a system of particles is the single point at which the entire mass of the system can be considered concentrated for the purpose of analysing its overall (translational) motion — it moves exactly as if all external forces acted directly on a single point particle of the system's total mass.
Similar formulas apply for $y_{cm}$ and $z_{cm}$ in three dimensions.
The centre of mass need not lie within the physical body itself — e.g., the centre of mass of a uniform ring or hollow sphere lies at its geometric centre, in empty space, not on the material of the object.
Two point masses, 2 kg and 3 kg, are placed at positions x = 0 m and x = 5 m respectively. Find the position of their centre of mass.
$x_{cm} = \dfrac{(2)(0) + (3)(5)}{2+3} = \dfrac{15}{5} = 3\ \text{m}$
The centre of mass of a system moves according to $\vec{F}_{ext} = M_{total}\vec{a}_{cm}$ — meaning only external forces affect its motion; internal forces between particles of the system (which always occur in equal-and-opposite pairs per Newton's third law) have no net effect on the centre of mass's motion, even though they can dramatically change the individual particles' motion relative to each other.
An artillery shell explodes mid-air into several fragments while in projectile motion. Explain what happens to the trajectory of the shell's centre of mass immediately after the explosion (ignoring air resistance).
The explosion is caused entirely by internal forces (the explosive charge pushing fragments apart) — no new external force is introduced by the explosion itself. Since only external forces (here, just gravity, exactly as before the explosion) affect the centre of mass's motion, the centre of mass of all the fragments continues along the exact same parabolic trajectory that the original, unexploded shell would have followed, even as the individual fragments scatter in different directions.
Torque ($\vec{\tau} = \vec{r} \times \vec{F}$) is the rotational analogue of force — it measures the tendency of a force to produce rotation about a chosen axis or pivot point. Magnitude: $\tau = rF\sin\theta = F \times (\text{perpendicular distance from axis to line of action of force})$.
A force applied far from the pivot, or applied perpendicular to the lever arm, produces greater torque than the same force applied close to the pivot or at an angle — this is why door handles are placed far from the hinge, and why pushing a door perpendicular to its surface is more effective than pushing at an angle.
A force of 20 N is applied at the end of a 0.5 m wrench, perpendicular to its length. Calculate the torque produced about the pivot (bolt).
$\tau = rF\sin\theta = (0.5)(20)\sin90° = (0.5)(20)(1) = 10\ \text{N m}$
A couple consists of two equal and opposite forces acting along different (parallel) lines, producing a net force of zero but a non-zero net torque — this is why a couple causes pure rotation without any translational motion, e.g., turning a steering wheel with both hands pushing in opposite directions.
Two forces of 10 N each, in opposite directions, act at the two ends of a steering wheel of diameter 0.4 m. Calculate the torque produced by this couple.
Torque of a couple $= F \times d$, where $d$ is the perpendicular distance between the two forces (here, the diameter):
$\tau = (10)(0.4) = 4\ \text{N m}$
Moment of inertia (I) is the rotational analogue of mass — it measures an object's resistance to a change in its rotational motion (angular acceleration) about a given axis. It depends not just on the mass of the object, but crucially on how that mass is distributed relative to the axis of rotation: $I = \sum m_ir_i^2$, where $r_i$ is each mass element's perpendicular distance from the axis.
| Object | Axis | Moment of Inertia |
|---|---|---|
| Thin ring | Through centre, perpendicular to plane | $MR^2$ |
| Solid disc/cylinder | Through centre, perpendicular to plane | $\tfrac{1}{2}MR^2$ |
| Solid sphere | Through centre (diameter) | $\tfrac{2}{5}MR^2$ |
| Thin rod | Through centre, perpendicular to length | $\tfrac{1}{12}ML^2$ |
| Thin rod | Through one end, perpendicular to length | $\tfrac{1}{3}ML^2$ |
Since $I \propto r^2$, mass located farther from the axis contributes disproportionately more to the moment of inertia than the same mass located closer in — this is why a ring (all mass at radius R) has a larger moment of inertia than a solid disc of the same mass and radius (mass spread from centre to R), even though both have identical total mass.
Compare the moment of inertia of a ring and a solid disc, both of mass 2 kg and radius 0.5 m, about an axis through the centre perpendicular to the plane.
Ring: $I_{ring} = MR^2 = (2)(0.5)^2 = 0.5\ \text{kg m}^2$
Disc: $I_{disc} = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(2)(0.5)^2 = 0.25\ \text{kg m}^2$
The ring has exactly twice the moment of inertia of the disc, despite identical mass and radius, since all the ring's mass sits at the maximum distance R from the axis, while the disc's mass is distributed from the centre outward.
Using the standard formulas, compare the moment of inertia of a thin rod of mass M and length L about an axis through its centre versus through one end.
Through centre: $I_{centre} = \tfrac{1}{12}ML^2$. Through one end: $I_{end} = \tfrac{1}{3}ML^2$
$\dfrac{I_{end}}{I_{centre}} = \dfrac{1/3}{1/12} = \dfrac{12}{3} = 4$ — the moment of inertia about the end is 4 times greater than about the centre, since more of the rod's mass is, on average, farther from an end-axis than from a centre-axis.
The radius of gyration (k) is the hypothetical distance from the axis at which the object's entire mass could be concentrated as a single point, while still producing the same moment of inertia: $I = Mk^2$, so $k = \sqrt{I/M}$. It provides a convenient single length-scale summarising how "spread out" an object's mass is relative to a given axis.
A disc of mass 4 kg has a moment of inertia of 2 kg m² about a given axis. Calculate its radius of gyration.
$k = \sqrt{I/M} = \sqrt{2/4} = \sqrt{0.5} \approx 0.71\ \text{m}$
The moment of inertia of a rigid body about any axis is equal to its moment of inertia about a parallel axis through its centre of mass, plus the product of its mass and the square of the perpendicular distance between the two axes.
where $I_{cm}$ = moment of inertia about the centre of mass axis, $d$ = distance between the two parallel axes.
A rod of mass 2 kg and length 1 m has a moment of inertia of $\tfrac{1}{12}ML^2$ about an axis through its centre. Use the parallel axis theorem to find its moment of inertia about an axis through one end.
$I_{cm} = \tfrac{1}{12}(2)(1)^2 = \tfrac{1}{6}\ \text{kg m}^2$. Distance from centre to end $= 0.5\ \text{m}$.
$I_{end} = I_{cm} + Md^2 = \tfrac{1}{6} + (2)(0.5)^2 = \tfrac{1}{6} + 0.5 = \tfrac{1}{6}+\tfrac{3}{6} = \tfrac{4}{6} = \tfrac{2}{3}\ \text{kg m}^2$
This matches the standard formula $\tfrac{1}{3}ML^2 = \tfrac{1}{3}(2)(1)^2 = \tfrac{2}{3}\ \text{kg m}^2$ — confirming the theorem.
For a planar (flat, 2D) object, the moment of inertia about an axis perpendicular to the plane of the object equals the sum of the moments of inertia about two mutually perpendicular axes lying within the plane, all three axes intersecting at the same point.
where the object lies in the x-y plane, and z is the axis perpendicular to this plane, all passing through a common point.
This theorem applies only to planar (flat, essentially 2D) objects — it cannot be used for 3D objects like solid spheres or cylinders, since it relies on the object lying entirely within a single plane.
A uniform circular disc has a moment of inertia of $\tfrac{1}{2}MR^2$ about an axis through its centre, perpendicular to its plane. Using the perpendicular axis theorem (and symmetry), find its moment of inertia about a diameter.
By symmetry, the moment of inertia about any diameter is the same: $I_x = I_y$. By the perpendicular axis theorem: $I_z = I_x + I_y = 2I_x$
$I_x = \dfrac{I_z}{2} = \dfrac{\tfrac{1}{2}MR^2}{2} = \tfrac{1}{4}MR^2$
Angular momentum ($\vec{L} = \vec{r} \times \vec{p} = I\vec{\omega}$) is the rotational analogue of linear momentum — for a rigid body rotating about a fixed axis, it equals the product of moment of inertia and angular velocity.
Just as force is the rate of change of linear momentum, torque is the rate of change of angular momentum: $\tau = \dfrac{dL}{dt}$ — this is the rotational analogue of Newton's second law.
A disc with moment of inertia 0.4 kg m² rotates with an angular velocity of 5 rad/s. Calculate its angular momentum.
$L = I\omega = (0.4)(5) = 2\ \text{kg m}^2\text{/s}$
Every linear kinematic quantity has a direct rotational counterpart, and the standard equations of motion carry over exactly, with $\theta$ (angular displacement), $\omega$ (angular velocity), and $\alpha$ (angular acceleration) replacing $s$, $v$, and $a$ respectively:
$$\omega = \omega_0 + \alpha t \qquad \theta = \omega_0t + \tfrac{1}{2}\alpha t^2 \qquad \omega^2 = \omega_0^2 + 2\alpha\theta$$A wheel starts from rest and reaches an angular velocity of 20 rad/s after undergoing an angular displacement of 50 rad. Calculate its angular acceleration.
Using $\omega^2 = \omega_0^2 + 2\alpha\theta$: $(20)^2 = 0 + 2\alpha(50) \Rightarrow 400 = 100\alpha \Rightarrow \alpha = 4\ \text{rad/s}^2$
In the absence of any net external torque, the total angular momentum of a system remains constant: $I_1\omega_1 = I_2\omega_2$. This is why a figure skater spins faster when pulling their arms in (decreasing I) and slower when extending their arms outward (increasing I) — angular momentum stays the same, so angular velocity adjusts inversely with moment of inertia.
A figure skater spinning at 2 rad/s with arms extended (moment of inertia 5 kg m²) pulls their arms in, reducing their moment of inertia to 2 kg m². Calculate their new angular velocity.
By conservation of angular momentum: $I_1\omega_1 = I_2\omega_2$
$(5)(2) = (2)\omega_2 \Rightarrow \omega_2 = \dfrac{10}{2} = 5\ \text{rad/s}$
When an object rolls without slipping, its centre of mass velocity is directly related to its angular velocity: $v_{cm} = R\omega$. The point of the rolling object in contact with the ground is instantaneously at rest at any given moment (its velocity relative to the ground is zero), even though the rest of the object is moving.
Explain why the point of a rolling wheel touching the ground is momentarily at rest, even as the wheel moves forward.
The contact point has two velocity contributions: the translational velocity of the centre of mass ($v_{cm}$, forward) and the rotational velocity due to spinning ($R\omega$, backward relative to the centre, at the bottom of the wheel). For rolling without slipping, $v_{cm} = R\omega$ exactly — meaning these two contributions are equal in magnitude but opposite in direction at the contact point, cancelling to give zero net velocity there relative to the ground at that instant.
A rolling object has both translational kinetic energy (motion of its centre of mass) and rotational kinetic energy (spinning about its centre of mass) simultaneously:
$$KE_{total} = \tfrac{1}{2}Mv_{cm}^2 + \tfrac{1}{2}I\omega^2$$A rigid body is in complete (mechanical) equilibrium only when both conditions are satisfied: (1) the net external force is zero (translational equilibrium, no linear acceleration), and (2) the net external torque about any point is zero (rotational equilibrium, no angular acceleration). Satisfying only one condition is not sufficient for full equilibrium.
A solid sphere of mass 2 kg and radius 0.1 m rolls without slipping with a centre-of-mass speed of 4 m/s. Calculate its total kinetic energy. (I for a solid sphere = ⅖MR²)
$\omega = v_{cm}/R = 4/0.1 = 40\ \text{rad/s}$
$I = \tfrac{2}{5}MR^2 = \tfrac{2}{5}(2)(0.1)^2 = 0.008\ \text{kg m}^2$
$KE_{total} = \tfrac{1}{2}Mv_{cm}^2 + \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}(2)(4)^2 + \tfrac{1}{2}(0.008)(40)^2 = 16 + 6.4 = 22.4\ \text{J}$