🔄 Rotation of rigid bodies

🎯 The angular quantities

To describe rotation we measure angles in radians (rad): one full revolution equals $2\pi$ rad $\approx 6.28$ rad.

• Rotation angle $\theta$ — how much the body has rotated (analogous to the displacement $s$).
• Angular velocity $\omega$ — how fast the angle changes (analogous to $v$).
• Angular acceleration $\alpha$ — how fast $\omega$ changes (analogous to $a$).

where: $\theta$ = angle (rad)  ·  $\omega$ = angular velocity (rad/s)  ·  $\alpha$ = angular acceleration (rad/s²)  ·  $\Delta t$ = time interval (s).

Uniformly accelerated rotational motion

If $\alpha$ is constant, the equations are identical to those of uniformly accelerated linear motion — with $\theta,\ \omega,\ \alpha$ in place of $s,\ v,\ a$:

where: $\omega_0$ = initial angular velocity  ·  $\theta_0$ = initial angle  ·  $t$ = time (s)  ·  $\Delta\theta=\theta-\theta_0$.

The link with linear quantities

A point of the body at distance $r$ from the axis travels an arc of length $s = r\,\theta$. Differentiating with respect to time gives the tangential velocity and acceleration:

where: $r$ = distance from the axis (m)  ·  $s$ = arc travelled (m)  ·  $v$ = tangential velocity (m/s)  ·  $a_t$ = tangential acceleration (m/s²).

Moreover, even when moving with constant $\omega$, the point has a centripetal acceleration directed toward the axis:

where: $a_c$ = centripetal acceleration (m/s²), directed from the point toward the axis of rotation.

Diagram: angle θ, tangential velocity v, centripetal acceleration a꜀ and the direction of ω.

⚙️ What makes a body rotate

The torque $M$ of a force

A force makes a body rotate only if it is applied «with leverage». The torque measures this ability: it depends on the force $F$, on the distance $r$ from the point of application to the axis and on the angle $\varphi$ between them.

where: $M$ = torque (N·m)  ·  $F$ = force (N)  ·  $r$ = distance from the axis (m)  ·  $\varphi$ = angle between $\vec r$ and $\vec F$  ·  $b = r\,\sin\varphi$ = lever arm (distance of the axis from the line of action, m).

The torque is maximum when the force is perpendicular ($\varphi = 90^\circ$), and zero if it is directed toward the axis.

The moment of inertia $I$

Rotational inertia does not depend only on the mass, but on how it is distributed around the axis: mass far from the axis «weighs» more.

where: $I$ = moment of inertia (kg·m²)  ·  $m_i$ = mass of the $i$-th element  ·  $r_i$ = its distance from the axis (m).

The 2nd law of rotational dynamics

The total torque produces a proportional angular acceleration, with $I$ as the «inertia». It is an exact copy of $F = m\,a$:

The angular momentum $L$

The analogue of momentum $p = m\,v$ is the angular momentum $L = I\,\omega$. And just as $F = \Delta p/\Delta t$, we have:

Finally, the rotational kinetic energy mirrors $\tfrac{1}{2}m\,v^{2}$:

where: $L$ = angular momentum (kg·m²/s)  ·  $E_{c,\text{rot}}$ = rotational kinetic energy (J)  ·  $\Delta t$ = time interval (s).

Diagram: a force F applied at distance r generates a torque M that accelerates the rotation (α).

⛰️ Bodies connected by a string over a rotating pulley

Two bodies are connected by an inextensible, massless string that runs over a pulley. If the pulley were ideal (massless) the tension would be the same at both ends. But a real pulley is a rotating disk: setting it spinning requires a torque, so the two tensions are not equal.

The three equations of the system

Consider a mass $m_1$ on an inclined plane (angle $\theta_1$, friction $\mu$) connected to a hanging mass $m_2$. If $m_2$ descends, the system accelerates with acceleration $a$ that is the same for the two bodies (inextensible string), and the pulley (disk, mass $M_p$, radius $R$) rotates with $\alpha = a/R$.

From the last equation: $\;T_2 - T_1 = \tfrac{1}{2}\,M_p\,a$. Adding the three equations the tension cancels out and we obtain:

where: $a$ = acceleration (m/s²)  ·  $T_1,T_2$ = string tensions (N)  ·  $f_a=\mu\,m_1 g\cos\theta_1$ = friction (N)  ·  $M_p$ = pulley mass (kg)  ·  $R$ = pulley radius (m)  ·  $g=9.81$ m/s².

The term $\tfrac{1}{2}M_p$ in the denominator is the pulley's inertia contribution: it is equivalent to adding «half its mass» to the system. With two inclined planes ($\theta_1$ and $\theta_2$) you just replace the drive from $m_2$ with $m_2\,g\sin\theta_2$.

Diagram: the difference T₂ − T₁ = ½M_p·a makes the pulley-disk rotate.

⏱️ What happens if a torque acts for a time $\Delta t$

A rigid body is already rotating with angular velocity $\omega_0$. If we apply a torque $M$ to it for a time interval $\Delta t$, its angular momentum changes. This is the angular impulse theorem, an exact copy of the impulse theorem ($F\,\Delta t = \Delta p$):

The quantity $M\,\Delta t$ is called the angular impulse. If $M$ is constant, dividing by $I$ we recover $\alpha = M/I$ and therefore:

where: $M$ = applied torque (N·m)  ·  $\Delta t$ = duration (s)  ·  $\Delta L$ = change in angular momentum (kg·m²/s)  ·  $\omega_0,\ \omega_f$ = initial and final angular velocity (rad/s)  ·  $I$ = moment of inertia (kg·m²).

What effects?

• If $M$ is in the same direction as the rotation → the body speeds up ($\omega$ increases, $L$ increases).
• If $M$ is opposite (e.g. a brake) → the body slows down, can stop and even reverse direction.
• To stop a flywheel with angular momentum $L$ in a time $\Delta t$ you need an average torque $M = L/\Delta t$: the longer $\Delta t$, the smaller the torque needed (like the airbag for momentum).

Diagram: a torque M applied for Δt changes the angular momentum by ΔL = M·Δt.

💫 When angular momentum is conserved

From the relation $M = \Delta L/\Delta t$ a very powerful principle follows immediately: if the total external torque is zero, the angular momentum is conserved.

The body can, however, change shape and thus its own moment of inertia $I$. Since $I\,\omega$ stays constant, if $I$ decreases then $\omega$ increases (and vice versa):

where: $L$ = angular momentum (kg·m²/s)  ·  $I$ = moment of inertia (kg·m²)  ·  $\omega$ = angular velocity (rad/s)  ·  the indices $1,2$ denote before and after the change of shape.

🎡 Person on a rotating platform

A person on a rotating platform (negligible axle friction) moves toward the center: their $I$ decreases (the mass is closer to the axis), so $\omega$ increases and the platform spins faster. If they move away, it slows down. The angular momentum does not change, but the kinetic energy $\tfrac{1}{2}I\,\omega^{2}$ increases: it is supplied by the muscular work done to pull inward toward the center.

💃 The dancer's pirouette

Same principle: the dancer (or the skater) starts with arms out ($I$ large, $\omega$ small) and, by pulling the arms in, reduces $I$ and speeds up the pirouette. By opening them again, they slow down to stop gracefully.

🎯 A rod-pendulum struck by a body (full treatment)

A rigid rod (mass $M$, length $L$), pivoted at one end and initially at rest, is struck at the free end by a body (mass $m$) with velocity $v$ perpendicular to the rod, and stays embedded in it.

① During the collision — the collision is very brief: the angular momentum about the pivot is conserved (the impulsive pivot forces have zero lever arm). Right after the collision:

② After the collision — the rod (with the body embedded) swings upward like a compound pendulum: now the torque of gravity acts, producing an angular acceleration. When the rod is tilted by $\varphi$ from the vertical:

where: $M,\ L$ = mass and length of the rod  ·  $m,\ v$ = mass and velocity of the body  ·  $\omega$ = angular velocity right after the collision  ·  $\varphi$ = angle of the rod from the vertical  ·  the «$-$» sign indicates that gravity slows the rise.

At the instant of the collision ($\varphi \approx 0$) the angular acceleration is almost zero; it grows as the rod rises, slowing the rotation up to the maximum height.

Left: platform/dancer (I·ω constant). Right: rod-pendulum struck at the end.

🎳 When a body translates and rotates at the same instant

A wheel rolling forward, a ball rolling: the body translates (its center of mass moves) and at the same time rotates about the center of mass. The velocity of any point is the sum:

Pure rolling (without slipping)

If the body rolls without slipping, the point in contact with the ground at that instant is at rest. From this come the two fundamental links between linear and angular quantities:

where: $v_\text{cm}$ = velocity of the center of mass (m/s)  ·  $a_\text{cm}$ = its acceleration (m/s²)  ·  $\omega$ = angular velocity (rad/s)  ·  $R$ = radius (m).

Consequence for the notable points: the contact point has velocity $0$, the center has velocity $v_\text{cm}$, the top has double velocity $2\,v_\text{cm}$. It is as if, at every instant, the body were rotating about the contact point (instantaneous axis of rotation).

Kinetic energy: two contributions

The kinetic energy is the sum of the translational and the rotational ones:

Writing $I_\text{cm} = c\,m\,R^{2}$ (with $c$ depending only on the shape) and using $v = \omega R$:

Rolling down an inclined plane

A body that rolls without slipping down a plane of angle $\theta$ has a center-of-mass acceleration:

Surprise: it depends neither on the mass nor on the radius, but only on the shape (that is, on $c$)! So in a «race» down the plane the body with the smallest $c$ arrives first:

With energy conservation (starting from a height drop $h$) we also find the velocity at the bottom:

Diagram: in pure rolling the contact point is at rest, the center goes at v, the top at 2·v.