Theory

Rigid-body dynamics

How the rotation of a rigid body is described, how torque, moment of inertia and angular momentum determine its motion, and when the body rolls without slipping or remains in equilibrium.

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Rigid body and angular quantities
Torque and angular momentum

Rigid body and angular quantities

1. Rigid body

A rigid body is a system of particles whose mutual distances remain constant during the motion. It is a model, adequate when the deformations of the real body are negligible. The mass can be concentrated at points or distributed with density $ρ$ ; in this case a volume element $d V$ has mass $d m = ρ d V$ and the sums over the points become integrals.

The motion is a translation if every segment joining two points of the body remains parallel to itself, and then all points have the same velocity. It is a rotation about a fixed axis if the points of a line remain at rest and every other point describes a circle centered on the axis. The most general motion decomposes into a translation of the center of mass and a rotation about an axis through the center of mass.

2. Angular position, velocity and acceleration

Consider a rigid body rotating about a fixed axis, chosen as the $z$ axis. The configuration of the body is determined by the angular position $θ$ , the angle in radians that a half-line fixed in the body and perpendicular to the axis forms with a fixed half-line. The angular velocity and the angular acceleration are

ω = \frac{d θ}{d t}, α = \frac{d ω}{d t} = \frac{d ^{2} θ}{d t ^{2}},

and they are measured in $rad/s$ and in $rad/ s^{2}$ . They are the same for all points of the body, so the rotation is described by a single equation of motion $θ (t)$ . We take $ω > 0$ for counterclockwise rotation seen from the positive side of the $z$ axis, and the angular velocity vector is $ω = ω \hat{k}$ .

When $α$ is constant, the laws of uniformly accelerated rectilinear motion hold, for which we refer to the topic on kinematics in one dimension, with $θ$ in place of $x$ :

ω = ω_{0} + α t, θ = θ_{0} + ω_{0} t + \frac{1}{2} α t^{2}, ω^{2} = ω_{0}^{2} + 2 α (θ - θ_{0}) .

A grinding wheel that starts from rest with $α = 4.0 rad/ s^{2}$ rotates after $5.0 s$ at $ω = 20 rad/s$ and has described the angle $θ - θ_{0} = 50 rad$ , that is $\frac{50}{2 π} \approx 8.0$ revolutions.

3. Velocity and acceleration of the points of the body

Let $P$ be a point of the body at distance $r$ from the axis of rotation. Then $P$ describes a circle of radius $r$ and

v = ω r, a_{t} = α r, a_{c} = ω^{2} r = \frac{v ^{2}}{r},

where $a_{t}$ is the tangential acceleration and $a_{c}$ the centripetal acceleration, directed toward the axis. The formulas follow from those of circular motion, treated in the topic on motion in the plane, because the arc traveled by $P$ is $s = r θ$ . In vector form, with the origin on the axis, $v = ω \times r$ . A disk of radius $0.20 m$ rotating at $30 rad/s$ has on its rim $v = 6.0 m/s$ and $a_{c} = 180 m/ s^{2}$ .

Torque and angular momentum

1. Torque of a force

Let $F$ be a force applied at the point $P$ and let $O$ be a fixed point, called the pole. The torque of the force about $O$ is

τ_{O} = r \times F, r = O P .

Its magnitude is $τ_{O} = r F sin φ = F b$ , where $φ$ is the angle between $r$ and $F$ and $b = r sin φ$ is the lever arm, that is the distance of $O$ from the line of action of the force; it is measured in $N \cdot m$ . For a given force the torque grows with the lever arm, and it is zero when the line of action passes through $O$ . A wrench $0.25 m$ long pushed at its end with $40 N$ perpendicular to the handle produces a torque of $10 N \cdot m$ ; with the force at $3 0^{\circ}$ to the handle the torque is $5.0 N \cdot m$ .

For a body rotating about a $z$ axis what matters is the axial torque $τ_{z} = \hat{k} \cdot τ_{O}$ , with $O$ on the axis; it does not depend on the position of $O$ on the axis, because moving the pole along the axis changes $r$ by a vector parallel to $\hat{k}$ . For a force in a plane perpendicular to the axis $τ_{z} = \pm F b$ , with the positive sign when the force tends to rotate the body counterclockwise.

The lever arm

b

is the distance of the pole

O

from the line of action of

F

, so

b = r sin φ

and

τ_{O} = F b

2. Angular momentum

The angular momentum about the pole $O$ of a particle of mass $m$ , position $r$ and velocity $v$ is

L_{O} = r \times m v = r \times p,

and it is measured in $kg \cdot m^{2} /s$ ; the angular momentum of a system of particles is the sum of the angular momenta of the individual particles. For a particle in uniform circular motion of radius $r$ , about the center, $r$ is perpendicular to $v$ , so $L_{O} = m v r = m r^{2} ω$ and $L_{O}$ is directed along the axis of the circle.

3. Angular momentum of a rotating rigid body

Let a rigid body, made of particles of mass $m_{i}$ , rotate with angular velocity $ω$ about the fixed axis $z$ , and let $R_{i}$ be the distance of the $i$ -th particle from the axis. Then the component along the axis of the angular momentum, computed about a pole on the axis, is

L_{z} = (i \sum m_{i} R_{i}^{2}) ω .

The coefficient of $ω$ depends only on how the mass is distributed around the axis and is the moment of inertia of the body about the axis. Two masses of $0.50 kg$ at the ends of a light rod $0.60 m$ long, rotating at $4.0 rad/s$ about the perpendicular axis through the midpoint, have $L_{z} = 2 \cdot 0.50 \cdot 0.3 0^{2} \cdot 4.0 = 0.36 kg \cdot m^{2} /s$ .

Proof

We choose the pole $O$ on the axis and write the position of the $i$ -th particle as $r_{i} = R_{i} + z_{i} \hat{k}$ , with $R_{i}$ perpendicular to the axis and of magnitude $R_{i}$ . The particle describes a circle of radius $R_{i}$ , so its velocity $v_{i}$ is perpendicular both to $\hat{k}$ and to $R_{i}$ and has magnitude $∣ ω ∣ R_{i}$ . Then

\hat{k} \cdot (r_{i} \times m_{i} v_{i}) = \hat{k} \cdot (R_{i} \times m_{i} v_{i}) + z_{i} \hat{k} \cdot (\hat{k} \times m_{i} v_{i}) .

The second term is zero, because $\hat{k} \times m_{i} v_{i}$ is orthogonal to $\hat{k}$ . The vector $R_{i} \times m_{i} v_{i}$ is parallel to $\hat{k}$ , since $R_{i}$ and $v_{i}$ are perpendicular to the axis, and has magnitude $m_{i} R_{i} \cdot ∣ ω ∣ R_{i}$ , since $R_{i}$ and $v_{i}$ are orthogonal; its direction is that of $\hat{k}$ for $ω > 0$ and the opposite one for $ω < 0$ . Therefore the contribution of the $i$ -th particle to $L_{z}$ is $m_{i} R_{i}^{2} ω$ , and summing over all particles we obtain the claim.

4. Torque of the weight

In a uniform gravitational field the resultant torque, about a pole $O$ , of the weights of the individual particles of a body of mass $M$ is equal to the torque of the total weight $M g$ applied at the center of mass $C$ :

i \sum r_{i} \times m_{i} g = r_{C} \times M g .

For this reason the center of mass is also called the center of gravity, and in the computation of torques the weight is treated as a single force applied at $C$ ; in a uniform rod $C$ is the midpoint.

Proof

The vector $g$ is the same for all particles, so it is factored out on the right: $\sum_{i} r_{i} \times m_{i} g = (\sum_{i} m_{i} r_{i}) \times g$ . By definition of the center of mass $\sum_{i} m_{i} r_{i} = M r_{C}$ , and therefore the torque is $r_{C} \times M g$ .

The theory continues with 6 more sections (moment of inertia, second cardinal equation and rotational motion, energy in rotational motion…), plus all worked examples, the solutions and the PDF.

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