Torque and Rotational Dynamics

Task 1 - Describing rotation

Hook

Every point on a spinning wheel turns through the same angle in the same time, yet the rim is moving much faster than the hub. How can one wheel have one rate of spin but many different speeds?

Intuition

Rotation is described by angles instead of distances. The angle turned through, $\theta$, is the same for every point on a rigid object, and so is the angular velocity $\omega$, how fast that angle changes. A point further from the axis travels round a bigger circle in the same time, so its linear speed is larger.

Physics measures angles in radians. One radian is the angle where the arc length equals the radius, and one full turn is $2\pi$ radians, about 6.28. Radians make the link between angle and distance very simple, as the equations below show.

Formalise

Angular velocity is in rad/s and angular acceleration in rad/s². The constant acceleration equations from the Kinematics worksheet carry over with $\theta$ for $s$, $\omega$ for $v$ and $\alpha$ for $a$.

Worked example

A wheel of radius 0.3 m spins up from rest with $\alpha = 2\text{ rad/s}^2$ for 5 s.

1. Final angular velocity. $\omega = 0 + 2 \times 5 = 10\text{ rad/s}$.
2. Angle turned. $\theta = \tfrac{1}{2} \times 2 \times 5^2 = 25\text{ rad}$, which is $25 / 2\pi = 3.98$ turns.
3. Rim speed. $v = 0.3 \times 10 = 3\text{ m/s}$.

The first rows of the translational to rotational table.

Your turn

Task 2 - Torque

Hook

Why are door handles always placed on the side furthest from the hinge? And why does pushing on the edge of a door, straight toward the hinge, do nothing at all, however hard you push?

Intuition

Torque depends on the force and on the lever arm, the perpendicular distance from the pivot to the line along which the force acts. Push far from the hinge and the lever arm is long. Push straight toward the hinge and the line of action passes through the pivot, so the lever arm is zero and so is the torque.

Drag the red point and change the angle below. Watch the gold lever arm.

Formalise

For a force $F$ applied at distance $r$ from the pivot, at angle $\phi$ between the force and the line from the pivot. Torque is measured in newton metres (N m). By convention, anticlockwise torque is positive.

Worked example

A 0.25 m spanner is used on a stiff bolt.

1. Perpendicular push of 40 N. $\tau = 0.25 \times 40 \times \sin 90^\circ = 10\text{ N m}$.
2. Same push at 30° to the handle. $\tau = 0.25 \times 40 \times \sin 30^\circ = 5\text{ N m}$. Half the turning effect.
3. Bolt needs 10 N m but the push is at 30°. Either double the force to 80 N or use a spanner twice as long.

Your turn

Task 3 - Moment of inertia and Newton's second law for rotation

Hook

A ring, a solid disc and a solid ball, all with the same mass and radius, are released together at the top of a slope and roll down without slipping. Which reaches the bottom first? Does it change if one of them is much heavier?

Intuition

A rolling object has to do two things at once. It has to speed up along the slope and it has to spin up. Some of the energy available goes into spinning. An object with more of its mass far from its centre, like a ring, needs more energy to spin up, so less is left for moving down the slope.

The moment of inertia $I$ measures this. Each small piece of mass contributes its mass times the square of its distance from the axis, so far-away mass counts four times as much when it is twice as far out.

Race the shapes below. Then change the mass and race again.

Formalise

The definition of moment of inertia, Newton's second law for rotation, and standard results for uniform shapes of mass $M$.

Worked example

A 2 kg solid disc of radius 0.5 m is free to spin on an axle. A torque of 1 N m is applied.

1. Moment of inertia. $I = \tfrac{1}{2} \times 2 \times 0.5^2 = 0.25\text{ kg m}^2$.
2. Angular acceleration. $\alpha = \tau / I = 1 / 0.25 = 4\text{ rad/s}^2$.

For rolling down a slope, the acceleration is $a = g\sin\theta / (1 + I/MR^2)$. Mass cancels, so only the shape factor $I/MR^2$ matters. The sphere, with the smallest factor, wins.

Your turn

Task 4 - Static equilibrium

Hook

A 30 kg child and a 40 kg child want to use a seesaw. If they both sit at the ends, the heavier child drops to the ground. Where should the heavier child sit so the seesaw balances?

Intuition

An object in static equilibrium does not move and does not turn. That needs two conditions. The forces must balance, so it does not accelerate, and the torques about any point must balance, so it does not start to spin. On a seesaw, each child's weight produces a torque about the pivot. One turns it clockwise and the other anticlockwise. Balance means they are equal.

Try it below. Drag the blocks to balance the beam, then change the masses and balance it again.

Formalise

Both conditions together. You may take torques about any point you like, so choose one where an unknown force acts. Its torque is then zero and it drops out of the equation.

Worked example 1

The seesaw. Torques about the pivot must balance.

$30g \times 2 = 40g \times d$, so $d = 60 / 40 = 1.5\text{ m}$. The heavier child sits 1.5 m from the pivot. The starting setup of the diagram shows exactly this.

Worked example 2

A uniform 5 m beam of mass 20 kg rests on supports at each end. A 60 kg load sits 1 m from the left end. Find the support forces $R_L$ and $R_R$.

1. Torques about the left support, so $R_L$ drops out. The beam's weight acts at its centre, 2.5 m along. $5R_R = 20g \times 2.5 + 60g \times 1 = 110g$, so $R_R = 22g = 216\text{ N}$.
2. Forces. $R_L + R_R = 80g$, so $R_L = 58g = 569\text{ N}$.

The support nearer the load carries most of it, as you would expect.

Your turn

Task 5 - Angular momentum

Hook

A spinning ice skater pulls their arms in and suddenly spins much faster. Nobody pushed them. Where did the extra spin come from?

Intuition

Angular momentum is the rotational partner of momentum. Just as momentum is conserved when no outside force acts, angular momentum is conserved when no outside torque acts. When the skater pulls their arms in, their moment of inertia drops. To keep the same angular momentum, their angular velocity must rise.

Their rotational kinetic energy goes up too. That energy comes from the work their muscles do pulling their arms inward against the tendency of the spinning arms to fly outward.

Formalise

Worked example

A skater with $I = 4\text{ kg m}^2$ spins at 2 rad/s, then pulls in to $I = 1.6\text{ kg m}^2$.

1. Angular momentum. $L = 4 \times 2 = 8\text{ kg m}^2\text{/s}$, which stays the same.
2. New spin rate. $\omega_2 = 8 / 1.6 = 5\text{ rad/s}$.
3. Kinetic energy. Before $\tfrac{1}{2} \times 4 \times 2^2 = 8\text{ J}$, after $\tfrac{1}{2} \times 1.6 \times 5^2 = 20\text{ J}$.

The completed translational to rotational table.

Your turn

Task 6 - Torque in a 2D game engine

Intuition

In a 2D game, rotation is always about the axis pointing out of the screen, so torque is a single number. When a force $\vec{F}$ is applied at a point offset by $\vec{r}$ from the body's centre of mass, the torque is the 2D cross product of the offset and the force. A force through the centre of mass gives zero torque and only pushes the body. The same force applied off centre both pushes it and spins it. This is why a car clipped at its corner spins out.

Create a new Console App (C++) project named RigidBody2D and replace its source file with the skeleton below, saved as RigidBody2D.cpp. The program runs two scenarios on the same 2 kg box, 1 m wide and 0.5 m tall. Both apply a 10 N upward push for 0.1 s. The first pushes through the centre, and the second pushes at the right-hand edge.

1. In applyForceAtPoint, calculate r as point minus position, then add cross(r, f) to torque.
2. In integrate, calculate the angular acceleration as torque / inertia, then update angularVelocity and then angle, mirroring the linear lines above them.

Success condition: both scenarios print the same upward velocity of 0.5 m/s. The centre push prints an angular velocity of 0. The edge push prints about 2.4 rad/s. Check this by hand. The torque is $0.5 \times 10 = 5\text{ N m}$, the moment of inertia is $2(1^2 + 0.5^2)/12 = 0.208\text{ kg m}^2$, and the angular impulse $\tau\Delta t = 0.5$ divided by $I$ gives the angular velocity.

#include <iomanip>
#include <iostream>
#include <string>

struct Vec2
{
    double x;
    double y;
};

Vec2 operator+(Vec2 a, Vec2 b) { return { a.x + b.x, a.y + b.y }; }
Vec2 operator-(Vec2 a, Vec2 b) { return { a.x - b.x, a.y - b.y }; }
Vec2 operator*(Vec2 a, double s) { return { a.x * s, a.y * s }; }
Vec2& operator+=(Vec2& a, Vec2 b) { a.x += b.x; a.y += b.y; return a; }

// The 2D cross product gives the single number that is the torque about z
double cross(Vec2 a, Vec2 b) { return a.x * b.y - a.y * b.x; }

struct RigidBody2D
{
    double mass = 1.0;
    double inertia = 1.0;
    Vec2 position{ 0.0, 0.0 };
    Vec2 velocity{ 0.0, 0.0 };
    double angle = 0.0;
    double angularVelocity = 0.0;
    Vec2 force{ 0.0, 0.0 };
    double torque = 0.0;

    void applyForceAtPoint(Vec2 f, Vec2 point)
    {
        force += f;

        // TODO: find r, the offset of point from the centre of mass, and add
        // the torque cross(r, f) into the torque accumulator
    }

    void integrate(double dt)
    {
        velocity += force * (1.0 / mass) * dt;
        position += velocity * dt;

        // TODO: angular acceleration is torque divided by inertia. Update
        // angularVelocity, then angle, the same way as the linear part

        force = { 0.0, 0.0 };
        torque = 0.0;
    }
};

void runScenario(const std::string& name, Vec2 offset)
{
    const double width = 1.0;
    const double height = 0.5;
    const double mass = 2.0;
    const double dt = 0.001;

    // A solid box rotating about its centre has I = m (w^2 + h^2) / 12
    RigidBody2D body;
    body.mass = mass;
    body.inertia = mass * (width * width + height * height) / 12.0;

    for (int step = 0; step < 1000; step++)
    {
        // A 10 N upward push for the first 0.1 s only
        if (step < 100)
        {
            body.applyForceAtPoint({ 0.0, 10.0 }, body.position + offset);
        }
        body.integrate(dt);
    }

    std::cout << name << "\n";
    std::cout << "  velocity         (" << body.velocity.x << ", " << body.velocity.y << ") m/s\n";
    std::cout << "  angular velocity " << body.angularVelocity << " rad/s\n";
    std::cout << "  angle after 1 s  " << body.angle << " rad\n\n";
}

int main()
{
    std::cout << std::fixed << std::setprecision(3);
    runScenario("Push through the centre", { 0.0, 0.0 });
    runScenario("Push at the right-hand edge", { 0.5, 0.0 });
    return 0;
}