Calculus and Particle Motion

Calculus becomes essential for describing particle motion whenever the acceleration changes continuously over time. While simple algebra is sufficient for constant velocity motion, and constant acceleration problems can be solved with the 'suvat' equations, more complicated scenarios require calculus.

                    When Do You Need Calculus?
                    Constant speed but changing direction: calculus can be used to find instantaneous velocity as a function of time or position
Varying force: Any changing force causes changing acceleration, requiring calculus to link position, velocity, and acceleration
Complex paths: Curved trajectories, circular motion, or oscillations demand differential equations
Time-varying forces: When forces change with time or position (like springs or gravity wells)

                

The fundamental relationship in particle kinematics connects position x(t), velocity v(t), and acceleration a(t) through derivatives:

v(t) = dxdt and a(t) = dvdt = d²xdt²

Conversely, if you know acceleration, integration lets you work backward to find velocity and position. This is crucial in physics where Newton's second law gives you acceleration from forces, and you need to determine where the particle actually goes.

Watch as a sphere oscillates horizontally under the influence of a perfect spring attached at the center. This demonstrates simple harmonic motion where the restoring force is proportional to displacement from equilibrium.

Variables added to oscillation

x = 0 (equilibrium)

x = −A

x = +A

                    Animation Details
                    Idealisation: The spring does not 'get in the way' of the motion of the object. As the object passes through the centre the 'spring' is always pulling the object back to the centre. There is no friction or energy losses so the object oscillates continuously, never stopping
Left Motion: Moves to the left, slowing down as the spring extends (2 seconds)
Reversal: Stops at maximum displacement, then accelerates back
Through Center: Passes through equilibrium at maximum speed
Right Motion: Continues to the right, stops at maximum displacement
Return: Springs back to center, completing one full cycle (8 seconds total)

                

The motion is determined by Hooke's Law: F = −kx, where the restoring force, F, is proportional, but in the opposite direction, to the displacement, x, from the centre of motion. A basic solution for the sphere's position, x, at any time, t, is:

x(t) = A sin(ωt)

where A is the amplitude or maximum displacement from the centre, and ω is the angular frequency.

To give some physical meaning to A and ω:

If A increases that means at its maximum displacement, the object is further away from the centre of motion. As t changes, the value of x(t) varies between +A and −A (its maximum displacement) because the value of sin varies between +1 and −1.

And finally if we think of values of ω resulting in sin(2t) or sin(3t), the object is oscillating around the centre faster as ω increases. We say the object's 'angular frequency' is increasing.

Much more mathematical detail about the solution is available in the following sub-page.

Watch as a sphere oscillates vertically under the combined influence of a spring force and gravity. The equilibrium position occurs where the spring force balances the weight, and the motion about this point exhibits simple harmonic motion.

                    Physics of Vertical SHM
                    Forces: Spring force (F = -kx) and gravitational force (F = -mg)
Equilibrium: At equilibrium, the spring force balances gravity: kxeq = mg
SHM about equilibrium: When displaced from equilibrium, the net restoring force is proportional to displacement
Period: The period of oscillation T = 2π√(m/k) is the same as horizontal SHM

                

The equation of motion from the natural spring length is:

y(t) = y_eq + A sin(ωt)

where y_eq is the equilibrium position (spring stretched by mg/k), and the sphere oscillates harmonically about this point. Gravity shifts the equilibrium but doesn't affect the frequency of oscillation.

Watch a disk rotating in a horizontal plane, viewed from 30 degrees above. The red spot on the circumference traces an elliptical path (eccentricity = 0.5). The disk completes one full rotation every 3 seconds.

Basics of Circular Motion

If a force acts in the direction the particle is already moving, only the speed will increase or decrease, the direction of motion will not change as a result of the force. If a (resultant) force acts at a non-zero angle to the direction of motion of the particle, both speed and direction of motion will change. If the force remains perpendicular (at right angles) to direction of motion at any moment, toward the centre of motion, then the speed does not change, but the object will move in a circle.

                    Circular Motion Kinematics
                    Constant speed: The spot moves at constant speed around the circle
Changing velocity: Although speed is constant, velocity direction continuously changes
Centripetal acceleration: Acceleration always points toward the center: a = v²/r
Angular velocity: The disk rotates with constant angular velocity ω = 2π/T

                

Angular Velocity

Angular velocity (ω) measures how quickly the angle changes with time. It is defined as the derivative of the angle with respect to time:

ω = dθdt

The angle θ increases as the object rotates. Angular velocity ω indicates the rate of rotation.

Linear Velocity and Angular Velocity Relationship

For a particle moving in a circle at constant distance r from the center with constant angular velocity ω, we can derive the linear (tangential) velocity v using calculus.

The arc length traveled by the particle is given by s = rθ, where θ is the angle in radians. The linear velocity is the rate of change of arc length:

v = dsdt = d(rθ)dt

Since r is constant (the distance from the center doesn't change), we can take it outside the derivative:

v = r dθdt = rω

This shows that the linear speed equals the radius times the angular velocity. Objects farther from the center move faster linearly, even though they have the same angular velocity.

Radial Acceleration Using Small Changes

The radial or centripetal acceleration is the acceleration to the centre or axis of rotation. Don't confuse with the angular acceleration which is a measure of how quickly the angular velocity is changing.

We can derive the centripetal acceleration by examining the change in velocity as the object moves through a small angle dθ.

As the particle moves from P to Q through angle dθ, the velocity vector rotates by the same angle. The triangle shows the component of velocity perpendicular to the original direction.

Radial Acceleration Detailed Explanation

The derivation of the equation for radial acceleration of an object travelling at constant (tangential) speed around a horizontal circle follows.

Although the diagram is of a vertical circle, always remember this derivation is only for an object travelling in a horizontal circular path.

A result derived earlier, to be used later — the relationship between the tangential velocity v, the angular velocity ω = dθdt, and the radius of the path r:

v = rω = r dθdt (*)

Start with (from the diagram) the object on the circumference of the circle at point P. Here the "horizontal" tangential velocity is v; there is no radial velocity towards the centre of the circle at this point. As the object moves around the circular path the radius traces out a small angle dθ of a sector, and the object is now at point Q. The tangential velocity is still v, but we can decompose it into a "horizontal" component and a "vertical" component. The vertical component of the tangential velocity (using the geometry of the triangle drawn at Q) is v sin δθ.

However, this "vertical" component of the tangential velocity at Q is parallel to the radial velocity at P, which was zero. So the velocity has changed by v sin δθ. Notice the object has moved in a "vertical" direction parallel to the direction from P to O, so it has accelerated in the direction P to centre — inwards along the radius to the centre.^†

Since this change in velocity has happened over a short time interval δt, the average acceleration along the radius OP is:

acceleration = change in velocitytime = v sin δθδt

Since this change in velocity has happened over a small time interval we can replace v sin δθ by v δθ (for small angles, sin δθ ≈ δθ). So the average acceleration in the radial direction PO is:

a_r = v δθδt = v δθδt

Since δθδt = ω, or replacing v by rω from (*), we have alternate expressions for radial acceleration:

a_r = vω = rω² or v²r

directed from the circumference towards the centre of the circle.

^† The "horizontal" component of velocity when the object is at P is v; at Q it is v cos δθ, which is approximately v(1) = v when δθ is small. So the "horizontal" velocity has not changed from P to Q — the object is not accelerating along the tangential direction, only along the radial direction, towards the centre of the circle.

Radial Acceleration Using Polar Coordinate Vectors

To find the acceleration of an object moving in a circle at constant radius r with constant angular velocity ω, we use polar coordinates with unit vectors: r̂ (pointing radially outward) and θ̂ (pointing tangentially).

The position vector is simply:

r = r r̂

As the object rotates, the unit vectors themselves rotate. The key insight from calculus is that these rotating unit vectors have time derivatives:

dr̂dt = ω θ̂ and dθ̂dt = −ω r̂

The velocity is found by differentiating the position vector (noting that r is constant):

v = drdt = r dr̂dt = rω θ̂

This shows the velocity is purely tangential with magnitude v = rω.

The acceleration is found by differentiating the velocity:

a = dvdt = rω dθ̂dt = rω (−ω r̂) = −rω² r̂

The negative sign shows the acceleration points radially inward (toward the center), opposite to r̂. The magnitude of this centripetal acceleration is:

a = rω² = v²r

This centripetal (center-seeking) acceleration is essential for circular motion. The beauty of using polar coordinates is that it clearly shows the acceleration is purely radial, with no tangential component when speed is constant.

Cartesian Coordinates of Position of Spot

The position of the red spot can be described using parametric equations:

x(t) = r cos(ωt) y(t) = r sin(ωt)

Taking derivatives gives the velocity components, which are perpendicular to the position vector and tangent to the circle. This is a perfect example where calculus reveals that constant speed combined with changing direction produces acceleration toward the center.

Watch a red dot moving around the perimeter of an ellipse (eccentricity = 0.5). The vertical axis is positioned at one focal point of the ellipse. As the dot traces the elliptical path, its distance from the focus varies continuously.

                    Elliptical Motion About a Focus
                    Fixed ellipse: The ellipse remains stationary with one focus at the vertical axis
Variable radius: The dot's distance from the focus varies between a(1-e) and a(1+e)
Focal property: At periapsis (closest point): r = 50px; at apoapsis (farthest): r = 150px
Kepler's 2nd Law: The dot sweeps out equal areas in equal times, moving faster near periapsis and slower near apoapsis

                

The position of the red dot relative to the focus can be described using polar coordinates:

r(θ) = a(1-e²)/(1 + e cos θ)

where a = 100px is the semi-major axis and e = 0.5 is the eccentricity. This animation uses Kepler's equation to solve for the true orbital position at each moment, demonstrating actual Keplerian motion. The dot moves fastest at periapsis and slowest at apoapsis, exactly as planets orbit the sun. This non-uniform motion requires calculus and differential equations to describe accurately.

In order to calculate the motion of an object that is subject to a resistance proportional to the velocity of the object at any moment, one needs to use calculus. The equation of motion is based on the standard Newtonian equation F = ma. Write acceleration as dvdt. For an object thrown vertically upwards, take upwards the initial direction of motion as the positive direction for our reference.

The standard mass multiplied by acceleration, ma, is written as mdvdt. This is equal to the sum of the forces acting on the object: 1) the weight acting downwards −mg, and 2) the resistance proportional to velocity as −mKv (to make later algebra easier since m and K are constants. K has dimensions s⁻¹).

mdvdt = −mg − mKv

Cancelling m throughout and separating the variables we get the following integral:

∫ dvg + Kv = −∫ dt

Using u for the initial velocity, when t = 0, we can rearrange and integrate this equation:

1K ln(g + Kv) = −t + C

where C is the constant of integration. Making the initial velocity represented by the letter u, when t = 0, the integrated expression becomes the equation:

t = 1K ln{g + Kug + Kv}

So if we have a value for K and the initial velocity, we can calculate how long it takes for the object to reach a certain velocity, v.

Maximum Height

As usual with objects moving under gravity, the object reaches its maximum height when v = 0. Writing time to reach maximum height as t_hmax:

t_hmax = 1K ln{g + Kug} = 1K ln{1 + Kug}

Rearranging the equation to make v, the upward velocity at any time, the subject:

v = ue^−Kt + gK(e^−Kt − 1)

As t increases (the object is travelling for a long time), e^−Kt gets close to zero, and the velocity becomes −gK, as the object is moving downwards after reaching its maximum height. The object is slowing on the journey upwards, so maximum velocity must occur when the object is moving down.

When the ball is travelling horizontally with constant speed it is not subject to a horizontal force or acceleration. When the object hits the wall the forces are complicated to model and depend on the material properties of the object and wall. In this situation conservation of linear momentum is the preferred technique to analyze the change in motion.

Watch as a ball travels horizontally at constant velocity, collides with a vertical wall, and bounces back with the same speed in the opposite direction. This demonstrates an elastic collision where kinetic energy is conserved.

                    Animation Details
                    Left to Right: The ball travels from left to right in 3 seconds
Collision: At the wall, the ball immediately reverses direction
Right to Left: The ball returns at the same speed, taking 3 seconds
Pause & Repeat: A brief 0.5-second pause before the cycle repeats

                

In this idealized scenario, we assume a perfectly elastic collision where the ball's velocity changes from +v to -v instantaneously. In reality, the collision would take a finite time and involve deformation of both the ball and wall.

Setting up the Unit Vectors

In polar coordinates, at any instant there are two perpendicular unit vectors that travel with the object:

r̂ — points outward from the centre along the radius
θ̂ — points tangentially, perpendicular to r̂, in the direction of increasing θ

In Cartesian components these are:

r̂ = (cos θ, sin θ) θ̂ = (−sin θ, cos θ)

Why does dθ̂/dt point inward?

Differentiating θ̂ with respect to time (using the chain rule, since θ changes with time):

dθ̂dt = (−cos θ, −sin θ) · dθdt = −ω (cos θ, sin θ) = −ω r̂

The result −ωr̂ points in the opposite direction to r̂, which is directly inward toward the centre. The negative sign is the mathematical statement of that inward direction.

The Geometric Picture

Here is the intuition without algebra. Both r̂ and θ̂ are unit vectors — their tips always sit on a unit circle centred at the origin. As the object moves around its circular path, θ increases, and θ̂ rotates steadily. The tip of θ̂ is therefore itself moving around a circle, and the rate of change of any vector whose tip moves on a circle always points toward the centre of that circle — which here is the origin, i.e. the inward radial direction.

You can see this concretely at a few key positions:

θ	r̂ (outward)	θ̂ (tangential)	dθ̂/dt direction
0°	→ (right)	↑ (up)	← (left = inward)
90°	↑ (up)	← (left)	↓ (down = inward)
180°	← (left)	↓ (down)	→ (right = inward)

In every case, the change in θ̂ points back toward the centre — always opposite to r̂.

Why this Matters for Acceleration

When you differentiate the velocity v = rω θ̂ to get acceleration, you use the product rule and the result dθ̂/dt = −ω r̂ appears. This is precisely what produces the centripetal (inward) acceleration −rω² r̂. So the inward direction of the acceleration is not imposed by hand — it emerges naturally from the geometry of how the tangential unit vector rotates as the object moves round the circle.

1. Start from Hooke's Law

Hooke's law states that the restoring force of a spring is proportional to the displacement from equilibrium and acts in the opposite direction:

F = −kx

where k is the spring constant and x is the displacement from the central (equilibrium) point. The negative sign means the force always pulls the object back toward the centre.

2. Apply Newton's Second Law

Newton's second law says:

F = m d²xdt²

Combining with Hooke's law and rearranging:

d²xdt² + kmx = 0

Now define ω² = k/m. The equation of motion becomes:

d²xdt² + ω²x = 0

This is the standard differential equation for simple harmonic motion (SHM).

3. Test the Proposed Solution x(t) = A sin(ωt)

We check whether this function satisfies the equation of motion.

First derivative:

dxdt = Aω cos(ωt)

Second derivative:

d²xdt² = −Aω² sin(ωt) = −ω²x(t)

since x(t) = A sin(ωt) by definition.

4. Substitute Back into the Equation of Motion

(−ω²x) + ω²x = 0

The equation is satisfied exactly. ✓

5. Physical Interpretation — Why Sine Appears

The acceleration is proportional to −x. Sine (and cosine) functions are the only functions whose second derivative is proportional to the negative of the function itself:

d²dt² sin(ωt) = −ω² sin(ωt)

That mathematical property perfectly matches the physics of a restoring force that always points toward the centre.

6. General Solution

The full solution to the SHM equation is:

x(t) = A sin(ωt) + B cos(ωt)

or equivalently:

x(t) = X₀ sin(ωt + φ)

where X₀ is the amplitude, ω = √(k/m) is the angular frequency, and φ is the phase constant determined by initial conditions. The form x(t) = A sin(ωt) is the valid solution when the object starts at x = 0 with maximum velocity.

7. Physical Meaning of ω² = k/m

How fast the system oscillates: k is the stiffness of the spring (strength of restoring force) and m is the mass (inertia resisting acceleration). So:

ω = √ km

is the natural angular frequency of the motion. A large k (stiff spring) means a stronger restoring force and faster oscillations; a large m (heavy object) means more inertia and slower oscillations.

Acceleration per unit displacement: Rewriting the equation of motion as

d²xdt² = −kmx

shows that k/m is the proportionality constant between acceleration and displacement — it tells you how strongly the system accelerates back toward equilibrium for a given displacement.

Connection to period and frequency:

T = 2πω = 2π √ mk

Doubling the mass increases the period (slower motion); doubling the spring constant decreases the period (faster motion).

Why the square appears: The differential equation involves a second derivative (acceleration). Since sine satisfies

d²dt² sin(ωt) = −ω² sin(ωt)

the physics (force ∝ displacement) naturally produces a second derivative proportional to −x, which mathematically introduces ω², not just ω. In short, ω² = k/m means "the square of the natural oscillation rate is set by the ratio of restoring force strength to inertia."

8. Meaning of ω in One-Dimensional Oscillation

How fast the object oscillates: ω measures how rapidly the motion repeats in time:

ω = 2πf = 2πT

where f is the frequency (oscillations per second) and T is the period. ω is the rate of change of the phase of the motion, in radians per second.

Why "angular" frequency for straight-line motion? Even though the mass moves in a straight line, its motion is mathematically equivalent to the projection of uniform circular motion onto one axis. A point moving in a circle at constant angular speed ω has horizontal position x(t) = A sin(ωt) — identical to SHM. So ω comes from the circular-motion analogy, not because the object is physically rotating.

Physical takeaway: In one-dimensional oscillation, ω is the natural angular frequency — the intrinsic rate at which the system oscillates about equilibrium. It is not about direction of motion, but about how quickly the system cycles through its back-and-forth motion in time, determined entirely by the balance between restoring force (k) and inertia (m). Units: rad s⁻¹.

Calculus & Particle Motion

When Do You Need Calculus?

Horizontal Motion Attached to Perfect Spring

Variables added to oscillation

Animation Details

Vertical Motion on a Perfect Spring

Physics of Vertical SHM

Horizontal Circular Motion

Basics of Circular Motion

Circular Motion Kinematics

Angular Velocity

Linear Velocity and Angular Velocity Relationship

Radial Acceleration Using Small Changes

Radial Acceleration Detailed Explanation

Radial Acceleration Using Polar Coordinate Vectors

Cartesian Coordinates of Position of Spot

Elliptical Orbit

Elliptical Motion About a Focus

Vertical Motion with Resistance

Maximum Height

Where Calculus Doesn't Help

Animation Details

Differentiation of Radial and Tangential Vectors