Why a Stone Whirled on a String Accelerates Even at Steady Speed

A cricket ball hit for a six, a car going round a roundabout and a satellite circling the Earth all move in two dimensions. To handle such motion, physics uses vectors — quantities with both size and direction — and splits every motion into simpler perpendicular parts.

This lesson covers adding, subtracting and resolving vectors, projectile motion, and uniform circular motion with centripetal acceleration.

**Vectors are added by the triangle or parallelogram law, subtracted by adding the negative vector, and resolved into perpendicular components $A\cos \theta$ and $A\sin \theta$, which can then be added separately.

Addition:

- Triangle law — place the tail of B at the head of A; the resultant runs from the tail of A to the head of B
- Parallelogram law** — for two vectors at an angle $\theta$, the resultant has magnitude and direction

$$R=\sqrt{A^2+B^2+2AB\cos \theta }\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }$$

Subtraction. $\vec{A}-\vec{B}=\vec{A}+(-\vec{B})$, with magnitude $\sqrt{A^2+B^2-2AB\cos \theta }$.

Resolution into components:

$$A_x=A\cos \theta A_y=A\sin \theta A=\sqrt{A_x^2+A_y^2}$$

Worked example 1 — the parallelogram law. Forces of 6.0 N and 8.0 N act at 60° to each other:

$$R=\sqrt{6.0^2+8.0^2+2(6.0)(8.0)\cos 60^{\circ }}=\sqrt{36+64+48}=\sqrt{148}=12.2\text{ N}$$

Worked example 2 — adding by components. A walker goes 5.0 km east, then 5.0 km at 60° north of east:

$$R_x=5.0+5.0\cos 60^{\circ }=7.5\text{ km}{R}_y=5.0\sin 60^{\circ }=4.33\text{ km}$$

$$R=\sqrt{7.5^2+4.33^2}=8.66\text{ km, at }{\tan }^{-1}\frac{4.33}{7.5}=30^{\circ }\text{ north of east}$$

An everyday example. A kite string pulled at an angle during Makar Sankranti exerts a force with a horizontal part that drags the kite along and a vertical part that holds it up.

The substance. Two vectors of fixed size give the largest resultant when parallel and the smallest when opposite — 6 N and 8 N can combine to anything from 2 N to 14 N.

**A projectile launched at speed u and angle $\theta$ keeps a constant horizontal velocity $u\cos \theta$ while accelerating downwards at g, so its path is a parabola with time of flight $\frac{2u\sin \theta }{g}$, maximum height $\frac{u^2{\sin }^2\theta }{2g}$ and range $\frac{u^2\sin 2\theta }{g}$.

Components of motion**, ignoring air resistance:

- Horizontal: $a_x=0$, so $x=(u\cos \theta )t$
- Vertical: $a_y=-g$, so $y=(u\sin \theta )t-\frac{1}{2}g{t}^2$

Trajectory. Substituting $t=\frac{x}{u\cos \theta }$:

$$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$$

This has the form $y=bx-c{x}^2$, a parabola.

Time of flight. Setting $y=0$: $(u\sin \theta )T-\frac{1}{2}g{T}^2=0$, so $T=\frac{2u\sin \theta }{g}$.

Maximum height. At the top the vertical velocity is zero, so $0=u^2{\sin }^2\theta -2gH$, giving $H=\frac{u^2{\sin }^2\theta }{2g}$.

Range.

$$R=(u\cos \theta )T=\frac{2u^2\sin \theta \cos \theta }{g}=\frac{u^2\sin 2\theta }{g}$$

The range is greatest at 45°, where $R_{\max }=u^2/g$.

Worked example — a football kick. A ball is kicked at 20 m s${}^{-1}$ at 30°, with $g=9.8$ m s${}^{-2}$:

$$T=\frac{2\times 20\times \sin 30^{\circ }}{9.8}=2.04\text{ s}H=\frac{20^2{\sin }^{2}30^{\circ }}{2\times 9.8}=5.10\text{ m}R=\frac{20^2\sin 60^{\circ }}{9.8}=35.3\text{ m}$$

An everyday example. Water from a garden hose tilted upwards traces a clear parabola, and tilting the nozzle towards 45° sends the stream farthest.

The substance. The horizontal velocity never changes during the flight — only the vertical velocity does, so even at the highest point the projectile still moves sideways at $u\cos \theta$.

**In uniform circular motion a body goes round a circle at constant speed, but its velocity keeps changing direction, so it has an acceleration $a_c=\frac{v^2}{r}={\omega }^{2}r$ directed towards the centre, called centripetal acceleration.

Describing circular motion:

- Angular velocity** $\omega =\frac{\Delta \theta }{\Delta t}$, in rad s${}^{-1}$, and linear speed $v=\omega r$
- Time period $T=\frac{2\pi }{\omega }$ and frequency $f=\frac{1}{T}$, so $\omega =2\pi f$

Deriving centripetal acceleration. In a short time $\Delta t$ the body moves from P to Q, turning through a small angle $\Delta \theta$:

- The velocity vectors at P and Q both have size v but differ in direction by $\Delta \theta$
- The triangle of these velocity vectors is similar to the triangle formed by the two radii and the chord PQ, so $\frac{|\Delta \vec{v}|}{v}=\frac{PQ}{r}$
- For a small angle, the chord PQ is nearly the arc length $v\Delta t$, so $|\Delta \vec{v}|=\frac{v^2\Delta t}{r}$
- Dividing by $\Delta t$ gives $a_c=\frac{v^2}{r}={\omega }^{2}r$, and as $\Delta t$ shrinks, $\Delta \vec{v}$ points towards the centre

Worked example 1 — a stone on a string. A stone whirled in a horizontal circle of radius 0.50 m makes 2.0 revolutions per second:

$$\omega =2\pi \times 2.0=12.6{\text{ rad s}}^{-1}v=12.6\times 0.50=6.28{\text{ m s}}^{-1}{a}_c=\frac{6.28^2}{0.50}=79{\text{ m s}}^{-2}$$

An everyday example. Riders on a giant wheel at a village mela move at a steady speed, yet their velocity turns constantly, so a centripetal acceleration acts on them throughout the ride.

The substance. Constant speed does not mean zero acceleration — in circular motion the speed stays fixed while the direction, and so the velocity, keeps changing.