MD4 - Circular motion with vectors

MD4 - Circular motion with vectors

Describe a circular path using vector components, differentiate to find velocity and acceleration, and prove that uniform circular motion has tangential velocity and inward acceleration. See why the centre's position matters when interpreting a position vector.

Position, velocity and acceleration vectors

A particle moves around a circle centred at (2,−1)(2,-1), rather than the origin. Does its acceleration point towards the origin or towards its own circle's centre? To answer, separate the position of the centre from the particle's displacement relative to that centre.

A particle 3 m3\,\mathrm{m} from the origin at angle θ\theta anticlockwise from the positive horizontal axis has coordinates (3cos⁡θ,3sin⁡θ)(3\cos\theta,3\sin\theta). These are the horizontal and vertical projections of its radius. The identity cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1 guarantees that its distance from the centre stays 3.

Position and relative position

The position vector r\mathbf r runs from the fixed origin to the particle. If the fixed centre has position vector c\mathbf c, then r−c\mathbf r-\mathbf c is the outward radius vector. Velocity v=dr/dt\mathbf v=\mathrm d\mathbf r/\mathrm dt and acceleration a=dv/dt\mathbf a=\mathrm d\mathbf v/\mathrm dt describe how position and velocity change with time.

For a fixed circle of radius R>0R>0 centred at (h,k)(h,k), one parametrisation is

r=(h+Rcos⁡θ)i+(k+Rsin⁡θ)j\mathbf r=(h+R\cos\theta)\mathbf i+(k+R\sin\theta)\mathbf j

It satisfies (x−h)2+(y−k)2=R2(x-h)^2+(y-k)^2=R^2. With θ=ωt+ϕ\theta=\omega t+\phi, positive constant ω\omega gives anticlockwise motion; a negative angular rate gives clockwise motion. Angular speed is the magnitude of that rate. The constant ϕ\phi sets the starting point.

Worked example 1: a circle at the origin

Question. A particle has position r=3cos⁡(2t)i+3sin⁡(2t)j\mathbf r=3\cos(2t)\mathbf i+3\sin(2t)\mathbf j metres, with tt in seconds. Find the centre, radius, starting position and position at t=π/4t=\pi/4.

Plan. Compare the components with a circle parametrisation, then substitute the two times.

x2+y2=9cos⁡2(2t)+9sin⁡2(2t)=9identify the circler(0)=3icos⁡0=1, sin⁡0=0r(π/4)=3j2t=π/2\begin{aligned} x^2+y^2&=9\cos^2(2t)+9\sin^2(2t)=9 && \text{identify the circle}\\ \mathbf r(0)&=3\mathbf i && \cos0=1,\ \sin0=0\\ \mathbf r(\pi/4)&=3\mathbf j && 2t=\pi/2 \end{aligned}

Answer. Centre (0,0)(0,0), radius 3 m3\,\mathrm{m}, starting at (3,0)(3,0) and reaching (0,3)(0,3) at t=π/4t=\pi/4.

Check. The angular rate is 2 rad s−12\,\mathrm{rad\,s^{-1}}, so π/4\pi/4 seconds is a quarter-turn. The two positions are a quarter-circle apart, and both lie 3 metres from the centre.

Translate the same motion without changing its radius or angular rate. Only the constant centre vector changes.

Worked example 2: translating the centre

Question. A particle has position r=(2+3cos⁡(2t))i+(−1+3sin⁡(2t))j\mathbf r=(2+3\cos(2t))\mathbf i+(-1+3\sin(2t))\mathbf j metres. Find the centre, radius, starting position and position at t=π/4t=\pi/4.

Plan. Subtract the constant centre vector, then use the same trigonometric values.

(x−2)2+(y+1)2=9subtract the fixed centrer(0)=5i−jadd the starting radiusr(π/4)=2i+2jadd the quarter-turn radius\begin{aligned} (x-2)^2+(y+1)^2&=9 && \text{subtract the fixed centre}\\ \mathbf r(0)&=5\mathbf i-\mathbf j && \text{add the starting radius}\\ \mathbf r(\pi/4)&=2\mathbf i+2\mathbf j && \text{add the quarter-turn radius} \end{aligned}

Answer. Centre (2,−1)(2,-1), radius 3 m3\,\mathrm{m}, starting at (5,−1)(5,-1) and reaching (2,2)(2,2).

Check. Subtracting (2,−1)(2,-1) from the two positions gives (3,0)(3,0) and (0,3)(0,3), exactly the relative positions in Worked example 1. Their distances from the origin need not equal 3.

Watch out

Wrong: ∣r∣=R|\mathbf r|=R for every circle of radius RR. That is true only when the origin is the centre. In general, ∣r−c∣=R|\mathbf r-\mathbf c|=R.

Differentiating trigonometric components

Differentiate each vector component separately. The fixed unit vectors do not change. The chain rule matters: differentiating cos⁡(2t)\cos(2t) gives −2sin⁡(2t)-2\sin(2t), and differentiating again introduces another factor of 2.

Vector derivatives for a uniform circle

For constant c\mathbf c, R>0R>0, constant signed angular rate ω\omega, and θ=ωt+ϕ\theta=\omega t+\phi in radians,

r=c+Rcos⁡θ i+Rsin⁡θ jv=−Rωsin⁡θ i+Rωcos⁡θ ja=−Rω2cos⁡θ i−Rω2sin⁡θ j\begin{aligned} \mathbf r&=\mathbf c+R\cos\theta\,\mathbf i+R\sin\theta\,\mathbf j\\ \mathbf v&=-R\omega\sin\theta\,\mathbf i+R\omega\cos\theta\,\mathbf j\\ \mathbf a&=-R\omega^2\cos\theta\,\mathbf i-R\omega^2\sin\theta\,\mathbf j \end{aligned}

These circular vector formulae are not in the formula booklet: obtain them by differentiation rather than memorising component signs.

The constants describing the centre disappear when differentiated. Translating a motion therefore changes position but leaves its velocity and acceleration unchanged.

Worked example 3: differentiating twice

Question. For r=(2+3cos⁡(2t))i+(−1+3sin⁡(2t))j\mathbf r=(2+3\cos(2t))\mathbf i+(-1+3\sin(2t))\mathbf j metres, find v\mathbf v and a\mathbf a, and evaluate them at t=π/4t=\pi/4 seconds.

Plan. Differentiate each component twice before substituting the time.

v=−6sin⁡(2t)i+6cos⁡(2t)jchain factor 2a=−12cos⁡(2t)i−12sin⁡(2t)jdifferentiate againv(π/4)=−6i m s−1at the uppermost pointa(π/4)=−12j m s−2towards the centre\begin{aligned} \mathbf v&=-6\sin(2t)\mathbf i+6\cos(2t)\mathbf j && \text{chain factor }2\\ \mathbf a&=-12\cos(2t)\mathbf i-12\sin(2t)\mathbf j && \text{differentiate again}\\ \mathbf v(\pi/4)&=-6\mathbf i\,\mathrm{m\,s^{-1}} && \text{at the uppermost point}\\ \mathbf a(\pi/4)&=-12\mathbf j\,\mathrm{m\,s^{-2}} && \text{towards the centre} \end{aligned}

Answer. The derivatives are as above; at t=π/4t=\pi/4, velocity is leftwards and acceleration is downwards.

Check. Geometry gives speed Rω=3×2=6R\omega=3\times2=6 and acceleration magnitude Rω2=12R\omega^2=12. The centre is directly below (2,2)(2,2), so the acceleration's direction also agrees.

Now double only the angular rate, retaining the same radius and centre. Compare at the same angular position, which is reached in half the time.

Worked example 4: doubling the angular rate

Question. For r=(2+3cos⁡(4t))i+(−1+3sin⁡(4t))j\mathbf r=(2+3\cos(4t))\mathbf i+(-1+3\sin(4t))\mathbf j metres, find v\mathbf v, a\mathbf a, and their values at t=π/8t=\pi/8 seconds.

Plan. Keep the chain factor 4 at each differentiation, then use 4t=π/24t=\pi/2.

v=−12sin⁡(4t)i+12cos⁡(4t)jfirst derivativea=−48cos⁡(4t)i−48sin⁡(4t)jsecond derivativev(π/8)=−12i m s−1a(π/8)=−48j m s−2\begin{aligned} \mathbf v&=-12\sin(4t)\mathbf i+12\cos(4t)\mathbf j && \text{first derivative}\\ \mathbf a&=-48\cos(4t)\mathbf i-48\sin(4t)\mathbf j && \text{second derivative}\\ \mathbf v(\pi/8)&=-12\mathbf i\,\mathrm{m\,s^{-1}}\\ \mathbf a(\pi/8)&=-48\mathbf j\,\mathrm{m\,s^{-2}} \end{aligned}

Answer. Velocity is −12i m s−1-12\mathbf i\,\mathrm{m\,s^{-1}} and acceleration is −48j m s−2-48\mathbf j\,\mathrm{m\,s^{-2}} at the specified time.

Check. Doubling the angular speed doubles R∣ω∣R|\omega| and quadruples Rω2R\omega^2. The derivative magnitudes have exactly those changes. The radius still has length 3.

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