3.3 - Horizontal circular motion

3.3 - Horizontal circular motion

A particle can have constant speed and still accelerate. Learn how angular speed describes a turn, why circular motion needs an inward resultant force, and how friction or a sloping string can supply that force. You will use these ideas for horizontal circles and conical pendulums.

Speed and angular speed

Two points on the same rotating wheel finish each turn together. Why does the point nearer the rim travel faster? Both sweep out the same angle, but the outer point travels along a longer arc.

Recall that an arc of radius rr and angle θ\theta radians has length s=rθs=r\theta. For steady rotation, divide by the elapsed time: distance per second is radius times radians per second.

Angular speed

The angular speed ω\omega of a particle moving steadily around a circle is the angle swept out per unit time, measured in rad s−1\mathrm{rad\,s^{-1}}. A complete revolution sweeps out 2π2\pi radians.

A period PP is the time for one complete turn. A frequency ff is the number of turns per second, so P=1/fP=1/f.

Linear and angular speed

For a circle of radius rr, with constant speed vv and angular speed ω\omega,

v=rω,ω=2πP=2πf.v=r\omega,\qquad \omega=\frac{2\pi}{P}=2\pi f.

Convert revolutions per minute to revolutions per second before using the final expression. These relations are not printed in the formula booklet: learn them. The arc-length relation s=rθs=r\theta is printed there.

Here speed means the magnitude of the velocity. Its direction will be important in the next part.

Worked example 1: a point near the rim

Question. A wheel makes one revolution every 4 s4\,\mathrm{s}. Find its angular speed and the speed of a point 0.6 m0.6\,\mathrm{m} from its centre.

Plan. Convert one turn to radians, then use the point's radius.

ω=2π4=π2 rad s−1v=0.6(π2)=0.3π m s−1\begin{aligned} \omega&=\frac{2\pi}{4}=\frac{\pi}{2}\,\mathrm{rad\,s^{-1}} \\ v&=0.6\left(\frac{\pi}{2}\right)=0.3\pi\,\mathrm{m\,s^{-1}} \end{aligned}

Answer. The angular speed is π/2 rad s−1\pi/2\,\mathrm{rad\,s^{-1}}, and the point's speed is 0.3π m s−10.3\pi\,\mathrm{m\,s^{-1}}.

Check. Its circumference is 2π(0.6)=1.2π m2\pi(0.6)=1.2\pi\,\mathrm{m}. Dividing by 4 s4\,\mathrm{s} gives the same speed.

Keep the wheel's rotation unchanged and move the point closer to the centre. The angular speed stays fixed; the linear speed changes.

Worked example 2: a point nearer the centre

Question. The same wheel takes 4 s4\,\mathrm{s} per revolution. A point is 0.2 m0.2\,\mathrm{m} from its centre. Find its speed and the distance it travels in 2.52.5 revolutions.

Plan. Use the same angular speed, then multiply the circumference by the number of turns.

v=0.2(π2)=0.1π m s−1s=2.5×2π(0.2)=π m\begin{aligned} v&=0.2\left(\frac{\pi}{2}\right)=0.1\pi\,\mathrm{m\,s^{-1}} \\ s&=2.5\times2\pi(0.2)=\pi\,\mathrm{m} \end{aligned}

Answer. Its speed is 0.1π m s−10.1\pi\,\mathrm{m\,s^{-1}}, and the distance is π m\pi\,\mathrm{m}.

Check. The motion takes 2.5×4=10 s2.5\times4=10\,\mathrm{s}. Speed times time gives 0.1π×10=π m0.1\pi\times10=\pi\,\mathrm{m}.

The two speeds are in the ratio 0.6:0.2=3:10.6:0.2=3:1, as the arc lengths suggest.

Watch out

Putting 2020 revolutions per minute straight into v=rωv=r\omega gives the wrong units. First convert to 20(2π)/60 rad s−120(2\pi)/60\,\mathrm{rad\,s^{-1}}.

Apply that conversion to a different wheel.

Acceleration towards the centre

At any instant a particle's velocity is tangent to its circular path. A little later the tangent points in a different direction. Even if the speed is unchanged, that change in velocity means there is an acceleration.

For constant speed, this acceleration points towards the centre. There is no acceleration along the tangent because the speed is not changing. A proof of the circular-acceleration formula is not needed here, but its direction is essential to choosing a force equation.

[DIAGRAM: asset_name: Velocity turns while speed stays fixed; asset_slug: cam9231-circular-velocity; caption: Tangential velocity changes direction while the acceleration points towards the centre; description: Exact circle with two marked positions on an anticlockwise path. Equal-length blue velocity arrows are tangent at the two positions. Lavender acceleration arrows point from each position towards the centre O. A dashed radius r is shown and a side panel distinguishes constant speed from changing velocity.]
Tangential velocity changes direction while the acceleration points towards the centre

Centre-directed acceleration

For constant-speed circular motion of radius r>0r>0,

a=v2r=rω2.a=\frac{v^2}{r}=r\omega^2.

The acceleration points towards the centre. Both expressions and this direction are in the formula booklet. They agree because v=rωv=r\omega.

If the speed varies around a circle, v2/rv^2/r is still the inward radial component of acceleration; a tangential component may also be present. We will use that distinction for vertical circles.

Worked example 3: the two acceleration formulae agree

Question. A particle moves at constant speed 6 m s−16\,\mathrm{m\,s^{-1}} around a circle of radius 2 m2\,\mathrm{m}. Find its angular speed and acceleration.

Plan. Find ω\omega from the speed, then calculate the inward acceleration.

ω=vr=62=3 rad s−1a=622=18 m s−2\begin{aligned} \omega&=\frac vr=\frac62=3\,\mathrm{rad\,s^{-1}} \\ a&=\frac{6^2}{2}=18\,\mathrm{m\,s^{-2}} \end{aligned}

Answer. The angular speed is 3 rad s−13\,\mathrm{rad\,s^{-1}}, and the acceleration is 18 m s−218\,\mathrm{m\,s^{-2}} towards the centre.

Check. The alternative formula gives rω2=2(32)=18 m s−2r\omega^2=2(3^2)=18\,\mathrm{m\,s^{-2}}. Also, v2/rv^2/r has units m s−2\mathrm{m\,s^{-2}}, as an acceleration must.

Change only the radius while keeping the linear speed fixed. A gentler turn needs less acceleration.

Worked example 4: a larger circle at the same speed

Question. A second particle moves at constant speed 6 m s−16\,\mathrm{m\,s^{-1}} around a circle of radius 3 m3\,\mathrm{m}. Find its angular speed and acceleration, and compare its acceleration with Worked example 3.

Plan. Keep v=6v=6 in v2/rv^2/r; do not keep the old angular speed.

ω=63=2 rad s−1a=363=12 m s−2\begin{aligned} \omega&=\frac63=2\,\mathrm{rad\,s^{-1}} \\ a&=\frac{36}{3}=12\,\mathrm{m\,s^{-2}} \end{aligned}

The acceleration is 12/18=2/312/18=2/3 of the earlier value.

Answer. ω=2 rad s−1\omega=2\,\mathrm{rad\,s^{-1}}, and the acceleration is 12 m s−212\,\mathrm{m\,s^{-2}} inwards, two-thirds of the previous acceleration.

Check. The period is 2πr/v=π s2\pi r/v=\pi\,\mathrm{s}, so ω=2π/π=2\omega=2\pi/\pi=2. Then rω2=3(4)=12r\omega^2=3(4)=12.

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