Hard Further Maths Horizontal circular motion Questions

Challenging, exam-style Further Maths Horizontal circular motion questions with worked solutions. Stretch yourself on the hardest banked-track, friction, limiting-speeds, smooth-cone problems.

banked-trackfrictionlimiting-speedssmooth-coneresolvingradial-equation
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
A particle PP is attached to two light inextensible strings APAP and BPBP, whose ends AA and BB are attached to a fixed vertical pole with AA vertically above BB. When PP moves in a horizontal circle about the pole with both strings taut, the point AA is a distance hh vertically above the plane of the circle. Which of the following expressions gives the least angular speed for which the string BPBP can remain taut?
Show worked solution

Worked solution

  1. Set up the geometry

    let h and k be the heights of A above and B below the plane of the circle\text{let }h\ \text{and}\ k\ \text{be the heights of }A\ \text{above and }B\ \text{below the plane of the circle}

    The circle is horizontal, so AA is a height hh above it and BB a depth kk below.

  2. Resolve vertically for PP

    hAPTAP−kBPTBP−mg=0\frac{h}{AP}T_{AP}-\frac{k}{BP}T_{BP}-mg=0

    APAP pulls upwards and inwards, BPBP pulls downwards and inwards.

  3. Resolve horizontally, towards the pole

    TAPAP+TBPBP=mω2\frac{T_{AP}}{AP}+\frac{T_{BP}}{BP}=m\omega^{2}

    Both tensions have a component towards the axis; the radius cancels.

  4. Impose the condition for BPBP to be taut

    TBP≥0T_{BP}\ge0

    A string can pull but it cannot push.

  5. Set TBP=0T_{BP}=0 for the limiting case

    hAPTAP=mg,TAPAP=mω2\frac{h}{AP}T_{AP}=mg,\qquad\frac{T_{AP}}{AP}=m\omega^{2}

    At the least angular speed the lower string is just about to go slack.

  6. Eliminate the tension

    h mω2=mg ⇒ ω2=ghh\,m\omega^{2}=mg\ \Rightarrow\ \omega^{2}=\frac{g}{h}

    Substituting TAP=m AP ω2T_{AP}=m\,AP\,\omega^{2} into the vertical equation removes TAPT_{AP}.

  7. Take the positive square root

    ω=gh\omega=\sqrt{\frac{g}{h}}

    Below this angular speed TBPT_{BP} would have to be negative, which is impossible.

  8. Recognise the limiting case

    the system becomes a conical pendulum on AP\text{the system becomes a conical pendulum on }AP

    With BPBP slack only one string acts, and hh is the depth of the circle below AA.

  9. Note that the answer does not involve APAP

    ωmin⁡=gh\omega_{\min}=\sqrt{\frac{g}{h}}

    Only the vertical height of AA above the circle matters.

  10. Note that the mass does not appear

    m cancelsm\ \text{cancels}

    As always in circular motion under gravity, the mass divides out.

  11. Recall the acceleration of a particle in a horizontal circle

    a=rω2=v2ra=r\omega^{2}=\frac{v^{2}}{r}

    The acceleration is directed towards the centre of the circle at every instant.

  12. Recall the link between the linear and the angular speed

    v=rωv=r\omega

    A point at distance rr from the axis sweeps out arc length rωr\omega each second.

  13. Recall the link between angular speed and period

    ω=2πT\omega=\frac{2\pi}{T}

    One complete revolution turns the radius through 2π2\pi radians.

  14. Recall the link between angular speed and revolutions per minute

    ω=2πN60\omega=\frac{2\pi N}{60}

    NN revolutions per minute is N60\frac{N}{60} revolutions per second.

  15. Note that the vertical acceleration is zero

    resolving vertically: ΣF↑=0\text{resolving vertically: }\Sigma F_{\uparrow}=0

    The circle is horizontal and the speed is constant, so nothing accelerates vertically.

  16. Select the correct expression

    gh\sqrt{\frac{g}{h}}

    This is the only option that satisfies both the vertical and the radial equations.

Answer
gh\sqrt{\frac{g}{h}}
Question 2
9 markschallenging
A car of mass mm travels round a bend on a road banked at a constant angle α\alpha to the horizontal. The bend is modelled as an arc of a horizontal circle of radius rr. The coefficient of friction between the tyres of the car and the road is μ\mu, where μ<tan⁡α\mu<\tan\alpha. Select the expression that gives the least speed at which the car can travel round the bend without slipping.
Show worked solution

Worked solution

  1. Identify the forces on the car

    weight mg; normal reaction N⊥road; friction F along the road surface\text{weight }mg;\ \text{normal reaction }N\perp\text{road};\ \text{friction }F\ \text{along the road surface}

    On a banked bend all three forces have a horizontal component or a vertical one.

  2. Recall the law of friction

    F≤μNF\le\mu N

    At a limiting speed the friction takes its greatest possible value μN\mu N.

  3. Decide the direction of the friction

    the car tends to slide down the bank, so F acts up the slope\text{the car tends to slide down the bank, so }F\ \text{acts up the slope}

    Friction always opposes the tendency to slide.

  4. Resolve vertically

    Ncos⁡α+μNsin⁡α−mg=0N\cos\alpha+\mu N\sin\alpha-mg=0

    The friction lies along the slope, so it has a vertical component as well.

  5. Resolve horizontally, towards the centre

    Nsin⁡α−μNcos⁡α=mv2rN\sin\alpha-\mu N\cos\alpha=\frac{mv^{2}}{r}

    The reaction and the friction both contribute to the centripetal force.

  6. Make NN the subject of the vertical equation

    N=mgcos⁡α+μsin⁡αN=\frac{mg}{\cos\alpha+\mu\sin\alpha}

    The normal reaction is now known in terms of the data.

  7. Substitute for NN in the radial equation

    mg(sin⁡α−μcos⁡α)cos⁡α+μsin⁡α=mv2r\frac{mg\left(\sin\alpha-\mu\cos\alpha\right)}{\cos\alpha+\mu\sin\alpha}=\frac{mv^{2}}{r}

    Eliminating NN leaves a single equation in vv.

  8. Cancel the mass and rearrange

    v2=gr(sin⁡α−μcos⁡α)cos⁡α+μsin⁡αv^{2}=\frac{gr\left(\sin\alpha-\mu\cos\alpha\right)}{\cos\alpha+\mu\sin\alpha}

    The mass divides out of both sides.

  9. Divide the numerator and the denominator by cos⁡α\cos\alpha

    v2=gr(tan⁡α−μ)1+μtan⁡αv^{2}=\frac{gr\left(\tan\alpha-\mu\right)}{1+\mu\tan\alpha}

    This puts the answer in terms of tan⁡α\tan\alpha, which is what is wanted.

  10. Take the positive square root

    v=gr(tan⁡α−μ)1+μtan⁡αv=\sqrt{\frac{gr\left(\tan\alpha-\mu\right)}{1+\mu\tan\alpha}}

    Speed is positive, so only the positive root is relevant.

  11. Note that the mass cancels

    m appears in every term and divides outm\ \text{appears in every term and divides out}

    The limiting speeds depend only on gg, rr, α\alpha and μ\mu.

  12. Note the design speed

    v0=grtan⁡αv_{0}=\sqrt{gr\tan\alpha}

    Setting μ=0\mu=0 in either limiting speed returns the design speed.

  13. Note the condition on μ\mu

    μtan⁡α<1\mu\tan\alpha<1

    Otherwise the denominator of the greatest speed vanishes and no upper limit exists.

  14. Note the condition for a positive least speed

    tan⁡α>μ\tan\alpha>\mu

    If μ≥tan⁡α\mu\ge\tan\alpha the car can be at rest on the bank, so the least speed is 00.

  15. Select the correct expression

    gr(tan⁡α−μ)1+μtan⁡α\sqrt{\frac{gr\left(\tan\alpha-\mu\right)}{1+\mu\tan\alpha}}

    This is the only option that satisfies both the vertical and the radial equations.

Answer
gr(tan⁡α−μ)1+μtan⁡α\sqrt{\frac{gr\left(\tan\alpha-\mu\right)}{1+\mu\tan\alpha}}
Question 3
9 markschallenging
A bend on a smooth road is modelled as an arc of a horizontal circle of radius rr, banked at a constant angle α\alpha to the horizontal. A second bend on the same road is banked at the same angle α\alpha but is modelled as an arc of a horizontal circle of radius 2r2r. By what factor is the speed at which a car can travel round the bend with no sideways frictional force multiplied?
Show worked solution

Worked solution

  1. Write down the design speed

    v=grtan⁡αv=\sqrt{gr\tan\alpha}

    On a smooth banked bend the design speed satisfies tan⁡α=v2rg\tan\alpha=\frac{v^{2}}{rg}.

  2. Replace rr by 2r2r at the same angle

    v′=g(2r)tan⁡α=2grtan⁡αv'=\sqrt{g\left(2r\right)\tan\alpha}=\sqrt{2}\sqrt{gr\tan\alpha}

    Only the radius changes, so only the factor 2\sqrt{2} appears.

  3. Form the ratio

    v′v=2\frac{v'}{v}=\sqrt{2}

    Doubling the radius multiplies the design speed by 2\sqrt{2}, not by 22.

  4. Note that the speed is constant

    ∣v∣=constant\left|v\right|=\text{constant}

    Only the direction of the velocity changes, and that is what the acceleration does.

  5. Note that the acceleration points at the centre

    a is directed along the inward radius\mathbf{a}\ \text{is directed along the inward radius}

    This is why the resultant force must also point at the centre.

  6. Reject any expression that is linear in ω\omega

    doubling ω must quadruple a\text{doubling }\omega\ \text{must quadruple }a

    The acceleration depends on the square of the angular speed.

  7. Check the effect of doubling the radius

    r→2r ⇒ a→2a at fixed ωr\rightarrow2r\ \Rightarrow\ a\rightarrow2a\ \text{at fixed}\ \omega

    The acceleration is directly proportional to the radius at fixed angular speed.

  8. Recall the acceleration of a particle in a horizontal circle

    a=rω2=v2ra=r\omega^{2}=\frac{v^{2}}{r}

    The acceleration is directed towards the centre of the circle at every instant.

  9. Recall the link between the linear and the angular speed

    v=rωv=r\omega

    A point at distance rr from the axis sweeps out arc length rωr\omega each second.

  10. Recall the link between angular speed and period

    ω=2πT\omega=\frac{2\pi}{T}

    One complete revolution turns the radius through 2π2\pi radians.

  11. Recall the link between angular speed and revolutions per minute

    ω=2πN60\omega=\frac{2\pi N}{60}

    NN revolutions per minute is N60\frac{N}{60} revolutions per second.

  12. Note that the vertical acceleration is zero

    resolving vertically: ΣF↑=0\text{resolving vertically: }\Sigma F_{\uparrow}=0

    The circle is horizontal and the speed is constant, so nothing accelerates vertically.

  13. Note that the tangential acceleration is zero

    speed constant ⇒ no tangential component\text{speed constant}\ \Rightarrow\ \text{no tangential component}

    The whole resultant force is therefore horizontal and points at the centre.

  14. Write Newton's second law towards the centre

    Fnet=mrω2=mv2rF_{\text{net}}=mr\omega^{2}=\frac{mv^{2}}{r}

    The resultant of all the forces, resolved towards the centre, supplies the acceleration.

  15. Select the correct expression

    2\sqrt{2}

    This is the only option that satisfies both the vertical and the radial equations.

Answer
2\sqrt{2}
Question 4
9 markschallenging
A chairoplane ride has chairs hanging from the rim of a horizontal disc of radius 3.53.5 m which rotates about a fixed vertical axis through its centre. Each chair hangs from a light chain of length 2.52.5 m attached to the rim of the disc. When the ride is running steadily, a chair carrying a rider of total mass 6060 kg moves in a horizontal circle and the chain makes a constant angle θ\theta with the vertical, where cos⁡θ=35\cos\theta=\frac{3}{5}. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the speed of the chair. Give your answer to 33 significant figures.
Show worked solution

Worked solution

  1. Find the radius of the circle described by the chair

    r=R+lsin⁡θ=3.5+2.5×0.8=5.5 mr=R+l\sin\theta=3.5+2.5\times 0.8=5.5\ \text{m}

    The chair swings out beyond the rim of the disc, so the radius is bigger than RR.

  2. Find sin⁡θ\sin\theta from cos⁡θ\cos\theta

    sin⁡θ=1−(35)2=0.8\sin\theta=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=0.8

    Use the Pythagorean identity.

  3. Resolve vertically for the chair

    Tcos⁡θ−mg=0T\cos\theta-mg=0

    The chair stays at the same height, so the vertical forces balance.

  4. Resolve horizontally, towards the axis

    Tsin⁡θ=mrω2T\sin\theta=mr\omega^{2}

    The horizontal component of the tension is the whole centripetal force.

  5. Divide the radial equation by the vertical equation

    tan⁡θ=rω2g\tan\theta=\frac{r\omega^{2}}{g}

    Dividing eliminates both the tension and the mass.

  6. Make ω2\omega^{2} the subject

    ω2=gtan⁡θr=9.8×1.3335.5\omega^{2}=\frac{g\tan\theta}{r}=\frac{9.8\times 1.333}{5.5}

    Note this is NOT the simple conical-pendulum formula: the radius is R+lsin⁡θR+l\sin\theta.

  7. Find the speed from the angular speed

    v=rω=5.5×1.541v=r\omega=5.5\times 1.541

    The chair travels round a circle of radius 5.55.5 m.

  8. Find the angular speed

    ω=1.541 rad s−1\omega=1.541\ \text{rad s}^{-1}

    Take the positive square root of ω2\omega^{2}.

  9. Find the tension in the chain

    T=mgcos⁡θ=980 NT=\frac{mg}{\cos\theta}=980\ \text{N}

    The chain must support more than the weight because it is not vertical.

  10. Check the radial equation

    Tsin⁡θ=784,mrω2=784T\sin\theta=784,\qquad mr\omega^{2}=784

    Both sides agree, so the solution is consistent.

  11. Check the vertical equation

    Tcos⁡θ=588≈mg=588T\cos\theta=588\approx mg=588

    The vertical component of the tension does support the chair.

  12. Find the period of the ride

    Tp=2πω=4.076 sT_{p}=\frac{2\pi}{\omega}=4.076\ \text{s}

    This is the time for one circuit of the ride.

  13. Note the angle of the chain

    θ=arccos⁡(35)=53.13∘\theta=\arccos\left(\frac{3}{5}\right)=53.13^{\circ}

    The faster the disc turns, the further out the chairs swing.

  14. Recall the acceleration of a particle in a horizontal circle

    a=rω2=v2ra=r\omega^{2}=\frac{v^{2}}{r}

    The acceleration is directed towards the centre of the circle at every instant.

  15. Recall the link between the linear and the angular speed

    v=rωv=r\omega

    A point at distance rr from the axis sweeps out arc length rωr\omega each second.

  16. State the answer

    v=8.48 m s−1v=8.48\ \text{m s}^{-1}

    This is the quantity the question asked for.

Answer
v=8.48 m s−1v=8.48\ \text{m s}^{-1}
Question 5
9 markschallenging
A chairoplane ride has chairs hanging from the rim of a horizontal disc of radius 55 m which rotates about a fixed vertical axis through its centre. Each chair hangs from a light chain of length 44 m attached to the rim of the disc. When the ride is running steadily, a chair carrying a rider of total mass 7070 kg moves in a horizontal circle and the chain makes a constant angle θ\theta with the vertical, where cos⁡θ=1213\cos\theta=\frac{12}{13}. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the tension in the chain. Give your answer to 33 significant figures.
Show worked solution

Worked solution

  1. Find the radius of the circle described by the chair

    r=R+lsin⁡θ=5+4×0.3846=6.538 mr=R+l\sin\theta=5+4\times 0.3846=6.538\ \text{m}

    The chair swings out beyond the rim of the disc, so the radius is bigger than RR.

  2. Find sin⁡θ\sin\theta from cos⁡θ\cos\theta

    sin⁡θ=1−(1213)2=0.3846\sin\theta=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=0.3846

    Use the Pythagorean identity.

  3. Resolve vertically for the chair

    Tcos⁡θ−mg=0T\cos\theta-mg=0

    The chair stays at the same height, so the vertical forces balance.

  4. Resolve horizontally, towards the axis

    Tsin⁡θ=mrω2T\sin\theta=mr\omega^{2}

    The horizontal component of the tension is the whole centripetal force.

  5. Divide the radial equation by the vertical equation

    tan⁡θ=rω2g\tan\theta=\frac{r\omega^{2}}{g}

    Dividing eliminates both the tension and the mass.

  6. Make ω2\omega^{2} the subject

    ω2=gtan⁡θr=9.8×0.41676.538\omega^{2}=\frac{g\tan\theta}{r}=\frac{9.8\times 0.4167}{6.538}

    Note this is NOT the simple conical-pendulum formula: the radius is R+lsin⁡θR+l\sin\theta.

  7. Use the vertical equation to find the tension

    T=mgcos⁡θ=70×9.81213T=\frac{mg}{\cos\theta}=\frac{70\times 9.8}{\frac{12}{13}}

    The tension is determined by the vertical equation alone.

  8. Find the angular speed

    ω=0.7903 rad s−1\omega=0.7903\ \text{rad s}^{-1}

    Take the positive square root of ω2\omega^{2}.

  9. Find the tension in the chain

    T=mgcos⁡θ=743.2 NT=\frac{mg}{\cos\theta}=743.2\ \text{N}

    The chain must support more than the weight because it is not vertical.

  10. Check the radial equation

    Tsin⁡θ=285.8,mrω2=285.8T\sin\theta=285.8,\qquad mr\omega^{2}=285.8

    Both sides agree, so the solution is consistent.

  11. Check the vertical equation

    Tcos⁡θ=686≈mg=686T\cos\theta=686\approx mg=686

    The vertical component of the tension does support the chair.

  12. Find the period of the ride

    Tp=2πω=7.951 sT_{p}=\frac{2\pi}{\omega}=7.951\ \text{s}

    This is the time for one circuit of the ride.

  13. Note the angle of the chain

    θ=arccos⁡(1213)=22.62∘\theta=\arccos\left(\frac{12}{13}\right)=22.62^{\circ}

    The faster the disc turns, the further out the chairs swing.

  14. Recall the acceleration of a particle in a horizontal circle

    a=rω2=v2ra=r\omega^{2}=\frac{v^{2}}{r}

    The acceleration is directed towards the centre of the circle at every instant.

  15. Recall the link between the linear and the angular speed

    v=rωv=r\omega

    A point at distance rr from the axis sweeps out arc length rωr\omega each second.

  16. Recall the link between angular speed and period

    ω=2πT\omega=\frac{2\pi}{T}

    One complete revolution turns the radius through 2π2\pi radians.

  17. State the answer

    T=743 NT=743\ \text{N}

    This is the quantity the question asked for.

Answer
T=743 NT=743\ \text{N}

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