Further Maths Horizontal circular motion Practice Questions

Free Further Maths Horizontal circular motion practice questions with full step-by-step worked solutions. Covers angular-speed, circular-motion, v-equals-r-omega, acceleration. Practise exam-style problems and check your method.

angular-speedcircular-motionv-equals-r-omegaaccelerationr-omega-squarednewton-second-law
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The rotor of a centrifuge spins about a fixed vertical axis at a constant rate of 24002400 revolutions per minute. Find the angular speed of the rotor, in rad s−1^{-1}. Give your answer to 33 significant figures.
Show worked solution

Worked solution

  1. Convert revolutions per minute to revolutions per second

    240060 rev s−1\frac{2400}{60}\ \text{rev s}^{-1}

    There are 6060 seconds in a minute.

  2. Convert revolutions per second to radians per second

    ω=2π×240060\omega=\frac{2\pi\times 2400}{60}

    One complete revolution is a turn of 2π2\pi radians.

  3. Evaluate the angular speed

    ω=251 rad s−1\omega=251\ \text{rad s}^{-1}

    This is the angular speed of the rotor.

Answer
ω=251 rad s−1\omega=251\ \text{rad s}^{-1}
Question 2
2 markseasy
A particle of mass mm moves in a horizontal circle of radius rr with constant angular speed ω\omega. The angular speed is now doubled, while the mass and the radius are unchanged. By what factor is the magnitude of the resultant force on the particle towards the centre multiplied?
Show worked solution

Worked solution

  1. Write down the radial equation

    F=mrω2F=mr\omega^{2}

    The resultant force towards the centre is mrω2mr\omega^{2}.

  2. Replace ω\omega by 2ω2\omega

    F′=mr(2ω)2=4mrω2F'=mr\left(2\omega\right)^{2}=4mr\omega^{2}

    Squaring the factor 22 produces a factor 44.

  3. Form the ratio

    F′F=4\frac{F'}{F}=4

    The force is four times as large.

  4. Select the correct expression

    44

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

Answer
44
Question 3
4 marksintermediate
A particle attached to a fixed point OO by a light inextensible string moves as a conical pendulum with constant angular speed ω\omega, and the centre of its circle is at a depth hh below OO. The angular speed is now doubled and the particle again moves as a conical pendulum on the same string. By what factor is the depth hh multiplied?
Show worked solution

Worked solution

  1. Write down the depth of the centre below OO

    h=lcos⁡θ=gω2h=l\cos\theta=\frac{g}{\omega^{2}}

    The vertical and radial equations combine to give cos⁡θ=glω2\cos\theta=\frac{g}{l\omega^{2}}.

  2. Replace ω\omega by 2ω2\omega

    h′=g(2ω)2=g4ω2h'=\frac{g}{\left(2\omega\right)^{2}}=\frac{g}{4\omega^{2}}

    The depth is inversely proportional to the square of the angular speed.

  3. Form the ratio

    h′h=14\frac{h'}{h}=\frac{1}{4}

    The circle rises: the faster the pendulum turns, the shallower the cone.

  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. Select the correct expression

    14\frac{1}{4}

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

Answer
14\frac{1}{4}
Question 4
6 markshard
A particle of mass mm moves in a horizontal circle on the smooth inner surface of a fixed hollow cone with its axis vertical and its vertex downwards. The semi-vertical angle of the cone is α\alpha. Which of the following expressions gives the magnitude of the normal reaction of the cone on the particle?
Show worked solution

Worked solution

  1. Identify the forces on the particle

    weight mg down; normal reaction N⊥surface\text{weight }mg\ \text{down};\ \text{normal reaction }N\perp\text{surface}

    The cone is smooth, so there is no friction along the surface.

  2. Fix the direction of the normal

    N makes an angle α with the horizontalN\ \text{makes an angle }\alpha\ \text{with the horizontal}

    The surface makes an angle α\alpha with the vertical axis, so its normal makes α\alpha with the horizontal.

  3. Resolve vertically

    Nsin⁡α−mg=0N\sin\alpha-mg=0

    The vertical component of the reaction supports the whole weight.

  4. Resolve horizontally, towards the axis

    Ncos⁡α=mv2rN\cos\alpha=\frac{mv^{2}}{r}

    The horizontal component of the reaction is the centripetal force.

  5. Make NN the subject of the vertical equation

    N=mgsin⁡αN=\frac{mg}{\sin\alpha}

    The reaction is now known in terms of the data.

  6. Note the corresponding angular speed

    ω=vr=grtan⁡α\omega=\frac{v}{r}=\sqrt{\frac{g}{r\tan\alpha}}

    Divide the speed by the radius of the circle.

  7. Note the effect of a steeper cone

    α↓ ⇒ tan⁡α↓ ⇒ v↑\alpha\downarrow\ \Rightarrow\ \tan\alpha\downarrow\ \Rightarrow\ v\uparrow

    A steep cone gives the surface very little vertical grip, so the particle must move fast.

  8. Note the limit of a very flat cone

    α→π2 ⇒ v→0\alpha\rightarrow\frac{\pi}{2}\ \Rightarrow\ v\rightarrow0

    A horizontal plane cannot hold a particle on a circle at all.

  9. Substitute for NN in the radial equation

    mgcos⁡αsin⁡α=mv2r\frac{mg\cos\alpha}{\sin\alpha}=\frac{mv^{2}}{r}

    Eliminating NN leaves a single equation in vv.

  10. Cancel the mass and rearrange

    v2=grcos⁡αsin⁡α=grtan⁡αv^{2}=\frac{gr\cos\alpha}{\sin\alpha}=\frac{gr}{\tan\alpha}

    The mass cancels, so the speed does not depend on the mass.

  11. Select the correct expression

    mgsin⁡α\frac{mg}{\sin\alpha}

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

Answer
mgsin⁡α\frac{mg}{\sin\alpha}
Question 5
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}}

Unlock 65 more Horizontal circular motion questions

Create a free account to work through every Further Maths Horizontal circular motion question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Horizontal circular motion practice

Hard QuestionsWorked Solutions

Related Mechanics topics

Elastic collisions in one dimensionElastic collisions in two dimensionsCentres of mass: solids and calculusCentres of mass: laminas and frameworksElastic strings and springsFurther dynamics