Further Maths Vertical circular motion Practice Questions

Free Further Maths Vertical circular motion practice questions with full step-by-step worked solutions. Covers vertical-circular-motion, conservation-of-energy, speed, radial-equation. Practise exam-style problems and check your method.

vertical-circular-motionconservation-of-energyspeedradial-equationtensionhighest-point
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A particle PP of mass 0.5 kg0.5\ \text{kg} is attached to one end of a light inextensible string of length 0.8 m0.8\ \text{m}. The other end of the string is attached to a fixed point OO. With the string taut, PP hangs in equilibrium vertically below OO and is then projected horizontally with speed 7 m s−17\ \text{m}\ \text{s}^{-1}. Take g=9.8 m s−2g=9.8\ \text{m}\ \text{s}^{-2}. Find the speed of PP at the highest point of the circle.
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Worked solution

  1. Apply conservation of energy between the lowest point and PP

    12(7)2=12v2+(9.8)(1.6)\frac{1}{2}\left(7\right)^{2}=\frac{1}{2}v^{2}+\left(9.8\right)\left(1.6\right)

    The tension is perpendicular to the velocity, so it does no work and the mechanical energy of PP is conserved; the mass cancels throughout.

  2. Solve for the square of the speed

    v2=17.64 m2 s−2v^{2}=17.64\ \text{m}^{2}\ \text{s}^{-2}

    This is positive, so PP does reach the position considered.

  3. State the speed of PP

    v=17.64=4.2 m s−1v=\sqrt{17.64}=4.2\ \text{m}\ \text{s}^{-1}

    This is the speed of PP at the position asked for.

Answer
4.2 m s−14.2\ \text{m}\ \text{s}^{-1}
Question 2
2 markseasy
A particle PP is placed at the highest point of a fixed smooth solid sphere of radius aa and centre OO. It is given a negligible disturbance and slides down the outer surface. Let θ\theta be the angle that OPOP makes with the upward vertical. Which of the following gives the value of cos⁡θ\cos\theta at the instant when PP leaves the surface of the sphere?
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Worked solution

  1. Use energy from the highest point

    v2=2ga(1−cos⁡θ)v^{2}=2ga\left(1-\cos\theta\right)

    The particle starts from rest and has fallen a height a(1−cos⁡θ)a\left(1-\cos\theta\right).

  2. Apply Newton's second law towards OO

    mgcos⁡θ−R=mv2amg\cos\theta-R=\frac{mv^{2}}{a}

    On the outside of the sphere the reaction points away from the centre.

  3. Set R=0R=0 and substitute for v2v^{2}

    gcos⁡θ=2g(1−cos⁡θ) ⇒ 3cos⁡θ=2g\cos\theta=2g\left(1-\cos\theta\right)\ \Rightarrow\ 3\cos\theta=2

    Contact is lost the instant the normal reaction falls to zero.

  4. Select the value

    cos⁡θ=23\cos\theta=\frac{2}{3}

    The particle leaves the sphere after turning through about 48.2∘48.2^{\circ}.

Answer
cos⁡θ=23\cos\theta=\frac{2}{3}
Question 3
4 marksintermediate
Two particles of equal mass are each projected horizontally with speed uu from the point vertically below a fixed point OO. One is attached to OO by a light inextensible string of length aa; the other is attached to OO by a light rigid rod of length aa which is freely pivoted at OO. Which of the following correctly compares the least speeds of projection needed for complete vertical circles?
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Worked solution

  1. State the condition at the top for the string

    T≥0 ⇒ vtop2≥gaT\geq 0\ \Rightarrow\ v_{\text{top}}^{2}\geq ga

    A string can only pull, so the tension may not become negative.

  2. State the condition at the top for the rod

    vtop2≥0v_{\text{top}}^{2}\geq 0

    A rod can push, so the particle only has to reach the top.

  3. Use energy to bring each condition to the lowest point

    u2=vtop2+4gau^{2}=v_{\text{top}}^{2}+4ga

    The top of the circle is a height 2a2a above the bottom.

  4. Write the two conditions

    string: u2≥5ga,rod: u2>4ga\text{string: }u^{2}\geq 5ga,\qquad\text{rod: }u^{2}>4ga

    Since 4ga<5ga4ga<5ga, the rod needs the smaller speed.

  5. State the acceleration of a particle moving on a circle

    aradial=v2ra_{\text{radial}}=\frac{v^{2}}{r}

    A particle moving on a circle of radius rr with speed vv has an acceleration of magnitude v2r\frac{v^{2}}{r} directed towards the centre.

  6. Write down Newton's second law towards the centre

    Ftowards O=mv2aF_{\text{towards }O}=\frac{mv^{2}}{a}

    Only the component of the resultant force along POPO produces the circular motion.

  7. Select the correct comparison

    4ga<5ga\sqrt{4ga}<\sqrt{5ga}

    The rod's least speed 2ga2\sqrt{ga} is smaller than the string's 5ga\sqrt{5ga}.

Answer
The string requires u2≥5gau^{2}\geq 5ga while the rod requires u2>4gau^{2}>4ga, so the least speed needed on the rod is the smaller of the two.
Question 4
6 markshard
A particle PP is attached to one end of a light inextensible string of length aa, the other end of which is attached to a fixed point OO. The particle is projected horizontally with speed uu from the point vertically below OO, where u2<2gau^{2}<2ga, so that PP oscillates. Which of the following gives cos⁡θ\cos\theta at the greatest angle θ\theta that OPOP makes with the downward vertical?
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Worked solution

  1. Identify the condition at the extreme position

    v=0v=0

    At the greatest angle the particle is instantaneously at rest.

  2. Apply conservation of energy

    12u2=ga(1−cos⁡θ)\frac{1}{2}u^{2}=ga\left(1-\cos\theta\right)

    All of the initial kinetic energy has become potential energy.

  3. Rearrange for cos⁡θ\cos\theta

    cos⁡θ=1−u22ga\cos\theta=1-\frac{u^{2}}{2ga}

    With u2<2gau^{2}<2ga this is positive, so θ<90∘\theta<90^{\circ} and the string is still taut.

  4. State the acceleration of a particle moving on a circle

    aradial=v2ra_{\text{radial}}=\frac{v^{2}}{r}

    A particle moving on a circle of radius rr with speed vv has an acceleration of magnitude v2r\frac{v^{2}}{r} directed towards the centre.

  5. Write down Newton's second law towards the centre

    Ftowards O=mv2aF_{\text{towards }O}=\frac{mv^{2}}{a}

    Only the component of the resultant force along POPO produces the circular motion.

  6. Note that the tension does no work

    T⋅v=0\mathbf{T}\cdot\mathbf{v}=0

    The tension acts along POPO while the velocity is along the tangent, so the mechanical energy of PP is conserved.

  7. State conservation of energy on the circle

    12mu2=12mv2+mgh\frac{1}{2}mu^{2}=\frac{1}{2}mv^{2}+mgh

    The weight is the only force that does work, so the total mechanical energy is constant.

  8. State the height risen above the lowest point

    h=a(1−cos⁡θ)h=a\left(1-\cos\theta\right)

    With θ\theta measured from the downward vertical, PP lies a distance acos⁡θa\cos\theta below OO.

  9. Note that the mass cancels in the energy equation

    12u2=12v2+ga(1−cos⁡θ)\frac{1}{2}u^{2}=\frac{1}{2}v^{2}+ga\left(1-\cos\theta\right)

    Every term carries a factor mm, so the speed at a given point does not depend on the mass.

  10. Recall that a string can only pull

    T≥0T\geq 0

    A string may go slack, but it can never push the particle outwards.

  11. Select the expression

    cos⁡θ=1−u22ga\cos\theta=1-\frac{u^{2}}{2ga}

    The string stays taut throughout an oscillation of this kind.

Answer
cos⁡θ=1−u22ga\cos\theta=1-\frac{u^{2}}{2ga}
Question 5
9 markschallenging
A particle PP of mass mm is attached to one end of a light rigid rod of length aa, freely pivoted at a fixed point OO, and is projected horizontally with speed uu from the point vertically below OO. Which of the following best describes the motion when 4ga<u2<5ga4ga<u^{2}<5ga?
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Worked solution

  1. Check that the circle is completed

    u2>4ga ⇒ vtop2=u2−4ga>0u^{2}>4ga\ \Rightarrow\ v_{\text{top}}^{2}=u^{2}-4ga>0

    A rod needs only u2>4gau^{2}>4ga, so PP does get round.

  2. Check what a string would do

    u2<5ga ⇒ Ttop=mu2a−5mg<0u^{2}<5ga\ \Rightarrow\ T_{\text{top}}=\frac{mu^{2}}{a}-5mg<0

    A string would have gone slack: the required force is outwards, which a string cannot supply.

  3. Write the force in the rod as a function of the angle

    F=mu2a−2mg+3mgcos⁡θF=\frac{mu^{2}}{a}-2mg+3mg\cos\theta

    The energy equation and the radial equation together eliminate vv.

  4. Find where the force changes sign

    F=0 ⇒ cos⁡θ=2ga−u23gaF=0\ \Rightarrow\ \cos\theta=\frac{2ga-u^{2}}{3ga}

    With 4ga<u2<5ga4ga<u^{2}<5ga this lies strictly between −1-1 and −23-\frac{2}{3}, so the change happens near the top.

  5. Interpret the sign of FF beyond that angle

    F<0 (a thrust)F<0\ \text{(a thrust)}

    Beyond that point the rod must push PP outwards to keep it on the circle.

  6. State the acceleration of a particle moving on a circle

    aradial=v2ra_{\text{radial}}=\frac{v^{2}}{r}

    A particle moving on a circle of radius rr with speed vv has an acceleration of magnitude v2r\frac{v^{2}}{r} directed towards the centre.

  7. Write down Newton's second law towards the centre

    Ftowards O=mv2aF_{\text{towards }O}=\frac{mv^{2}}{a}

    Only the component of the resultant force along POPO produces the circular motion.

  8. Note that the tension does no work

    T⋅v=0\mathbf{T}\cdot\mathbf{v}=0

    The tension acts along POPO while the velocity is along the tangent, so the mechanical energy of PP is conserved.

  9. State conservation of energy on the circle

    12mu2=12mv2+mgh\frac{1}{2}mu^{2}=\frac{1}{2}mv^{2}+mgh

    The weight is the only force that does work, so the total mechanical energy is constant.

  10. State the height risen above the lowest point

    h=a(1−cos⁡θ)h=a\left(1-\cos\theta\right)

    With θ\theta measured from the downward vertical, PP lies a distance acos⁡θa\cos\theta below OO.

  11. Note that the mass cancels in the energy equation

    12u2=12v2+ga(1−cos⁡θ)\frac{1}{2}u^{2}=\frac{1}{2}v^{2}+ga\left(1-\cos\theta\right)

    Every term carries a factor mm, so the speed at a given point does not depend on the mass.

  12. Recall that a string can only pull

    T≥0T\geq 0

    A string may go slack, but it can never push the particle outwards.

  13. Recall the condition for complete circles on a string

    u2≥5gau^{2}\geq 5ga

    The tension must stay non-negative all the way to the highest point.

  14. Recall the condition at the highest point for a string

    vtop2≥gav_{\text{top}}^{2}\geq ga

    At the top the weight alone must not exceed the force needed to hold PP on the circle.

  15. Recall that a rod can push as well as pull

    T<0 is allowed (a thrust)T<0\ \text{is allowed (a thrust)}

    A rigid rod can exert a thrust, so the only requirement is that PP reaches the top.

  16. Select the correct description

    cos⁡θ=2ga−u23ga\cos\theta=\frac{2ga-u^{2}}{3ga}

    The rod carries a thrust from this angle up to the highest point.

Answer
PP performs complete vertical circles, but near the top the rod exerts a thrust rather than a tension; the force in the rod is zero where cos⁡θ=2ga−u23ga\cos\theta=\frac{2ga-u^{2}}{3ga}.

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