Vertical circular motion Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Vertical circular motion questions. See exactly how to solve problems on vertical-circular-motion, conservation-of-energy, speed, radial-equation.

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.

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 of mass 0.6 kg0.6\ \text{kg} is attached to one end of a light inextensible string of length 1.2 m1.2\ \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 6 m s−16\ \text{m}\ \text{s}^{-1}. Take g=9.8 m s−2g=9.8\ \text{m}\ \text{s}^{-2}. Find the speed of PP when OPOP makes an angle of 60∘60^{\circ} with the downward vertical. Give your answer to 33 significant figures.

Worked solution

  1. Write down the data

    m=0.6 kg,a=1.2 m,u=6 m s−1m=0.6\ \text{kg},\quad a=1.2\ \text{m},\quad u=6\ \text{m}\ \text{s}^{-1}

    aa is the radius of the vertical circle and uu is the speed of PP at the lowest point.

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

    12(6)2=12v2+(9.8)(0.6)\frac{1}{2}\left(6\right)^{2}=\frac{1}{2}v^{2}+\left(9.8\right)\left(0.6\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.

  3. Solve for the square of the speed

    v2=24.24 m2 s−2v^{2}=24.24\ \text{m}^{2}\ \text{s}^{-2}

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

  4. State the speed of PP

    v=24.24=4.92 m s−1v=\sqrt{24.24}=4.92\ \text{m}\ \text{s}^{-1}

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

Answer
4.92 m s−14.92\ \text{m}\ \text{s}^{-1}
Question 3
2 markseasy
A particle PP of mass 0.8 kg0.8\ \text{kg} moves on the inside of a smooth circular track of radius 0.5 m0.5\ \text{m} which is fixed in a vertical plane with centre OO. The particle passes through the lowest point of the track with speed 5 m s−15\ \text{m}\ \text{s}^{-1}. Take g=9.8 m s−2g=9.8\ \text{m}\ \text{s}^{-2}. Find the speed of PP when OPOP makes an angle of 90∘90^{\circ} with the downward vertical. Give your answer to 33 significant figures.

Worked solution

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

    12(5)2=12v2+(9.8)(0.5)\frac{1}{2}\left(5\right)^{2}=\frac{1}{2}v^{2}+\left(9.8\right)\left(0.5\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=15.2 m2 s−2v^{2}=15.2\ \text{m}^{2}\ \text{s}^{-2}

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

  3. State the speed of PP

    v=15.2=3.90 m s−1v=\sqrt{15.2}=3.90\ \text{m}\ \text{s}^{-1}

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

Answer
3.90 m s−13.90\ \text{m}\ \text{s}^{-1}
Question 4
2 markseasy
A particle PP of mass 0.4 kg0.4\ \text{kg} is attached to one end of a light rigid rod of length 0.5 m0.5\ \text{m}, the other end of which is freely pivoted at a fixed point OO, so that PP can move on a vertical circle with centre OO. With PP hanging vertically below OO, the particle is projected horizontally with speed 6 m s−16\ \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. Give your answer to 33 significant figures.

Worked solution

  1. Write down the data

    m=0.4 kg,a=0.5 m,u=6 m s−1m=0.4\ \text{kg},\quad a=0.5\ \text{m},\quad u=6\ \text{m}\ \text{s}^{-1}

    aa is the radius of the vertical circle and uu is the speed of PP at the lowest point.

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

    12(6)2=12v2+(9.8)(1)\frac{1}{2}\left(6\right)^{2}=\frac{1}{2}v^{2}+\left(9.8\right)\left(1\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.

  3. Solve for the square of the speed

    v2=16.4 m2 s−2v^{2}=16.4\ \text{m}^{2}\ \text{s}^{-2}

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

  4. State the speed of PP

    v=16.4=4.05 m s−1v=\sqrt{16.4}=4.05\ \text{m}\ \text{s}^{-1}

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

Answer
4.05 m s−14.05\ \text{m}\ \text{s}^{-1}
Question 5
2 markseasy
A particle PP of mass 2 kg2\ \text{kg} is attached to one end of a light inextensible string of length 1.5 m1.5\ \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 6 m s−16\ \text{m}\ \text{s}^{-1}. Take g=9.8 m s−2g=9.8\ \text{m}\ \text{s}^{-2}. Find the tension in the string immediately after PP is projected.

Worked solution

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

    12(6)2=12v2+(9.8)(0)\frac{1}{2}\left(6\right)^{2}=\frac{1}{2}v^{2}+\left(9.8\right)\left(0\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=36 m2 s−2v^{2}=36\ \text{m}^{2}\ \text{s}^{-2}

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

  3. State the required force

    T=67.6 NT=67.6\ \text{N}

    This is the tension in the string at the position asked for.

Answer
67.6 N67.6\ \text{N}

Unlock 65 more Vertical circular motion questions

Create a free account to work through every Further Maths Vertical 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 Vertical circular motion practice

Hard QuestionsPractice Questions

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