Horizontal circular motion Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Horizontal circular motion questions. See exactly how to solve problems on angular-speed, circular-motion, v-equals-r-omega, acceleration.

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.

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 fairground big wheel turns at a constant rate, completing one revolution every 4040 seconds. Find the angular speed of the wheel, in rad s−1^{-1}. Give your answer to 33 significant figures.

Worked solution

  1. Write down the period

    T=40 sT=40\ \text{s}

    The period is the time taken for one complete revolution.

  2. Use the relation between angular speed and period

    ω=2πT=2π40\omega=\frac{2\pi}{T}=\frac{2\pi}{40}

    One revolution is 2π2\pi radians, completed in TT seconds.

  3. Note that the direction of the acceleration is towards the centre

    a points along the radius, at the centrea\ \text{points along the radius, at the centre}

    The speed is constant, so there is no tangential component.

  4. Evaluate the angular speed

    ω=0.157 rad s−1\omega=0.157\ \text{rad s}^{-1}

    This is the angular speed of the wheel.

Answer
ω=0.157 rad s−1\omega=0.157\ \text{rad s}^{-1}
Question 3
2 markseasy
A grinding wheel turns about a fixed axis with constant angular speed 88 rad s−1^{-1}. Find the rate of rotation of the wheel, in revolutions per minute. Give your answer to 33 significant figures.

Worked solution

  1. Find the number of revolutions per second

    ω2π=82π\frac{\omega}{2\pi}=\frac{8}{2\pi}

    Each revolution is 2π2\pi radians, so divide the angular speed by 2π2\pi.

  2. Multiply by 6060 to get revolutions per minute

    N=60×82πN=\frac{60\times 8}{2\pi}

    There are 6060 seconds in a minute.

  3. Note that the direction of the acceleration is towards the centre

    a points along the radius, at the centrea\ \text{points along the radius, at the centre}

    The speed is constant, so there is no tangential component.

  4. Evaluate the rate of rotation

    N=76.4 rev min−1N=76.4\ \text{rev min}^{-1}

    This is the rate of rotation in revolutions per minute.

Answer
N=76.4 rev min−1N=76.4\ \text{rev min}^{-1}
Question 4
2 markseasy
The tip of a fan blade moves in a horizontal circle of radius 0.250.25 m with constant angular speed 6060 rad s−1^{-1}. Find the speed of the tip of the blade, in m s−1^{-1}.

Worked solution

  1. Write down the relation between speed and angular speed

    v=rωv=r\omega

    The speed is the arc length swept out each second.

  2. Substitute the radius and the angular speed

    v=0.25×60v=0.25\times 60

    The radius is in metres and the angular speed in radians per second.

  3. Evaluate the speed

    v=15 m s−1v=15\ \text{m s}^{-1}

    This is the speed of the tip of the blade.

Answer
v=15 m s−1v=15\ \text{m s}^{-1}
Question 5
2 markseasy
A cyclist rides round a horizontal circular track of radius 4040 m at a constant speed of 1212 m s−1^{-1}. Find the angular speed of the cyclist, in rad s−1^{-1}.

Worked solution

  1. Write down the relation between speed and angular speed

    v=rω ⇒ ω=vrv=r\omega\ \Rightarrow\ \omega=\frac{v}{r}

    Rearrange v=rωv=r\omega to make the angular speed the subject.

  2. Substitute the speed and the radius

    ω=1240\omega=\frac{12}{40}

    Both quantities are already in SI units.

  3. Evaluate the angular speed

    ω=0.3 rad s−1\omega=0.3\ \text{rad s}^{-1}

    This is the angular speed of the cyclist.

Answer
ω=0.3 rad s−1\omega=0.3\ \text{rad s}^{-1}

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