Menu
Sign up Sign in

AI HL · AHL 4

Vectors

0 of 160 skills done

0
AHL 4.26

Circular motion

AHL 4.26

Circular motion

Target: I can integrate a velocity vector with sine and cosine components, and use the starting position to find the position vector. (+5 more)

By the end of this section

Key points

Circular motion about the origin

  • An object moving in a circle about the origin keeps a constant distance from the origin, \(|\mathbf{r}|\). Its velocity changes direction all the time.
  • Starting at \((r, 0)\) and going anticlockwise, \(\mathbf{r} = \begin{pmatrix} r\cos\omega t \\ r\sin\omega t \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} -r\omega\sin\omega t \\ r\omega\cos\omega t \end{pmatrix}\). The speed \(r\omega\) is constant.
  • The velocity is perpendicular to the position vector: \(\mathbf{r}\cdot\mathbf{v} = 0\). Draw a circle with \(\mathbf{r}\) from the centre and \(\mathbf{v}\) along the tangent.

Integrating trigonometric components

  • \(\cos\) integrates to \(\sin\), and \(\sin\) integrates to \(−\cos\). Divide by the number multiplying \(t\): \(\int \cos kt\, dt = \dfrac{1}{k}\sin kt\).
  • Use \(\mathbf{r}(0)\) to find the constants. Put \(t = 0\) into your answer and check that it gives the starting position.
  • Work in radians. Set the calculator to radian mode when you put in a value of \(t\).

Showing that the path is a circle

  • The distance from the origin is \(|\mathbf{r}|\). Square and add the components, take out the common factor, then use \(\sin^2\theta + \cos^2\theta = 1\).
  • If \(|\mathbf{r}|\) is the same number for every \(t\), the object moves in a circle about the origin, and that number is the radius.
  • Write the identity down. It is the reason the distance does not depend on \(t\).

Centre, direction and period

  • A circle with centre \((a, b)\) has \(\mathbf{r} = \begin{pmatrix} a + R\cos\omega t \\ b + R\sin\omega t \end{pmatrix}\). Take away the centre first, then find the distance. The velocity does not contain \(a\) or \(b\).
  • Starting at the right-hand point, a plus sign on \(\sin\omega t\) goes anticlockwise and a minus sign goes clockwise. To check, find \(\mathbf{v}(0)\): at the right-hand point, up is anticlockwise and down is clockwise.
  • One full turn takes \(T = \dfrac{2\pi}{\omega}\), where \(\omega\) is the number multiplying \(t\) (not the radius). The speed is \(r\omega\).
  • To write a position vector from a description, use the centre, the radius, where it starts, which way it goes, and \(\omega = \dfrac{2\pi}{T}\).

Explore: circular motion

The object has position r = (r cos ωt, r sin ωt). Choose the radius r and ω, then move the time slider round one full circle.

Distance from the centre, |r|
Speed, |v| = rω
Time for one circle, 2π/ω

Ink arrow: the position r from the centre. Teal arrow: the velocity v. Gold arrow: the acceleration a. The arrows are drawn to a convenient length, so read the sizes from the numbers. The calculator is in radians.

Move the time slider to move the object.

  1. Move the time slider. What stays the same and what changes: the distance from the centre, the speed, the direction of the velocity?
  2. Make ω twice as big. What happens to the speed, to the time for one circle and to the size of the acceleration?
  3. Make r twice as big and keep ω. What happens to the speed and to the time for one circle? Why does the time not change?
  4. Look at r · v. Why is it always 0? What does that say about the angle between v and r?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Position from a trigonometric velocity

An object has velocity \(\mathbf{v} = \begin{pmatrix} −9\sin 3t \\ 9\cos 3t \end{pmatrix}\) m s⁻¹ at time \(t\) seconds. When \(t = 0\) its position vector is \(\begin{pmatrix} 3 \\ 0 \end{pmatrix}\) m. Find its position vector at time \(t\).

Working
  1. Integrate

    \(\int −9\sin 3t\, dt = 3\cos 3t + c_1\), \(\int 9\cos 3t\, dt = 3\sin 3t + c_2\)

    \(\cos\) integrates to \(\sin\), and \(\sin\) to \(−\cos\). Divide by 3 because of the \(3t\).

  2. Use the start

    \(3\cos 0 + c_1 = 3\), \(3\sin 0 + c_2 = 0\)

    At \(t = 0\) the position is \(\begin{pmatrix} 3 \\ 0 \end{pmatrix}\). Both constants are zero.

  3. xyvOt = 0
    Position

    \(\mathbf{r} = \begin{pmatrix} 3\cos 3t \\ 3\sin 3t \end{pmatrix}\)

    Check: put \(t = 0\) in and you get \(\begin{pmatrix} 3 \\ 0 \end{pmatrix}\).

Video walkthrough coming
Show the path is a circle

An object has position vector \(\mathbf{r} = \begin{pmatrix} 5\cos 4t \\ 5\sin 4t \end{pmatrix}\) m at time \(t\) seconds. Find its distance from the origin, and describe its path.

Working
  1. The magnitude

    \(|\mathbf{r}| = \sqrt{(5\cos 4t)^2 + (5\sin 4t)^2}\)

    The distance from the origin is the magnitude of the position vector.

  2. Factor

    \(= \sqrt{25(\cos^2 4t + \sin^2 4t)}\)

    Take 25 out of both terms.

  3. Use the identity

    \(= \sqrt{25} = 5\)

    \(\sin^2\theta + \cos^2\theta = 1\) for any angle.

  4. xyrvOP
    The path

    A circle of radius 5 m about the origin

    The distance from the origin is the same at every time.

Video walkthrough coming
A circle that starts at the top

An object has velocity \(\mathbf{v} = \begin{pmatrix} 4\cos 2t \\ −4\sin 2t \end{pmatrix}\) m s⁻¹ at time \(t\) seconds. When \(t = 0\) its position vector is \(\begin{pmatrix} 0 \\ 2 \end{pmatrix}\) m. Find its position vector, show that it moves in a circle about the origin, and state the radius.

Working
  1. Integrate

    \(\int 4\cos 2t\, dt = 2\sin 2t + c_1\), \(\int −4\sin 2t\, dt = 2\cos 2t + c_2\)

    Divide by 2 because of the \(2t\). The sign of \(\sin\) integrates to \(−\cos\), and the minus in the velocity cancels it.

  2. Use the start

    \(2\sin 0 + c_1 = 0\), \(2\cos 0 + c_2 = 2\)

    At \(t = 0\) the position is \(\begin{pmatrix} 0 \\ 2 \end{pmatrix}\). Both constants are zero.

  3. xyvO
    Position

    \(\mathbf{r} = \begin{pmatrix} 2\sin 2t \\ 2\cos 2t \end{pmatrix}\)

    It starts at the top and moves to the right: clockwise.

  4. Distance

    \(|\mathbf{r}| = \sqrt{4(\sin^2 2t + \cos^2 2t)} = 2\)

    The distance is 2 m at every time: a circle of radius 2 m.

Video walkthrough coming
A slower turn: the argument t/3

An object has velocity \(\mathbf{v} = \begin{pmatrix} −2\sin \dfrac{t}{3} \\ 2\cos \dfrac{t}{3} \end{pmatrix}\) m s⁻¹ at time \(t\) seconds. When \(t = 0\) its position vector is \(\begin{pmatrix} 6 \\ 0 \end{pmatrix}\) m. Find its position vector, show that it moves in a circle about the origin, and find the time it takes to go once round.

Working
  1. Integrate

    \(\int −2\sin \dfrac{t}{3}\, dt = 6\cos \dfrac{t}{3} + c_1\), \(\int 2\cos \dfrac{t}{3}\, dt = 6\sin \dfrac{t}{3} + c_2\)

    The number multiplying \(t\) is \(\frac{1}{3}\). Dividing by it is the same as multiplying by 3.

  2. Use the start

    \(6\cos 0 + c_1 = 6\), \(6\sin 0 + c_2 = 0\)

    Both constants are zero.

  3. Position

    \(\mathbf{r} = \begin{pmatrix} 6\cos \dfrac{t}{3} \\ 6\sin \dfrac{t}{3} \end{pmatrix}\)

    At \(t = 0\) this is \(\begin{pmatrix} 6 \\ 0 \end{pmatrix}\).

  4. Distance

    \(|\mathbf{r}| = \sqrt{36(\cos^2 \frac{t}{3} + \sin^2 \frac{t}{3})} = 6\)

    A circle of radius 6 m about the origin.

  5. xyvOt = 0
    Once round

    \(\dfrac{t}{3} = 2\pi\), \(t = 6\pi = 18.8\) s

    One full turn is an angle of \(2\pi\).

Video walkthrough coming
A circle that is not centred at the origin

An object moves so that its position vector at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix} 6 + 2.5\cos 2t \\ 3 + 2.5\sin 2t \end{pmatrix}\) m. Show that it moves in a circle, and state the centre and the radius.

Working
  1. Take away the centre

    \(\mathbf{r} − \begin{pmatrix} 6 \\ 3 \end{pmatrix} = \begin{pmatrix} 2.5\cos 2t \\ 2.5\sin 2t \end{pmatrix}\)

    The numbers added on are the centre. What is left is a circle about the origin.

  2. Distance from the centre

    \(\sqrt{2.5^{2}\cos^{2} 2t + 2.5^{2}\sin^{2} 2t} = 2.5\)

    State that \(\sin^{2}\theta + \cos^{2}\theta = 1\). The distance from the point (6, 3) is the same at every time.

  3. xyvOCt = 0
    Conclusion

    A circle with centre \((6, 3)\) and radius 2.5 m

    The distance from the origin is not constant: \(|\mathbf{r}|\) is 9.01 at t = 0 and 4.61 when \(t = \frac{\pi}{2}\). Measure from the centre.

  4. Velocity and speed

    \(\mathbf{v} = \begin{pmatrix} −5\sin 2t \\ 5\cos 2t \end{pmatrix}\), \(|\mathbf{v}| = 5\) m s⁻¹

    The constants 6 and 3 disappear when you differentiate. The speed is \(r\omega = 2.5 \times 2\).

Video walkthrough coming
Clockwise or anticlockwise, and the time for one turn

An object has position vector \(\mathbf{r} = \begin{pmatrix} 3\cos 1.5t \\ −3\sin 1.5t \end{pmatrix}\) m at time \(t\) seconds. Find the direction in which it goes round the circle, and the time it takes to go round once.

Working
  1. Start

    \(\mathbf{r}(0) = \begin{pmatrix} 3 \\ 0 \end{pmatrix}\)

    The object starts at the right-hand point of the circle.

  2. xyvOt = 0
    Velocity at the start

    \(\mathbf{v} = \begin{pmatrix} −4.5\sin 1.5t \\ −4.5\cos 1.5t \end{pmatrix}\), \(\mathbf{v}(0) = \begin{pmatrix} 0 \\ −4.5 \end{pmatrix}\)

    At the right-hand point the velocity points down. Anticlockwise motion would go up here, so this is clockwise.

  3. Time for one turn

    \(1.5T = 2\pi\), \(T = 4.19\) s

    One full turn is an angle of \(2\pi\). Divide by the number multiplying t.

Video walkthrough coming
Writing the position vector of a wheel

The diagram shows an observation wheel of radius 20 m. Its centre is 22 m above level ground. Take the origin O on the ground directly below the centre, with \(\mathbf{i}\) horizontal and \(\mathbf{j}\) vertically up. A seat starts at the right-hand point of the wheel when \(t = 0\), and the wheel turns anticlockwise once every 36 seconds. Write down the position vector of the seat at time \(t\) seconds, and find its height above the ground when \(t = 6\).

xyOcentrestart
Working
  1. Angular speed

    \(\omega = \dfrac{2\pi}{36} = \dfrac{\pi}{18}\) rad s⁻¹

    One turn, an angle of 2π, takes 36 seconds.

  2. Position vector

    \(\mathbf{r} = \begin{pmatrix} 20\cos\dfrac{\pi t}{18} \\ 22 + 20\sin\dfrac{\pi t}{18} \end{pmatrix}\) m

    Start at the right-hand point and go anticlockwise: cosine in the x-component, sine in the y-component. The centre (0, 22) is added.

  3. Height at t = 6

    In radians: \(y = 22 + 20\sin\dfrac{6\pi}{18} = 39.3\) m

    Put t = 6 into the y-component. The calculator must be in radian mode.

Video walkthrough coming