Sign in to jmaths to print this page or save it as a PDF: jmaths-8ce638778fd6.herokuapp.com/maths/ai-hl/vectors/ahl-4-26/problems/

jmaths

IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.26

Circular motion Problem set A Worked solutions

Skills

  1. A bead is on a spinning wheel. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} −12\sin 4t \\ 12\cos 4t \end{pmatrix}\) m s⁻¹. When \(t = 0\) its position vector is \(\begin{pmatrix} 3 \\ 0 \end{pmatrix}\) m. Find its position vector at time \(t\).

    • Integrate each component. \(\int −12\sin 4t\, dt = 3\cos 4t + c_1\) and \(\int 12\cos 4t\, dt = 3\sin 4t + c_2\).
    • At \(t = 0\) the position is \(\begin{pmatrix} 3 \\ 0 \end{pmatrix}\), which gives \(3\cos 0 + c_1 = 3\) and \(3\sin 0 + c_2 = 0\), so \(c_1 = c_2 = 0\).
    • \(\mathbf{r} = \begin{pmatrix} 3\cos 4t \\ 3\sin 4t \end{pmatrix}\)

  2. A dot moves round a dial. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 6\cos 3t \\ −6\sin 3t \end{pmatrix}\) m s⁻¹. When \(t = 0\) its position vector is \(\begin{pmatrix} 0 \\ 2 \end{pmatrix}\) m. Find its position vector at time \(t\).

    • Integrate each component. \(\int 6\cos 3t\, dt = 2\sin 3t + c_1\) and \(\int −6\sin 3t\, dt = 2\cos 3t + c_2\).
    • At \(t = 0\) the position is \(\begin{pmatrix} 0 \\ 2 \end{pmatrix}\), which gives \(2\sin 0 + c_1 = 0\) and \(2\cos 0 + c_2 = 2\), so \(c_1 = c_2 = 0\).
    • \(\mathbf{r} = \begin{pmatrix} 2\sin 3t \\ 2\cos 3t \end{pmatrix}\)

  3. A satellite model and a toy car are tracked. Each position vector gives the position of an object at time \(t\) seconds, in metres. Show that each object moves in a circle about the origin, and state the radius.

    1. (a) \(\mathbf{r} = \begin{pmatrix} 5\cos 2t \\ 5\sin 2t \end{pmatrix}\)

      • \(|\mathbf{r}|^2 = (5\cos 2t)^2 + (5\sin 2t)^2 = 25(\sin^2 2t + \cos^2 2t)\)
      • Use \(\sin^2\theta + \cos^2\theta = 1\): \(|\mathbf{r}| = \sqrt{25} = 5\)
      • The distance from the origin is always 5 m: a circle of radius 5 m about the origin.

    2. (b) \(\mathbf{r} = \begin{pmatrix} 1.5\sin 7t \\ 1.5\cos 7t \end{pmatrix}\)

      • \(|\mathbf{r}|^2 = (1.5\sin 7t)^2 + (1.5\cos 7t)^2 = 2.25(\sin^2 7t + \cos^2 7t)\)
      • Use \(\sin^2\theta + \cos^2\theta = 1\): \(|\mathbf{r}| = \sqrt{2.25} = 1.5\)
      • The distance from the origin is always 1.5 m: a circle of radius 1.5 m about the origin.

  4. Each position vector gives the position of an object at time \(t\) seconds, in metres. Show that each object moves in a circle, and state the centre and the radius.

    1. (a) \(\mathbf{r} = \begin{pmatrix} 8 + 1.5\cos 4t \\ −2 + 1.5\sin 4t \end{pmatrix}\)

      • \(\mathbf{r} − \begin{pmatrix} 8 \\ −2 \end{pmatrix} = \begin{pmatrix} 1.5\cos 4t \\ 1.5\sin 4t \end{pmatrix}\).
      • \(\sqrt{1.5^{2}\cos^{2} 4t + 1.5^{2}\sin^{2} 4t} = 1.5\), using \(\sin^{2}\theta + \cos^{2}\theta = 1\).
      • The distance from \((8, −2)\) is always 1.5 m: a circle with that centre and radius.

    2. (b) \(\mathbf{r} = \begin{pmatrix} −3 + 2\sin 3t \\ 5 + 2\cos 3t \end{pmatrix}\)

      • \(\mathbf{r} − \begin{pmatrix} −3 \\ 5 \end{pmatrix} = \begin{pmatrix} 2\sin 3t \\ 2\cos 3t \end{pmatrix}\).
      • \(\sqrt{2^{2}\sin^{2} 3t + 2^{2}\cos^{2} 3t} = 2\), using \(\sin^{2}\theta + \cos^{2}\theta = 1\).
      • The distance from \((−3, 5)\) is always 2 m: a circle with that centre and radius.

  5. Each position vector gives the position of an object at time \(t\) seconds, in metres, about the origin. State whether the object goes round the circle clockwise or anticlockwise, and find the time it takes to go round once.

    1. (a) \(\mathbf{r} = \begin{pmatrix} 2\cos 3t \\ 2\sin 3t \end{pmatrix}\)

      • \(\mathbf{r}(0) = \begin{pmatrix} 2 \\ 0 \end{pmatrix}\), \(\mathbf{v}(0) = \begin{pmatrix} 0 \\ 6 \end{pmatrix}\) (from \(\mathbf{v} = \begin{pmatrix} −6\sin 3t \\ 6\cos 3t \end{pmatrix}\)).
      • From \((2, 0)\) the object moves upwards: anticlockwise.
      • \(3T = 2\pi\), \(T = 2.09\) s.

    2. (b) \(\mathbf{r} = \begin{pmatrix} 5\cos \dfrac{t}{2} \\ −5\sin \dfrac{t}{2} \end{pmatrix}\)

      • \(\mathbf{r}(0) = \begin{pmatrix} 5 \\ 0 \end{pmatrix}\), \(\mathbf{v}(0) = \begin{pmatrix} 0 \\ −2.5 \end{pmatrix}\) (from \(\mathbf{v} = \begin{pmatrix} −2.5\sin \dfrac{t}{2} \\ −2.5\cos \dfrac{t}{2} \end{pmatrix}\)).
      • From \((5, 0)\) the object moves downwards: clockwise.
      • \(\dfrac{T}{2} = 2\pi\), \(T = 12.6\) s.

    3. (c) \(\mathbf{r} = \begin{pmatrix} 1.5\sin 4t \\ 1.5\cos 4t \end{pmatrix}\)

      • \(\mathbf{r}(0) = \begin{pmatrix} 0 \\ 1.5 \end{pmatrix}\), \(\mathbf{v}(0) = \begin{pmatrix} 6 \\ 0 \end{pmatrix}\) (from \(\mathbf{v} = \begin{pmatrix} 6\cos 4t \\ −6\sin 4t \end{pmatrix}\)).
      • From \((0, 1.5)\) the object moves to the right: clockwise.
      • \(4T = 2\pi\), \(T = 1.57\) s.

In context

  1. A small light is fixed to the edge of a turntable. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} −10\sin 5t \\ 10\cos 5t \end{pmatrix}\) m s⁻¹. When \(t = 0\) its position vector is \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\) m, relative to the centre of the wheel.

    1. (a) Find the position vector of the light at time \(t\).

      • Integrate: \(\int −10\sin 5t\, dt = 2\cos 5t + c_1\), \(\int 10\cos 5t\, dt = 2\sin 5t + c_2\).
      • \(2\cos 0 + c_1 = 2\) and \(2\sin 0 + c_2 = 0\), so \(c_1 = c_2 = 0\).
      • \(\mathbf{r} = \begin{pmatrix} 2\cos 5t \\ 2\sin 5t \end{pmatrix}\)

    2. (b) Find its distance from the centre of the wheel, and describe its path.

      • \(|\mathbf{r}|^2 = (2\cos 5t)^2 + (2\sin 5t)^2 = 4(\sin^2 5t + \cos^2 5t)\)
      • Use \(\sin^2\theta + \cos^2\theta = 1\): \(|\mathbf{r}| = \sqrt{4} = 2\)
      • The distance from the origin is always 2 m: a circle of radius 2 m about the origin.

    3. (c) Find the first time at which it is back at its starting position.

      • It is back at the start when the angle \(5t\) reaches \(2\pi\).
      • \(5t = 2\pi\), so \(t = \dfrac{2\pi}{5} = 1.26\) s

  2. A seat on a fairground ride moves in a vertical plane. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 15\cos 3t \\ −15\sin 3t \end{pmatrix}\) m s⁻¹. When \(t = 0\) its position vector is \(\begin{pmatrix} 0 \\ 5 \end{pmatrix}\) m, relative to the centre of the ride.

    1. (a) Find the position vector of the seat at time \(t\).

      • Integrate: \(\int 15\cos 3t\, dt = 5\sin 3t + c_1\), \(\int −15\sin 3t\, dt = 5\cos 3t + c_2\).
      • \(5\sin 0 + c_1 = 0\) and \(5\cos 0 + c_2 = 5\), so \(c_1 = c_2 = 0\).
      • \(\mathbf{r} = \begin{pmatrix} 5\sin 3t \\ 5\cos 3t \end{pmatrix}\)

    2. (b) Show that it moves in a circle about the centre and state the radius.

      • \(|\mathbf{r}|^2 = (5\sin 3t)^2 + (5\cos 3t)^2 = 25(\sin^2 3t + \cos^2 3t)\)
      • Use \(\sin^2\theta + \cos^2\theta = 1\): \(|\mathbf{r}| = \sqrt{25} = 5\)
      • The distance from the origin is always 5 m: a circle of radius 5 m about the origin.

    3. (c) Find its position when \(t = 1\). Give the coordinates to 3 significant figures.

      • Put \(t = 1\) in, with the calculator in radians: \(\mathbf{r} = \begin{pmatrix} 5\sin 3 \\ 5\cos 3 \end{pmatrix}\)
      • \(\mathbf{r} = \begin{pmatrix} 0.706 \\ −4.95 \end{pmatrix}\)

  3. The diagram shows an observation wheel of radius 15 m. Its centre is 17 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 30 seconds.

    xyOcentrestart

    1. (a) Write down the position vector of the seat at time \(t\) seconds.

      • \(\omega = \dfrac{2\pi}{30} = \dfrac{\pi}{15}\) rad s⁻¹.
      • Start at the right-hand point of the circle, so cosine goes in the x-component and sine in the y-component. The centre \((0, 17)\) is added.
      • Anticlockwise: a plus sign on the sine. \(\mathbf{r} = \begin{pmatrix} 15\cos \dfrac{\pi t}{15} \\ 17 + 15\sin \dfrac{\pi t}{15} \end{pmatrix}\).

    2. (b) Find its height above the ground when \(t = 5\).

      • In radians: \(y = 17 + 15\sin\dfrac{5\pi}{15} = 30.0\) m.

    3. (c) Find the speed of the seat.

      • \(|\mathbf{v}| = r\omega = 15 \times 0.2094 = 3.14\) m s⁻¹.

Stretch

  1. A sensor is fixed to a slowly turning dish. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} −3\sin \dfrac{t}{2} \\ 3\cos \dfrac{t}{2} \end{pmatrix}\) m s⁻¹. When \(t = 0\) its position vector is \(\begin{pmatrix} 6 \\ 0 \end{pmatrix}\) m.

    1. (a) Find its position vector at time \(t\).

      • Integrate: \(\int −3\sin \dfrac{t}{2}\, dt = 6\cos \dfrac{t}{2} + c_1\), \(\int 3\cos \dfrac{t}{2}\, dt = 6\sin \dfrac{t}{2} + c_2\). Dividing by \(\frac{1}{2}\) is the same as multiplying by 2.
      • \(6\cos 0 + c_1 = 6\) and \(6\sin 0 + c_2 = 0\), so \(c_1 = c_2 = 0\).
      • \(\mathbf{r} = \begin{pmatrix} 6\cos \dfrac{t}{2} \\ 6\sin \dfrac{t}{2} \end{pmatrix}\)

    2. (b) Show that it moves in a circle about the origin and state the radius.

      • \(|\mathbf{r}|^2 = (6\cos \dfrac{t}{2})^2 + (6\sin \dfrac{t}{2})^2 = 36(\sin^2 \dfrac{t}{2} + \cos^2 \dfrac{t}{2})\)
      • Use \(\sin^2\theta + \cos^2\theta = 1\): \(|\mathbf{r}| = \sqrt{36} = 6\)
      • The distance from the origin is always 6 m: a circle of radius 6 m about the origin.

    3. (c) Show that its speed is constant and state it.

      • \(|\mathbf{v}| = \sqrt{(3\sin \dfrac{t}{2})^2 + (3\cos \dfrac{t}{2})^2} = \sqrt{9(\sin^2 \dfrac{t}{2} + \cos^2 \dfrac{t}{2})}\)
      • \(|\mathbf{v}| = \sqrt{9} = 3\) m s⁻¹, which does not depend on \(t\).

jmaths.xyz · AI HL · AHL 4.26 Circular motion