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/b/

jmaths

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

Circular motion Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. A marker is on a rotating disc. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} −14\sin 7t \\ 14\cos 7t \end{pmatrix}\) m s⁻¹. When \(t = 0\) its position vector is \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\) m. Find its position vector at time \(t\).

    • Integrate each component. \(\int −14\sin 7t\, dt = 2\cos 7t + c_1\) and \(\int 14\cos 7t\, dt = 2\sin 7t + c_2\).
    • At \(t = 0\) the position is \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\), which gives \(2\cos 0 + c_1 = 2\) and \(2\sin 0 + c_2 = 0\), so \(c_1 = c_2 = 0\).
    • \(\mathbf{r} = \begin{pmatrix} 2\cos 7t \\ 2\sin 7t \end{pmatrix}\)

  2. A point moves round a clock face. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 20\cos 4t \\ −20\sin 4t \end{pmatrix}\) m s⁻¹. When \(t = 0\) its position vector is \(\begin{pmatrix} 0 \\ 5 \end{pmatrix}\) m. Find its position vector at time \(t\).

    • Integrate each component. \(\int 20\cos 4t\, dt = 5\sin 4t + c_1\) and \(\int −20\sin 4t\, dt = 5\cos 4t + c_2\).
    • At \(t = 0\) the position is \(\begin{pmatrix} 0 \\ 5 \end{pmatrix}\), which gives \(5\sin 0 + c_1 = 0\) and \(5\cos 0 + c_2 = 5\), so \(c_1 = c_2 = 0\).
    • \(\mathbf{r} = \begin{pmatrix} 5\sin 4t \\ 5\cos 4t \end{pmatrix}\)

  3. A drone and a model train 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) \(\mathbf{r} = \begin{pmatrix} 9\cos 3t \\ 9\sin 3t \end{pmatrix}\)

      • \(|\mathbf{r}|^2 = (9\cos 3t)^2 + (9\sin 3t)^2 = 81(\sin^2 3t + \cos^2 3t)\)
      • Use \(\sin^2\theta + \cos^2\theta = 1\): \(|\mathbf{r}| = \sqrt{81} = 9\)
      • The distance from the origin is always 9 m: a circle of radius 9 m about the origin.

    2. (b) \(\mathbf{r} = \begin{pmatrix} 2.5\sin 4t \\ 2.5\cos 4t \end{pmatrix}\)

      • \(|\mathbf{r}|^2 = (2.5\sin 4t)^2 + (2.5\cos 4t)^2 = 6.25(\sin^2 4t + \cos^2 4t)\)
      • Use \(\sin^2\theta + \cos^2\theta = 1\): \(|\mathbf{r}| = \sqrt{6.25} = 2.5\)
      • The distance from the origin is always 2.5 m: a circle of radius 2.5 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) \(\mathbf{r} = \begin{pmatrix} −5 + 2.4\cos 3t \\ 6 + 2.4\sin 3t \end{pmatrix}\)

      • \(\mathbf{r} − \begin{pmatrix} −5 \\ 6 \end{pmatrix} = \begin{pmatrix} 2.4\cos 3t \\ 2.4\sin 3t \end{pmatrix}\).
      • \(\sqrt{2.4^{2}\cos^{2} 3t + 2.4^{2}\sin^{2} 3t} = 2.4\), using \(\sin^{2}\theta + \cos^{2}\theta = 1\).
      • The distance from \((−5, 6)\) is always 2.4 m: a circle with that centre and radius.

    2. (b) \(\mathbf{r} = \begin{pmatrix} 7 + 4\sin 2t \\ −1 + 4\cos 2t \end{pmatrix}\)

      • \(\mathbf{r} − \begin{pmatrix} 7 \\ −1 \end{pmatrix} = \begin{pmatrix} 4\sin 2t \\ 4\cos 2t \end{pmatrix}\).
      • \(\sqrt{4^{2}\sin^{2} 2t + 4^{2}\cos^{2} 2t} = 4\), using \(\sin^{2}\theta + \cos^{2}\theta = 1\).
      • The distance from \((7, −1)\) is always 4 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) \(\mathbf{r} = \begin{pmatrix} 3\cos 2t \\ 3\sin 2t \end{pmatrix}\)

      • \(\mathbf{r}(0) = \begin{pmatrix} 3 \\ 0 \end{pmatrix}\), \(\mathbf{v}(0) = \begin{pmatrix} 0 \\ 6 \end{pmatrix}\) (from \(\mathbf{v} = \begin{pmatrix} −6\sin 2t \\ 6\cos 2t \end{pmatrix}\)).
      • From \((3, 0)\) the object moves upwards: anticlockwise.
      • \(2T = 2\pi\), \(T = 3.14\) s.

    2. (b) \(\mathbf{r} = \begin{pmatrix} 4\cos \dfrac{t}{3} \\ −4\sin \dfrac{t}{3} \end{pmatrix}\)

      • \(\mathbf{r}(0) = \begin{pmatrix} 4 \\ 0 \end{pmatrix}\), \(\mathbf{v}(0) = \begin{pmatrix} 0 \\ −1.33 \end{pmatrix}\) (from \(\mathbf{v} = \begin{pmatrix} −1.33\sin \dfrac{t}{3} \\ −1.33\cos \dfrac{t}{3} \end{pmatrix}\)).
      • From \((4, 0)\) the object moves downwards: clockwise.
      • \(\dfrac{T}{3} = 2\pi\), \(T = 18.8\) s.

    3. (c) \(\mathbf{r} = \begin{pmatrix} 0.8\sin 5t \\ 0.8\cos 5t \end{pmatrix}\)

      • \(\mathbf{r}(0) = \begin{pmatrix} 0 \\ 0.8 \end{pmatrix}\), \(\mathbf{v}(0) = \begin{pmatrix} 4 \\ 0 \end{pmatrix}\) (from \(\mathbf{v} = \begin{pmatrix} 4\cos 5t \\ −4\sin 5t \end{pmatrix}\)).
      • From \((0, 0.8)\) the object moves to the right: clockwise.
      • \(5T = 2\pi\), \(T = 1.26\) s.

In context

  1. A reflector is fixed to the rim of a bicycle wheel. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} −18\sin 6t \\ 18\cos 6t \end{pmatrix}\) m s⁻¹. When \(t = 0\) its position vector is \(\begin{pmatrix} 3 \\ 0 \end{pmatrix}\) m, relative to the centre of the wheel.

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

      • Integrate: \(\int −18\sin 6t\, dt = 3\cos 6t + c_1\), \(\int 18\cos 6t\, dt = 3\sin 6t + c_2\).
      • \(3\cos 0 + c_1 = 3\) and \(3\sin 0 + c_2 = 0\), so \(c_1 = c_2 = 0\).
      • \(\mathbf{r} = \begin{pmatrix} 3\cos 6t \\ 3\sin 6t \end{pmatrix}\)

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

      • \(|\mathbf{r}|^2 = (3\cos 6t)^2 + (3\sin 6t)^2 = 9(\sin^2 6t + \cos^2 6t)\)
      • Use \(\sin^2\theta + \cos^2\theta = 1\): \(|\mathbf{r}| = \sqrt{9} = 3\)
      • The distance from the origin is always 3 m: a circle of radius 3 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 \(6t\) reaches \(2\pi\).
      • \(6t = 2\pi\), so \(t = \dfrac{2\pi}{6} = 1.05\) s

  2. A cabin on a giant wheel moves in a vertical plane. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 24\cos 4t \\ −24\sin 4t \end{pmatrix}\) m s⁻¹. When \(t = 0\) its position vector is \(\begin{pmatrix} 0 \\ 6 \end{pmatrix}\) m, relative to the centre of the ride.

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

      • Integrate: \(\int 24\cos 4t\, dt = 6\sin 4t + c_1\), \(\int −24\sin 4t\, dt = 6\cos 4t + c_2\).
      • \(6\sin 0 + c_1 = 0\) and \(6\cos 0 + c_2 = 6\), so \(c_1 = c_2 = 0\).
      • \(\mathbf{r} = \begin{pmatrix} 6\sin 4t \\ 6\cos 4t \end{pmatrix}\)

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

      • \(|\mathbf{r}|^2 = (6\sin 4t)^2 + (6\cos 4t)^2 = 36(\sin^2 4t + \cos^2 4t)\)
      • 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) Find its position when \(t = 1\). Give the coordinates to 3 significant figures.

      • Put \(t = 1\) in, with the calculator in radians: \(\mathbf{r} = \begin{pmatrix} 6\sin 4 \\ 6\cos 4 \end{pmatrix}\)
      • \(\mathbf{r} = \begin{pmatrix} −4.54 \\ −3.92 \end{pmatrix}\)

  3. The diagram shows a water wheel of radius 2.4 m. Its centre is 3.1 m above the bed of a stream. Take the origin O on the bed directly below the centre, with \(\mathbf{i}\) horizontal and \(\mathbf{j}\) vertically up. A paddle starts at the right-hand point of the wheel when \(t = 0\), and the wheel turns clockwise once every 8 seconds.

    xyOcentrestart

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

      • \(\omega = \dfrac{2\pi}{8} = \dfrac{\pi}{4}\) 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 \((0, 3.1)\) is added.
      • Clockwise: a minus sign on the sine. \(\mathbf{r} = \begin{pmatrix} 2.4\cos \dfrac{\pi t}{4} \\ 3.1 − 2.4\sin \dfrac{\pi t}{4} \end{pmatrix}\).

    2. (b) Find its height above the bed when \(t = 1\).

      • In radians: \(y = 3.1 − 2.4\sin\dfrac{\pi}{4} = 1.40\) m.

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

      • \(|\mathbf{v}| = r\omega = 2.4 \times 0.7854 = 1.88\) m s⁻¹.

Stretch

  1. A pointer on a large, slow-turning dial. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} −2\sin \dfrac{t}{4} \\ 2\cos \dfrac{t}{4} \end{pmatrix}\) m s⁻¹. When \(t = 0\) its position vector is \(\begin{pmatrix} 8 \\ 0 \end{pmatrix}\) m.

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

      • Integrate: \(\int −2\sin \dfrac{t}{4}\, dt = 8\cos \dfrac{t}{4} + c_1\), \(\int 2\cos \dfrac{t}{4}\, dt = 8\sin \dfrac{t}{4} + c_2\). Dividing by \(\frac{1}{4}\) is the same as multiplying by 4.
      • \(8\cos 0 + c_1 = 8\) and \(8\sin 0 + c_2 = 0\), so \(c_1 = c_2 = 0\).
      • \(\mathbf{r} = \begin{pmatrix} 8\cos \dfrac{t}{4} \\ 8\sin \dfrac{t}{4} \end{pmatrix}\)

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

      • \(|\mathbf{r}|^2 = (8\cos \dfrac{t}{4})^2 + (8\sin \dfrac{t}{4})^2 = 64(\sin^2 \dfrac{t}{4} + \cos^2 \dfrac{t}{4})\)
      • Use \(\sin^2\theta + \cos^2\theta = 1\): \(|\mathbf{r}| = \sqrt{64} = 8\)
      • The distance from the origin is always 8 m: a circle of radius 8 m about the origin.

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

      • \(|\mathbf{v}| = \sqrt{(2\sin \dfrac{t}{4})^2 + (2\cos \dfrac{t}{4})^2} = \sqrt{4(\sin^2 \dfrac{t}{4} + \cos^2 \dfrac{t}{4})}\)
      • \(|\mathbf{v}| = \sqrt{4} = 2\) m s⁻¹, which does not depend on \(t\).

jmaths.xyz · AI HL · AHL 4.26 Circular motion