Circular motion with changing speed Problem set B One question at a time

A bucket is swung on a rope. Its position vector at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}0.7\cos(0.4t^2)\\0.7\sin(0.4t^2)\end{pmatrix}\) m. Show that the object moves in a circle with its centre at the origin and state the radius.

• \(|\mathbf{r}| = \sqrt{0.7^2\cos^2(0.4t^2) + 0.7^2\sin^2(0.4t^2)}\)
• \(= 0.7\sqrt{\cos^2(0.4t^2) + \sin^2(0.4t^2)} = 0.7\), using \(\sin^2\theta + \cos^2\theta = 1\)
• The distance from the origin is the constant 0.7, so the object moves in a circle with centre O and radius 0.7 m.

A toy aeroplane is flown on a wire. Its position vector at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}1.4\cos(0.25t^2)\\1.4\sin(0.25t^2)\end{pmatrix}\) m. Find an expression for its velocity.

• Differentiate each component with the chain rule. The derivative of \(0.25t^2\) is \(0.5t\).
• \(\dfrac{d}{dt}\left(1.4\cos(0.25t^2)\right) = −1.4\times0.5t\sin(0.25t^2) = −0.7t\sin(0.25t^2)\)
• \(\mathbf{v} = \begin{pmatrix}-0.7t\sin(0.25t^2)\\0.7t\cos(0.25t^2)\end{pmatrix}\)

A satellite dish pointer moves in a circle. Its position vector is \(\mathbf{r} = \begin{pmatrix}2.5\cos(0.6t^2)\\2.5\sin(0.6t^2)\end{pmatrix}\) m and its velocity is \(\mathbf{v} = \begin{pmatrix}-3t\sin(0.6t^2)\\3t\cos(0.6t^2)\end{pmatrix}\) m s⁻¹. Show that the velocity is always perpendicular to the position vector.

• Two vectors are perpendicular if their scalar product is zero.
• \(\mathbf{v}\cdot\mathbf{r} = (2.5\cos(0.6t^2))(−3t\sin(0.6t^2)) + (2.5\sin(0.6t^2))(3t\cos(0.6t^2))\)
• \(= −7.5t\sin(0.6t^2)\cos(0.6t^2) + 7.5t\sin(0.6t^2)\cos(0.6t^2) = 0\)
• The scalar product is zero, so \(\mathbf{v}\) is perpendicular to \(\mathbf{r}\) at every time.

A weight on a spring arm moves in a circle. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix}-0.18t\sin(0.1t^2)\\0.18t\cos(0.1t^2)\end{pmatrix}\) m s⁻¹. Find an expression for its acceleration.

• Each component is a product of \(t\) and a trigonometric function, so use the product rule and the chain rule.
• \(\dfrac{d}{dt}\left(−0.18t\sin(0.1t^2)\right) = −0.18\sin(0.1t^2) − 0.18t\times0.2t\cos(0.1t^2) = −0.18\sin(0.1t^2) − 0.036t^2\cos(0.1t^2)\)
• \(\mathbf{a} = \begin{pmatrix}-0.18\sin(0.1t^2) − 0.036t^2\cos(0.1t^2)\\0.18\cos(0.1t^2) − 0.036t^2\sin(0.1t^2)\end{pmatrix}\)

A tether ball moves around its pole. Its acceleration at time \(t\) seconds is \(\mathbf{a} = \begin{pmatrix}-0.4\sin(0.1t^2) − 0.08t^2\cos(0.1t^2)\\0.4\cos(0.1t^2) − 0.08t^2\sin(0.1t^2)\end{pmatrix}\) m s⁻². Use your calculator to find the value of \(t\) for which the magnitude of the acceleration is 5 m s⁻².

• \(|\mathbf{a}| = 5\), so \((−0.4\sin(0.1t^2) − 0.08t^2\cos(0.1t^2))^2 + (0.4\cos(0.1t^2) − 0.08t^2\sin(0.1t^2))^2 = 25\)
• From the calculator: solve this equation for \(t\) (use a graph of the left side and the line \(y = 25\), and read the intersection with \(t > 0\)).
• \(t = 7.89\) s

A record turntable speeds up. It turns through an angle of \(0.12t^2\) radians in \(t\) seconds. How many complete revolutions does it make in the first 22 seconds?

• \(\text{angle turned} = 0.12\times22^2 = 58.1\) radians
• \(\dfrac{58.1}{2\pi} = 9.24\) revolutions
• Only whole turns count: 9 complete revolutions.

A ball is attached to the end of a rope and spun horizontally. Its position relative to a point \(O\) at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}0.9\cos(0.2t^2)\\0.9\sin(0.2t^2)\end{pmatrix}\), where all displacements are in metres and \(t \geq 0\).

1. (a) Show that the object moves in a circle with its centre at \(O\) and state the radius.

   • \(|\mathbf{r}| = \sqrt{0.9^2\cos^2(0.2t^2) + 0.9^2\sin^2(0.2t^2)} = 0.9\sqrt{\cos^2(0.2t^2) + \sin^2(0.2t^2)}\)
   • \(= 0.9\) because \(\sin^2\theta + \cos^2\theta = 1\).
   • The distance from \(O\) is constant, so the path is a circle with centre \(O\).
   • Radius = 0.9 m

2. (b) Find an expression for the velocity at time \(t\).

   • Chain rule on each component: \(\mathbf{v} = \begin{pmatrix}-0.36t\sin(0.2t^2)\\0.36t\cos(0.2t^2)\end{pmatrix}\)

3. (c) Hence show that the velocity is always perpendicular to the position vector.

   • \(\mathbf{v}\cdot\mathbf{r} = (0.9\cos(0.2t^2))(−0.36t\sin(0.2t^2)) + (0.9\sin(0.2t^2))(0.36t\cos(0.2t^2)) = −0.324t\sin(0.2t^2)\cos(0.2t^2) + 0.324t\sin(0.2t^2)\cos(0.2t^2) = 0\)
   • The scalar product is zero, so \(\mathbf{v}\) and \(\mathbf{r}\) are perpendicular.

4. (d) Find an expression for the acceleration at time \(t\).

   • Product rule and chain rule on each component:
   • \(\mathbf{a} = \begin{pmatrix}-0.36\sin(0.2t^2) − 0.144t^2\cos(0.2t^2)\\0.36\cos(0.2t^2) − 0.144t^2\sin(0.2t^2)\end{pmatrix}\)

5. (e) The rope breaks when the magnitude of the acceleration exceeds 12 m s⁻². Find the value of \(t\) at the instant the rope breaks.

   • \(|\mathbf{a}| = 12\): \((−0.36\sin(0.2t^2) − 0.144t^2\cos(0.2t^2))^2 + (0.36\cos(0.2t^2) − 0.144t^2\sin(0.2t^2))^2 = 144\)
   • From the calculator, solving for \(t > 0\):
   • \(t = 9.13\) s

6. (f) How many complete revolutions has the object made from \(t = 0\) to the instant the rope breaks?

   • Angle turned through: \(0.2\times9.13^2 = 16.7\) radians
   • Revolutions: \(\dfrac{16.7}{2\pi} = 2.65\)
   • 2 complete revolutions

A weight on a rope is swung clockwise. Its position vector at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}2\sin(0.3t^2)\\2\cos(0.3t^2)\end{pmatrix}\) m.

1. (a) Show that the object moves in a circle about the origin and state the radius.

   • \(|\mathbf{r}| = \sqrt{2^2\sin^2(0.3t^2) + 2^2\cos^2(0.3t^2)} = 2\sqrt{\sin^2(0.3t^2) + \cos^2(0.3t^2)} = 2\)
   • using \(\sin^2\theta + \cos^2\theta = 1\). The distance from the origin is constant, so the path is a circle of radius 2 m.

2. (b) Find the velocity vector.

   • Chain rule: \(\mathbf{v} = \begin{pmatrix}1.2t\cos(0.3t^2)\\−1.2t\sin(0.3t^2)\end{pmatrix}\)

3. (c) Show that the velocity is perpendicular to the position vector.

   • \(\mathbf{v}\cdot\mathbf{r} = (1.2t\cos(0.3t^2))(2\sin(0.3t^2)) + (−1.2t\sin(0.3t^2))(2\cos(0.3t^2)) = 2.4t\sin(0.3t^2)\cos(0.3t^2) − 2.4t\sin(0.3t^2)\cos(0.3t^2) = 0\)
   • The scalar product is zero, so the vectors are perpendicular.

4. (d) Find the time at which the speed is 6 m s⁻¹.

   • \(|\mathbf{v}| = \sqrt{1.2^2t^2(\cos^2(0.3t^2) + \sin^2(0.3t^2))} = 1.2t\)
   • \(1.2t = 6 \Rightarrow t = 5\) s