Circular motion with changing speed Problem set A Worked solutions

A stone is whirled on a string. Its position vector at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}4.5\cos(0.15t^2)\\4.5\sin(0.15t^2)\end{pmatrix}\) m. Show that the object moves in a circle with its centre at the origin and state the radius.

• \(|\mathbf{r}| = \sqrt{4.5^2\cos^2(0.15t^2) + 4.5^2\sin^2(0.15t^2)}\)
• \(= 4.5\sqrt{\cos^2(0.15t^2) + \sin^2(0.15t^2)} = 4.5\), using \(\sin^2\theta + \cos^2\theta = 1\)
• The distance from the origin is the constant 4.5, so the object moves in a circle with centre O and radius 4.5 m.

A ball is spun on a string. Its position vector at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}0.8\cos(0.5t^2)\\0.8\sin(0.5t^2)\end{pmatrix}\) m. Find an expression for its velocity.

• Differentiate each component with the chain rule. The derivative of \(0.5t^2\) is \(t\).
• \(\dfrac{d}{dt}\left(0.8\cos(0.5t^2)\right) = −0.8\times t\sin(0.5t^2) = −0.8t\sin(0.5t^2)\)
• \(\mathbf{v} = \begin{pmatrix}-0.8t\sin(0.5t^2)\\0.8t\cos(0.5t^2)\end{pmatrix}\)

A car on a ride moves in a circle. Its position vector is \(\mathbf{r} = \begin{pmatrix}6\cos(0.04t^2)\\6\sin(0.04t^2)\end{pmatrix}\) m and its velocity is \(\mathbf{v} = \begin{pmatrix}-0.48t\sin(0.04t^2)\\0.48t\cos(0.04t^2)\end{pmatrix}\) m s⁻¹. Show that the velocity is always perpendicular to the position vector.

• Two vectors are perpendicular if their scalar product is zero.
• \(\mathbf{v}\cdot\mathbf{r} = (6\cos(0.04t^2))(−0.48t\sin(0.04t^2)) + (6\sin(0.04t^2))(0.48t\cos(0.04t^2))\)
• \(= −2.88t\sin(0.04t^2)\cos(0.04t^2) + 2.88t\sin(0.04t^2)\cos(0.04t^2) = 0\)
• The scalar product is zero, so \(\mathbf{v}\) is perpendicular to \(\mathbf{r}\) at every time.

A mass on a cord moves in a circle. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix}-0.44t\sin(0.2t^2)\\0.44t\cos(0.2t^2)\end{pmatrix}\) 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.
• \(\dfrac{d}{dt}\left(−0.44t\sin(0.2t^2)\right) = −0.44\sin(0.2t^2) − 0.44t\times0.4t\cos(0.2t^2) = −0.44\sin(0.2t^2) − 0.176t^2\cos(0.2t^2)\)
• \(\mathbf{a} = \begin{pmatrix}-0.44\sin(0.2t^2) − 0.176t^2\cos(0.2t^2)\\0.44\cos(0.2t^2) − 0.176t^2\sin(0.2t^2)\end{pmatrix}\)

A hammer is swung on a chain. Its acceleration at time \(t\) seconds is \(\mathbf{a} = \begin{pmatrix}-0.6\sin(0.2t^2) − 0.24t^2\cos(0.2t^2)\\0.6\cos(0.2t^2) − 0.24t^2\sin(0.2t^2)\end{pmatrix}\) m s⁻². Use your calculator to find the value of \(t\) for which the magnitude of the acceleration is 7 m s⁻².

• \(|\mathbf{a}| = 7\), so \((−0.6\sin(0.2t^2) − 0.24t^2\cos(0.2t^2))^2 + (0.6\cos(0.2t^2) − 0.24t^2\sin(0.2t^2))^2 = 49\)
• From the calculator: solve this equation for \(t\) (use a graph of the left side and the line \(y = 49\), and read the intersection with \(t > 0\)).
• \(t = 5.39\) s

A fairground cup spins. It turns through an angle of \(0.3t^2\) radians in \(t\) seconds. How many complete revolutions does it make in the first 15 seconds?

• \(\text{angle turned} = 0.3\times15^2 = 67.5\) radians
• \(\dfrac{67.5}{2\pi} = 10.7\) revolutions
• Only whole turns count: 10 complete revolutions.

A conker is attached to the end of a cord and spun horizontally. Its position relative to a point \(O\) at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}2.4\cos(0.05t^2)\\2.4\sin(0.05t^2)\end{pmatrix}\), 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.

   • \(|\mathbf{r}| = \sqrt{2.4^2\cos^2(0.05t^2) + 2.4^2\sin^2(0.05t^2)} = 2.4\sqrt{\cos^2(0.05t^2) + \sin^2(0.05t^2)}\)
   • \(= 2.4\) because \(\sin^2\theta + \cos^2\theta = 1\).
   • The distance from \(O\) is constant, so the path is a circle with centre \(O\).
   • Radius = 2.4 m

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

   • Chain rule on each component: \(\mathbf{v} = \begin{pmatrix}-0.24t\sin(0.05t^2)\\0.24t\cos(0.05t^2)\end{pmatrix}\)

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

   • \(\mathbf{v}\cdot\mathbf{r} = (2.4\cos(0.05t^2))(−0.24t\sin(0.05t^2)) + (2.4\sin(0.05t^2))(0.24t\cos(0.05t^2)) = −0.576t\sin(0.05t^2)\cos(0.05t^2) + 0.576t\sin(0.05t^2)\cos(0.05t^2) = 0\)
   • 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:
   • \(\mathbf{a} = \begin{pmatrix}-0.24\sin(0.05t^2) − 0.024t^2\cos(0.05t^2)\\0.24\cos(0.05t^2) − 0.024t^2\sin(0.05t^2)\end{pmatrix}\)

5. (e) The cord breaks when the magnitude of the acceleration exceeds 10 m s⁻². Find the value of \(t\) at the instant the cord breaks.

   • \(|\mathbf{a}| = 10\): \((−0.24\sin(0.05t^2) − 0.024t^2\cos(0.05t^2))^2 + (0.24\cos(0.05t^2) − 0.024t^2\sin(0.05t^2))^2 = 100\)
   • From the calculator, solving for \(t > 0\):
   • \(t = 20.4\) s

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

   • Angle turned through: \(0.05\times20.4^2 = 20.8\) radians
   • Revolutions: \(\dfrac{20.8}{2\pi} = 3.31\)
   • 3 complete revolutions

A model plane is flown clockwise on a control line. Its position vector at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}1.6\sin(0.3t^2)\\1.6\cos(0.3t^2)\end{pmatrix}\) m.

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

   • \(|\mathbf{r}| = \sqrt{1.6^2\sin^2(0.3t^2) + 1.6^2\cos^2(0.3t^2)} = 1.6\sqrt{\sin^2(0.3t^2) + \cos^2(0.3t^2)} = 1.6\)
   • using \(\sin^2\theta + \cos^2\theta = 1\). The distance from the origin is constant, so the path is a circle of radius 1.6 m.

2. (b) Find the velocity vector.

   • Chain rule: \(\mathbf{v} = \begin{pmatrix}0.96t\cos(0.3t^2)\\−0.96t\sin(0.3t^2)\end{pmatrix}\)

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

   • \(\mathbf{v}\cdot\mathbf{r} = (0.96t\cos(0.3t^2))(1.6\sin(0.3t^2)) + (−0.96t\sin(0.3t^2))(1.6\cos(0.3t^2)) = 1.536t\sin(0.3t^2)\cos(0.3t^2) − 1.536t\sin(0.3t^2)\cos(0.3t^2) = 0\)
   • The scalar product is zero, so the vectors are perpendicular.

4. (d) Find the time at which the speed is 4.8 m s⁻¹.

   • \(|\mathbf{v}| = \sqrt{0.96^2t^2(\cos^2(0.3t^2) + \sin^2(0.3t^2))} = 0.96t\)
   • \(0.96t = 4.8 \Rightarrow t = 5\) s