Showing a circle
- Find \(|\mathbf{r}|\) and show it is constant. Write the identity \(\sin^2\theta + \cos^2\theta = 1\) in your working: the mark is for stating it.
- A constant distance from a fixed point is a circle. The radius is that distance.
AHL 4.29
Exam practice: circular motion with changing speed
AHL 4.29
Exam practice: circular motion with changing speed
By the end of this section
- I can show from a position vector that an object moves in a circle, stating \(\sin^2\theta + \cos^2\theta = 1\), and give the radius.
- I can differentiate a position vector with \(t^2\) inside the sine and cosine, using the chain rule, to find the velocity.
- I can show that the velocity is perpendicular to the position vector using the scalar product.
- I can differentiate a velocity vector with the product rule and chain rule to find the acceleration.
- I can set the magnitude of the acceleration to a given value and solve for \(t\) with the calculator.
- I can find the number of complete revolutions from the angle turned through.
Key points
Velocity and acceleration when the angle is \(kt^2\)
- The angle \(kt^2\) is not a multiple of \(t\), so the chain rule leaves a factor: \(\dfrac{d}{dt}\cos(kt^2) = −2kt\sin(kt^2)\).
- The velocity components then contain \(t\) times a trigonometric function. Differentiating them needs the product rule as well as the chain rule.
- To show \(\mathbf{v}\perp\mathbf{r}\), write out \(\mathbf{v}\cdot\mathbf{r}\) in full and show the two products cancel to 0.
- Here the speed is not constant. It grows with \(t\), so the acceleration is not just towards the centre: it also has a part along the path.
Solving and counting
- To find when \(|\mathbf{a}|\) equals a value, write \(\sqrt{a_x^2 + a_y^2}\) equal to the value (or square both sides) and solve for \(t\) with the calculator.
- The angle turned through is the angle inside the sine and cosine, for example \(kt^2\). Divide it by \(2\pi\) for the number of revolutions.
- A revolution counts only when it is complete: round down.
On the calculator
Solving an equation
- Enter the left side of the equation as one function of \(t\) and the right side as a second function.
- Graph both for \(t \geq 0\) and find the intersection, or use the calculator's numerical solver.
- Read the value of \(t\), and check it by substituting it back.
Screenshot and video to come
Worked examples
Problems like the ones in class. Pick one, try it, then open the working.Showing the path is a circle
A ball is spun on a string. Its position relative to a point \(O\) at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}3.2\cos(0.2t^2)\\3.2\sin(0.2t^2)\end{pmatrix}\) m, where \(t \geq 0\). Show that the ball moves in a circle with its centre at \(O\) and state the radius.
Working
Have a go first. Then open the working one line at a time.
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The magnitude
\(|\mathbf{r}| = \sqrt{3.2^2\cos^2(0.2t^2) + 3.2^2\sin^2(0.2t^2)}\)
Square each component and add.
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Take out the factor
\(= 3.2\sqrt{\cos^2(0.2t^2) + \sin^2(0.2t^2)} = 3.2\)
State the identity \(\sin^2\theta + \cos^2\theta = 1\) in words or symbols: the mark scheme wants it written.
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Conclude
Constant distance 3.2 m from O
A point at a constant distance from a fixed point moves on a circle. Radius 3.2 m.
Velocity by the chain rule, and perpendicular to r
A ball is spun on a string. Its position relative to \(O\) is \(\mathbf{r} = \begin{pmatrix}1.2\cos(0.3t^2)\\1.2\sin(0.3t^2)\end{pmatrix}\) m. (a) Find the velocity at time \(t\). (b) Show that the velocity is always perpendicular to the position vector.
Working
Have a go first. Then open the working one line at a time.
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Chain rule
\(\dfrac{d}{dt}\cos(0.3t^2) = −0.6t\sin(0.3t^2)\)
The inner function \(0.3t^2\) has derivative \(0.6t\), which stays in the answer.
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Velocity
\(\mathbf{v} = \begin{pmatrix}-0.72t\sin(0.3t^2)\\0.72t\cos(0.3t^2)\end{pmatrix}\)
Multiply the \(0.6t\) by 1.2 for each component.
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Scalar product
\(\mathbf{v}\cdot\mathbf{r} = −0.864t\sin(0.3t^2)\cos(0.3t^2) + 0.864t\sin(0.3t^2)\cos(0.3t^2) = 0\)
Write out both products. They cancel.
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Conclude
v · r = 0
The scalar product is zero, so the vectors are perpendicular at every time \(t\).
Acceleration by the product rule
The velocity of a ball is \(\mathbf{v} = \begin{pmatrix}-0.72t\sin(0.3t^2)\\0.72t\cos(0.3t^2)\end{pmatrix}\) m s⁻¹. Find an expression for its acceleration, and find the magnitude of the acceleration when \(t = 0\).
Working
Have a go first. Then open the working one line at a time.
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Product rule
\(\dfrac{d}{dt}\left(−0.72t\sin(0.3t^2)\right) = −0.72\sin(0.3t^2) − 0.72t\times0.6t\cos(0.3t^2)\)
Differentiate \(t\) (gives 1), then \(\sin(0.3t^2)\) (chain rule gives \(0.6t\cos\)).
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Acceleration
\(\mathbf{a} = \begin{pmatrix}-0.72\sin(0.3t^2) − 0.432t^2\cos(0.3t^2)\\0.72\cos(0.3t^2) − 0.432t^2\sin(0.3t^2)\end{pmatrix}\)
Do the same for the second component.
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At t = 0
\(\mathbf{a}(0) = \begin{pmatrix}0\\0.72\end{pmatrix}, \quad |\mathbf{a}| = 0.72\) m s⁻²
Every \(t^2\) term is zero and \(\sin 0 = 0\).
When the acceleration reaches a given size
The acceleration of the ball is \(\mathbf{a} = \begin{pmatrix}-0.72\sin(0.3t^2) − 0.432t^2\cos(0.3t^2)\\0.72\cos(0.3t^2) − 0.432t^2\sin(0.3t^2)\end{pmatrix}\) m s⁻². The string breaks when the magnitude of the acceleration first exceeds 15 m s⁻². Find the time at which the string breaks.
Working
Have a go first. Then open the working one line at a time.
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The equation
\(|\mathbf{a}|^2 = (−0.72\sin(0.3t^2) − 0.432t^2\cos(0.3t^2))^2 + (0.72\cos(0.3t^2) − 0.432t^2\sin(0.3t^2))^2 = 225\)
Set the magnitude to 15. Squaring both sides removes the square root.
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Solve
From the calculator: \(t = 5.89\) s
Graph the left side and the line \(y = 225\), then read the first intersection for \(t > 0\). Check that the value is sensible by testing it in the original equation.
Complete revolutions
A ball on a string turns through an angle of \(0.25t^2\) radians in \(t\) seconds. How many complete revolutions does it make in the first 9.7 seconds?
Working
Have a go first. Then open the working one line at a time.
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The angle
\(0.25\times9.7^2 = 23.5\) radians
Substitute the time into the angle.
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Revolutions
\(\dfrac{23.5}{2\pi} = 3.74\)
One revolution is \(2\pi\) radians.
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Whole turns
3 complete revolutions
The last part is unfinished, so round down.