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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2018

Paper 4

For constant angular velocity, the <equation of motion in a rotating frame> is
$$
m\left(\ddot{\mathbf x}+2\boldsymbol\omega\times\dot{\mathbf x}
+\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf x)\right)
=-9m|\boldsymbol\omega|^2\mathbf x.
$$
For $\boldsymbol\omega=(0,0,\omega)$ and $\xi=x+iy$, the planar equations combine to
$$
\ddot\xi+2i\omega\dot\xi+8\omega^2\xi=0.
$$
The initial conditions give
$$
\boxed{\xi(t)=\frac23e^{2i\omega t}+\frac13e^{-4i\omega t},\qquad z(t)=0.}
$$
Therefore
$$
|\dot{\mathbf x}|=|\dot\xi|
=\frac{8|\omega|}{3}|\sin(3\omega t)|.
$$
The maximum speed and, for $\omega>0$, its times are
$$
\boxed{v_{\max}=\frac{8\omega}{3},\qquad
t=\frac{(2k+1)\pi}{6\omega}\quad(k=0,1,2,\ldots).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 4 / 12A / Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 4 / 12A / Solution

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Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 4 / 12A / Solution

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