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I'm stuck on the second part of this question.

Suppose a particle moves in a plane with its trajectory given by the polar equation $r=2b\sin(\theta)$ for some constant $b>0$.

(i) Show that this can be written in Cartesian coordinates as $x^2+(y-b)^2=b^2$.

This is the equation for a circle of centre $(0,b)$ and radius $b$.

[Hint: recall that $r^2=x^2+y^2$ and $y=r\sin(\theta)$]

(ii) Suppose that the transverse component of the acceleration is zero.

(a) Prove that $r^2\dot{\theta}=h$ is constant.

(b) Assuming that $r\ne0$, show that $\dot{r}=2bhr^{-2}\cos(\theta)$ and hence find $\ddot{r}$.

(c) Use your answers to (b) to show that the radial component of the acceleration is $-8b^2h^2r^{-5}$.

So far, I've got:

$r=2b\sin(\theta)$

$\dot{r}=2b\cos(\theta)\dot{\theta}$

So, the transverse coordinate is $4b\cos(\theta)\dot{\theta}^2+2b\sin(\theta)\dot{ \theta}$.

Suppose a particle moves in a plane with its trajectory given by the polar equation $r=2b\sin(\theta)$ for some constant $b>0$.

(i) Show that this can be written in Cartesian coordinates as $x^2+(y-b)^2=b^2$.

This is the equation for a circle of centre $(0,b)$ and radius $b$.

[Hint: recall that $r^2=x^2+y^2$ and $y=r\sin(\theta)$]

(ii) Suppose that the transverse component of the acceleration is zero.

(a) Prove that $r^2\dot{\theta}=h$ is constant.

(b) Assuming that $r\ne0$, show that $\dot{r}=2bhr^{-2}\cos(\theta)$ and hence find $\ddot{r}$.

(c) Use your answers to (b) to show that the radial component of the acceleration is $-8b^2h^2r^{-5}$.

So far, I've got:

$r=2b\sin(\theta)$

$\dot{r}=2b\cos(\theta)\dot{\theta}$

So, the transverse coordinate is $4b\cos(\theta)\dot{\theta}^2+2b\sin(\theta)\dot{ \theta}$.

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