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I am doing a little study of the question found in the Engineering mechanics Dynamics book.

enter image description here

I tried to define the position of the point P on the $\mathbf{x}$ and $\mathbf{y}$ axes with respect to the point 0 using the circle equation.

$x^2+y^2=r^2 \Rightarrow y=\sqrt{r^2-x^2}$

Using this relationship I have tried to relate the ratio between $\theta$ and $x$.

$tan(\theta)=\frac{x}{y} \Rightarrow \theta(x)=arctan(\frac{x}{r^2-x^2})$

I made the derivative of this function to obtain a relation between the velocity of x and the angular velocity

$\frac{d(arctan(\frac{x}{r^2-x^2}))}{dx}=\frac{1}{\sqrt{r^2-x^2}}$

Therefore:

$\theta(\dot{x})=\frac{1}{\sqrt{r^2-x^2}}$

I made the derivative of this function to obtain relation between the acceleration of x with the angular acceleration

$\frac{d(\frac{1}{\sqrt{r^2-x^2}})}{dx}=\frac{x}{(r^2-x^2)^\frac{3}{2}}$

Therefore:

$\theta(\ddot{x})=\frac{x}{(r^2-x^2)^\frac{3}{2}}$

The question is this:

With these two functions found, can I get the answer to the image question or am I in the wrong way? I have this doubt ...

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According with the setup

$$ x = \bar{OP}\sin\theta $$

and also

$$ \dot x = \bar{OP}\dot\theta\cos\theta $$

and

$$ \ddot x = \bar{OP}\left(\ddot \theta\cos\theta-\dot\theta^2\sin\theta\right) $$

etc.

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  • $\begingroup$ You have $\theta = \arcsin\left(\frac{x}{\bar{OP}}\right)$ $\endgroup$ – Cesareo Apr 20 '18 at 9:48

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