Angular velocity of a particle in uniform circular motion about a general point

This problem was given by our professor.

Consider a particle P executing uniform circular motion wrt the point O with uniform angular velocity $$\omega$$ anticlockwise whose cordinate is $$(2R,0)$$ in a circle of radius $$R$$ . Consider the instant when the particle is at the point A whose cordinate is $$(2R,R)$$

My Attempt: Let's consider a general time when the line joining O and P makes angle $$\theta$$ with the line OA and the vector joining the origin and P can be written as $$\overrightarrow r = (2R+R\cos\theta)\hat i + (R\sin\theta)\hat j$$ Now consider a small time interval $$dt$$ after which the vector joining becomes $$\overrightarrow r+\overrightarrow{dr}$$ where $$\overrightarrow{dr}$$ is the change in vector position wrt origin to find which we can differentiate $$\overrightarrow{r}$$ wrt time which gives $$\overrightarrow{dr}=(R\omega\cos\theta)dt\hat{i} - (R\sin\theta\omega)dt\hat{j}$$ Now from the following figure Let the angle between $$\overrightarrow{r}$$ and $$\overrightarrow {r} + \overrightarrow {dr}$$ be $$d\alpha$$( as shown in following figure). Hence |$$\overrightarrow{dr}$$|=|$$\overrightarrow{r}|d\alpha$$ |$$\overrightarrow(r)$$|=$$\sqrt{(4R^2\sin\theta + 5R^2)}$$ from geometry and |$$\overrightarrow{dr}$$|=$$R\omega dt$$. which gives $$d\alpha\over{dt}=\omega\over\sqrt{5+4\sin\theta}$$. Now the question asks for $$\theta=0$$ so it gives answer $$\omega\over{\sqrt{5}}$$. But the answer is $$\omega\over5$$.

I have no idea where I am going wrong. Any help would be appreciated.

1 Answer

You can't say $$dr = r d\alpha$$ in this case because relative to the origin, the particle has radial velocity. You need to find the tangential component $$v_t$$ and use $$\frac{d\alpha}{dt} r=v_t$$.