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 $(2R,R)$enter image description here

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 $\omega\over5$. enter image description here

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


1 Answer 1


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 $\frac{d\alpha}{dt} r=v_t$.


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