Why is $R$ not the Radius of Curvature, here? Or is it?

Question

Consider this question,

Two masses A and B are connected by a massless string. A rest in equilibrium on a rough horizontal table and limiting friction is acting on it. B hang vertically at a distance R below the edge of the table. B is projected horizontally with velocity V, what is initial radius of curvature of B? coefficient of friction is μ.

The answer comes out to be $$ ((μ+1)/μ)R $$ In the solution we use the equation: $$T- m'g= m'( v^2/R - a),$$ where $v^2/R - a$ is the net acceleration of B.

My doubt is, what is $v^2/R$, here? Centripetal acceleration? If so, then in the formula R should represent radius of curvature.

If it isn't centripetal acceleration, then what is it? and why are we using this in the equation stated above.

Edit:I tried to analyze it with the frame of an observer moving down with acceleration $a$, and now I'm now even more confused. Like, for this observer the object seems to have radius of curvature $R$. Is the radius of curvature frame dependent?

Answer 1

I will first find the acceleration, and then explain how to obtain the radius of curvature $R_{curv}$ and why $R_{curv} \neq R$. For reference, here's my drawing

The equation of motion for block A is simply $$ T - \mu m g = m a \quad \equiv \quad T = m\left( \mu g + a\right), $$ while for block B it is $$ T - m'g - F_{centr}= -m'a \quad\equiv\quad T - m'g - m' \dfrac{v^2}{R}= -m'a \\ T - m'g = m'\left(\dfrac{v^2}{R} -a \right), $$ where the minus sign in the acceleration corresponds to the fact that it "points down".

Now, we also know that, initially, limiting friction is acting on A, so $T_0 = \mu m g = m'g$ (Note: $T_0 \neq T$). So $$ m = \dfrac{m'}{\mu}. $$ We combine these three equations to get $$ m\left( \mu g + a\right) - m'g = m'\left(\dfrac{v^2}{R} -a \right)\\ \dfrac{m'}{\mu}\left( {\mu g} + a\right) - \bar{m'g} = m'\left(\dfrac{v^2}{R} -a \right)\\ \dfrac{m'}{\mu} a = m'\left(\dfrac{v^2}{R} -a \right), $$ which can be reordered to get $$ a = \dfrac{\mu}{1 + \mu} \dfrac{v^2}{R}. $$

Now, this means that the motion described by B will be the one described by the red dashed lines in my drawing, which is approximately circular. Now, what is the radius of this imaginary circle? It can be calculated as $$ \dfrac{|\mathbf{v}|^3}{|\mathbf{v}\times\dot{\mathbf{v}}|} = \dfrac{v^2}{a} = \dfrac{1 + \mu}{\mu} R, $$ and it is called the radius of curvature, $R_{curv}$, which is observer independent (as it's just a length and we're doing classical mechanics). For an explanation of why this formula is valid, I refer you to the Wikipedia page.

Answer 2

Perhaps, answers to my recent questions have cleared this doubt. Anyway, I am writing the answer just in case if anyone needed. Note: The answer I am writing might not really be an answer but it satisfies me and was something that I was seeking.

Acceleration in polar coordinates is :

$$\\ \ddot{\mathbf{r}} = \left(\ddot{r} - r \dot{\theta}^2\right)\hat{\mathbf{r}} + \left(r \ddot{\theta} + 2\dot{r}\dot{\theta}\right)\hat{\boldsymbol{\theta}},$$

where the $\left(\ddot{r} - r \dot{\theta}^2\right)\hat{\mathbf{r}}$, term represent radial acceleration, the "$\ddot{r} $" represents linear acceleration in radial direction, and $ - r \dot{\theta}^2\hat{\mathbf{r}}$, represents centripetal acceleration in radial direction. In my question, linear radial acceleration is $a$ and centripetal acceleration is $v^2/R$.

And Radius of curvature is $v_{net}^2/a_{net} $ where $a_{net}$ is component of resultant acceleration perpendicular to $v_{net}$