Question

In a fixed quarter circular track of radius $R$ which lies in a vertical plane, a block is released from point $A$ and it leaves the path at point $B$ . The radius of curvature of its trajectory when it just leaves the path will be

• A. $R$
• B. $\frac{\text{R}}{4}$
• C. $\frac{\text{R}}{2}$
• D. None of these

Answer 1

Answer: (C)

By energy conservation between A and B
$\Rightarrow \text{Mg}\frac{2\text{R}}{5}+0=\frac{\text{MgR}}{5}+\frac{1}{2}{\text{Mv}}^2$
$\text{v}=\sqrt{\frac{2\text{gR}}{5}}$

Now, radius of curvature after just leaving B is
$\text{r}=\frac{{\text{v}}_{\perp }^2}{{\text{a}}_{\text{r}}}=\frac{2\text{gR}/5}{\text{g cos 37}}=\frac{\text{R}}{2}$