Question

A section of fixed smooth circular track of radius $R$ in vertical plane is shown in the figure. A block is released from position A and leaves the track at B. The radius of curvature of its trajectory just after it leaves the track at B is ?

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

Answer 1

The correct option is C $\frac{R}{2}$


$x=R(1-\cos 53^{\circ })=0.4R$
$y=R(1-\cos {37}^{\circ })=0.2R$
Apply conservation of energy at $A$ and $B$,
$mgx=mgy+\frac{1}{2}m{v}^2$
$mg\left(0.4R\right)=mg\left(0.2R\right)+\frac{1}{2}m{v}^2$
$v^2=0.4gR$

Also as block leaves the block at $B$, thus Normal reaction from contact $=0$

$\frac{m{v}^2}{r}=mg\cos {37}^{\circ }$
$\frac{0.4gR}{r}=g\left(0.8\right)$
$\Rightarrow r=\frac{R}{2}$