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. $\frac{R}{3}$

Answer 1

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

From geometry, $x=R(1-cos53)=0.4R$

$y=R(1-cos37)=0.2R$

Apply conservation of energy at A and B, $mgx=mgy+\frac{1}{2}m{v}^2$

$mg\left(.4R\right)=mg\left(0.2R\right)+\frac{1}{2}m{v}^2$

$v^2=0.4gR$

Also as block leaves the block at B, thus $N=0$

$m\frac{v^2}{r}=mgcos37$

$m\frac{0.4gR}{r}=mg\left(0.8\right)⟹r=\frac{R}{2}$