Question

Figure shows a smooth vertical circular track of radius $R$. A block slides along the surface $AB$ when it is given a velocity equal to $\sqrt{6gR}$ at point $A$. The ratio of the force exerted by the track on the block at point $A$ to that at point $B$ is

• A. $0.25$
• B. $0.35$
• C. $0.45$
• D. $0.55$

Answer 1

The correct option is D $0.55$

Free body diagram at $A$ and $B$ shown is figure


Solving from block's frame of reference, we have to assume a centrifugal force of $\frac{m{v}^2}{R}$.
At point $A$: $N_A+mg\cos {60}^{\circ }=\frac{m{v}_A^2}{R}$
$\Rightarrow N_A=m(\sqrt{6gR}{)}^2/R-mg\cos {60}^{\circ }=5.5mg$

Using energy conservation between $A$ and $B$, taking $B$ as a reference
$\frac{1}{2}m{v}_A^2+mgR(1+\cos {60}^{\circ })=\frac{1}{2}m{v}_B^2$
$\Rightarrow v_B^2=6gR+3gR\Rightarrow v_B^2=9gR$

At point $B$, $N_B=mg+\frac{m{v}_B^2}{R}$
$\Rightarrow N_B=10mg$
Therefore, $\frac{N_A}{N_B}=\frac{5.5mg}{10mg}=0.55$