Question

In the figure shown $ADB$ and $BEF$ are two fixed circular paths. A block of mass $m$ enters in the tube $ADB$ through point $A$ with minimum velocity to reach point $B$. From there the block moves on another circular path of radius $R^{{}^{\prime }}$, where it is just able to complete the circle.
Choose the correct statements.

• A. Velocity at $A$ must be $\sqrt{4Rg}$
• B. Velocity at $A$ must be $\sqrt{2Rg}$
• C. $\frac{R^{{}^{\prime }}}{R}=\frac{2}{3}$
• D. The normal reaction at point $E$ is $6\text{mg}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B Velocity at $A$ must be $\sqrt{2Rg}$
C $\frac{R^{{}^{\prime }}}{R}=\frac{2}{3}$
D The normal reaction at point $E$ is $6\text{mg}$

Since the block reaches $B$ with minimum speed, it should just reach $D\Rightarrow (V_D=0)$
Work energy theorem between points $A$ and $D$ gives
$\therefore \frac{1}{2}m{V}_A^2=mgR\Rightarrow V_A=\sqrt{2gR}$

For the block to just complete the vertical circle, $V_E=\sqrt{5g{R}^{\prime }}$

Applying work energy theorem between points $E$ and $D$
$\frac{1}{2}m{V}_E^2=mg(R+R^{{}^{\prime }})$
$5g{R}^{{}^{\prime }}=2gR+2g{R}^{{}^{\prime }}$
$3R^{{}^{\prime }}=2R$
$\frac{R^{{}^{\prime }}}{R}=\frac{2}{3}$

At $E$, in the rest frame of the block, FBD is given by

$\Rightarrow N=mg+\frac{m}{R^{{}^{\prime }}}\left(5g{R}^{{}^{\prime }}\right)$
$\Rightarrow N=6\text{mg}$