Question

An object slides without friction from the height $25\text{cm}$ and then goes around the vertical loop of radius $10\text{cm}$ from which a symmetrical section of angle $2\alpha$ has been removed. Find the angle $\alpha$ such that after losing contact at $A$ and flying through air, the object will reach point $B$.

• A. ${90}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is D ${60}^{\circ }$


Let $R$ be the radius of curvature.
So from the figure given in the problem,
$H=h+R+R\cos \alpha$
$\Rightarrow H=h+R(1+\cos \alpha )$
$\Rightarrow h=H-R(1+\cos \alpha )$

Applying energy conservation between the top and point $A$,
$mgh=\frac{1}{2}m{v}_A^2$
[Taking line $AB$ as a reference line]
$\Rightarrow gh=\frac{v_A^2}{2}\Rightarrow v_A^2=2gh$
$\Rightarrow v_A^2=2g[H-R(1+\cos \alpha \left)\right]$........(1)

From figure, at point A
Required range to complete the motion is $AB$
$\Rightarrow \text{Range}=2R\sin \alpha$
Applying formula of projectile motion,
$\text{Range}=\frac{v_A^2\sin 2\alpha }{g}$
$\Rightarrow 2R\sin \alpha =\frac{v_A^2\sin 2\alpha }{g}$
$\Rightarrow 2Rg\sin \alpha =v_A^2\times 2\sin \alpha \cos \alpha$
$\Rightarrow v_A^2=\frac{Rg}{\cos \alpha }$.............(2)

From eq (1) and (2)
$\frac{Rg}{\cos \alpha }=2g[H-R(1+\cos \alpha \left)\right]$
$\Rightarrow \frac{2g}{Rg}\cos \alpha [H-R(1+\cos \alpha \left)\right]=1$
$\Rightarrow 2\cos \alpha [\frac{H}{R}-(1+\cos \alpha \left)\right]=1$
$\Rightarrow 2\cos \alpha [2.5-1-\cos \alpha ]=1$
$[\because H=25\text{cm}\&R=10\text{cm}]$

$\Rightarrow 3\cos \alpha -2{\cos }^2\alpha =1$
$\Rightarrow 2{\cos }^2\alpha -3\cos \alpha +1=0$

Solving the quadratic equation,
$\cos \alpha =\frac{3\pm \sqrt{9-4\times 2}}{2.2}=\frac{3\pm 1}{4}$
$\Rightarrow \cos \alpha =1$ or $\frac{1}{2}$
i.e $\alpha =0$ or ${60}^{\circ }$
$\because \alpha \ne 0$
$\therefore \alpha ={60}^{\circ }$
Hence, option $\left(d\right)$ is the correct answer.