Question

A small object slides without friction from the height H=50cm and then loops the vertical loop of radius r=20cm from which a symmetrical section of angle 2 $\alpha$ has been removed. Find angle $\alpha$ such that after losing contact at A and flying through air, the object will reach point B.

Answer 1

$\frac{{mv}_A^2}{2}=mg[2.5r-r(1+\cos \alpha )]$

$h=2.5r-r(1+\cos \alpha )$

$v_A^2=gr(3-2\cos \alpha )$

$sin\alpha =\frac{v_A^{2}sin2\alpha }{g}$

$v_A^{2}cos\alpha =gr$

$\frac{gr}{\cos \alpha }=gr[3-2\cos \alpha ]$

$2{cos}^2\alpha -3\cos \alpha +1=0$
$\left(\cos \alpha -1\right)(2cos\alpha -1)=0or\cos \alpha =1$

$cos\alpha =1/2$

$\alpha =0$ or $\alpha ={60}^{\circ }$

since, $\alpha \ne 0$
$\therefore \alpha$ = ${60}^{\circ }$