Question

A small block is projected with a speed $v_0$ on a horizontal track placed on a sufficiently rough surface which turns into a semi-circle (vertical) of radius $R$. Find the main value of $V_o$ os that the body will hit the point A after leaving the track at its highest point. The arrangement is shown in the figure, given that the straight part is rough and the curved path is smooth. The coefficient of friction is $\mu$

Answer 1

Let the block escape the point at $C$ with a velocity $V$ horizontally. since it hits the initial spot A after falling through a height $2R$ we can write
$\left(2R\right)=\frac{1}{2}g{t}^2\Rightarrow t=2\sqrt{R/g}$
the distance
$AB=2v\sqrt{R/g}\Rightarrow d=2v_x\sqrt{R/g}...\left(a\right)$
Work energy theorem is applied to the motion of the body from A to B leads $\Delta KE=W_f$
$\Rightarrow \left(\frac{1}{2}\right)m{v}_0^2-\left(\frac{1}{2}\right)m{v}_1^2=\mu mgd\Rightarrow v_0=\sqrt{v_1^2+2\mu gd}....\left(b\right)$
Energy conservation between B and C yields
$\left(\frac{1}{2}\right)m{v}_1^2-\left(\frac{1}{2}\right)m{v}^2=mg\left(2R\right)\Rightarrow v_1=\sqrt{v^2+4gR}...\left(c\right)$
When the block escapes C, its minimum speed $v$ can be given as $\frac{m{v}^2}{R}=mg\Rightarrow v=\sqrt{gR}....\left(d\right)$
By using c and d we obtain
$v_1=\sqrt{5gR}....\left(e\right)$
using a and d we obtain
$d=\left(\sqrt{gR}\right)2\left(\sqrt{\frac{R}{g}}\right)=2R...\left(f\right)$
Putting the values of $v_1$ and d in (b) we obtain

$v_0=\sqrt{5gR+2\mu g\left(2R\right)}=\sqrt{(5+4\mu )gR}$