Answer:
Explanation:
Let T be the tension in the swing
At top point [tex]mg-T=\frac{mv^2}{r}[/tex]
where v=velocity needed to complete circular path
r=distance between point of rotation to the ball center=L+\frac{d}{2} (d=diameter of ball)
Th-resold velocity is given by [tex]mg-0=\frac{mv^2}{r}[/tex]
To get the velocity at bottom conserve energy at Top and bottom
At top [tex]E_T=mg\times 2L+\frac{mv^2}{2}[/tex]
Energy at Bottom [tex]E_b=\frac{mv_0^2}{2}[/tex]
Comparing two as energy is conserved
[tex]v_0^2=4gr+gr[/tex]
[tex]v_0^2=5gr[/tex]
[tex]v_0=\sqrt{5gr}[/tex]
[tex]v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}[/tex]