Question Number 185490 by ajfour last updated on 22/Jan/23
Commented by ajfour last updated on 23/Jan/23
$${If}\:{the}\:{solid}\:{ball}\:{is}\:{released}\:{at} \\ $$$${the}\:{heighest}\:{level}\:{touching}\:{both} \\ $$$${the}\:{touching}\:{fixed}\:{hemispheres}, \\ $$$${and}\:{if}\:{there}\:{be}\:{no}\:{friction}, \\ $$$${find}\:{b}/{a}\:{such}\:{that}\:{the}\:{solid} \\ $$$${ball}\:{loses}\:{contact}\:{with}\:{hemispheres} \\ $$$${just}\:{as}\:{it}\:{gains}\:{contact}\:{with}\:{the} \\ $$$${ground}. \\ $$
Answered by mr W last updated on 24/Jan/23
Commented by mr W last updated on 24/Jan/23
$${r}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} ={b}\left(\mathrm{2}{a}−{b}\right) \\ $$$${when}\:{contact}\:{gets}\:{lost},\:{N}=\mathrm{0}\:{and} \\ $$$$\frac{{mv}^{\mathrm{2}} }{{r}}={mg}\:\mathrm{sin}\:\theta \\ $$$${v}^{\mathrm{2}} ={rg}\:\mathrm{sin}\:\theta \\ $$$${on}\:{the}\:{other}\:{side}, \\ $$$${v}^{\mathrm{2}} =\mathrm{2}{gr}\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow{rg}\:\mathrm{sin}\:\theta=\mathrm{2}{gr}\:\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${at}\:{the}\:{instant}\:{when}\:{ball}\:{touches}\:{the} \\ $$$${ground},\:\mathrm{sin}\:\theta=\frac{{b}}{{r}} \\ $$$$\Rightarrow\frac{{b}}{{r}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{b}}{\:\sqrt{{b}\left(\mathrm{2}{a}−{b}\right)}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{b}}{{a}}=\frac{\mathrm{8}}{\mathrm{13}}\:\checkmark \\ $$
Commented by ajfour last updated on 24/Jan/23
$${Indeed}\:{very}\:{beautiful}\:{solution}\:{to} \\ $$$${a}\:{pretty}\:{question}.\:{Thanks}\:{Sir}. \\ $$