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Question-185490




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

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