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Question Number 143666 by ajfour last updated on 16/Jun/21

Commented by ajfour last updated on 19/Jun/21

If the upper disc is released at angle α, find α such that disc loses contact with the fixed hemisphere just when it hits the ground. (assume friction coeff. μ)

$${If}\:{the}\:{upper}\:{disc}\:{is}\:{released} \\ $$$${at}\:{angle}\:\alpha,\:{find}\:\alpha\:{such}\:{that} \\ $$$${disc}\:{loses}\:{contact}\:{with} \\ $$$${the}\:{fixed}\:{hemisphere}\:{just} \\ $$$${when}\:{it}\:{hits}\:{the}\:{ground}. \\ $$$$\left({assume}\:{friction}\:{coeff}.\:\mu\right) \\ $$

Answered by mr W last updated on 17/Jun/21

Commented by mr W last updated on 19/Jun/21

ϕ=((a+b)/b)θ (dϕ/dt)=((a+b)/b)×(dθ/dt)=(((a+b)ω)/b) mgsin θ−N=m(a+b)ω^2 ⇒(N/(mg))=sin θ−(((a+b)ω^2 )/g) (1/2)I((dϕ/dt))^2 +(1/2)m(a+b)^2 ω^2 =mg(a+b)(sin α−sin θ) (1/2)×((mb^2 )/2)(((a+b)^2 ω^2 )/b^2 )+((m(a+b)^2 ω^2 )/2)=mg(a+b)(sin α−sin θ) I=((mb^2 )/2) (as disc, not sphere) ⇒ω^2 =((4g)/(3(a+b)))(sin α−sin θ) (N/(mg))=sin θ−(((a+b)ω^2 )/g)=0 (loss of contact) sin θ=(4/3)(sin α−sin θ) ⇒sin α=(7/4)sin θ sin θ=(b/(a+b)) when ball hits the ground ⇒sin α=((7b)/(4(a+b)))

$$\varphi=\frac{{a}+{b}}{{b}}\theta \\ $$$$\frac{{d}\varphi}{{dt}}=\frac{{a}+{b}}{{b}}×\frac{{d}\theta}{{dt}}=\frac{\left({a}+{b}\right)\omega}{{b}} \\ $$$${mg}\mathrm{sin}\:\theta−{N}={m}\left({a}+{b}\right)\omega^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{N}}{{mg}}=\mathrm{sin}\:\theta−\frac{\left({a}+{b}\right)\omega^{\mathrm{2}} }{{g}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}\left(\frac{{d}\varphi}{{dt}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}\left({a}+{b}\right)^{\mathrm{2}} \omega^{\mathrm{2}} ={mg}\left({a}+{b}\right)\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{{mb}^{\mathrm{2}} }{\mathrm{2}}\frac{\left({a}+{b}\right)^{\mathrm{2}} \omega^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{m}\left({a}+{b}\right)^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{2}}={mg}\left({a}+{b}\right)\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$${I}=\frac{{mb}^{\mathrm{2}} }{\mathrm{2}}\:\left({as}\:{disc},\:{not}\:{sphere}\right) \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{4}{g}}{\mathrm{3}\left({a}+{b}\right)}\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$$\frac{{N}}{{mg}}=\mathrm{sin}\:\theta−\frac{\left({a}+{b}\right)\omega^{\mathrm{2}} }{{g}}=\mathrm{0}\:\left({loss}\:{of}\:{contact}\right) \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{7}}{\mathrm{4}}\mathrm{sin}\:\theta \\ $$$$\mathrm{sin}\:\theta=\frac{{b}}{{a}+{b}}\:{when}\:{ball}\:{hits}\:{the}\:{ground} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{7}{b}}{\mathrm{4}\left({a}+{b}\right)} \\ $$

Answered by ajfour last updated on 18/Jun/21

Energy conservation mg{(a+b)sin θ−b} = ((mb^2 ω^2 )/4)+((mv^2 )/2) v=ωb=−(a+b)(dθ/dt) ⇒ g{(a+b)sin θ−b}=((3b^2 ω^2 )/4) N=0 when at bottom ⇒ mgsin φ=((mω^2 b^2 )/((a+b))) sin φ=(b/(a+b)) ⇒ ω^2 =(g/b) sin θ=((((3b)/4)+b)/(a+b))=((7b)/(4(a+b)))

$${Energy}\:{conservation} \\ $$$${mg}\left\{\left({a}+{b}\right)\mathrm{sin}\:\theta−{b}\right\} \\ $$$$\:\:=\:\frac{{mb}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{4}}+\frac{{mv}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${v}=\omega{b}=−\left({a}+{b}\right)\frac{{d}\theta}{{dt}} \\ $$$$\Rightarrow\:{g}\left\{\left({a}+{b}\right)\mathrm{sin}\:\theta−{b}\right\}=\frac{\mathrm{3}{b}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{4}} \\ $$$${N}=\mathrm{0}\:\:{when}\:{at}\:{bottom}\:\Rightarrow \\ $$$${mg}\mathrm{sin}\:\phi=\frac{{m}\omega^{\mathrm{2}} {b}^{\mathrm{2}} }{\left({a}+{b}\right)} \\ $$$$\mathrm{sin}\:\phi=\frac{{b}}{{a}+{b}} \\ $$$$\Rightarrow\:\:\omega^{\mathrm{2}} =\frac{{g}}{{b}} \\ $$$$\mathrm{sin}\:\theta=\frac{\frac{\mathrm{3}{b}}{\mathrm{4}}+{b}}{{a}+{b}}=\frac{\mathrm{7}{b}}{\mathrm{4}\left({a}+{b}\right)} \\ $$

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