Question-186692

Question Number 186692 by ajfour last updated on 08/Feb/23

Commented by ajfour last updated on 08/Feb/23

(i) Find u, given the rest. (ii) Repeat (i) if R_1 =a , R_2 =b.

$$\left({i}\right)\:{Find}\:{u},\:{given}\:{the}\:{rest}. \\ $$$$\left({ii}\right)\:{Repeat}\:\left({i}\right)\:\:{if}\:{R}_{\mathrm{1}} ={a}\:,\:{R}_{\mathrm{2}} ={b}. \\ $$

Answered by mr W last updated on 08/Feb/23

Commented by mr W last updated on 09/Feb/23

g_l =g sin θ g_r =g cos θ L=vt+((g_l t^2 )/2) t=((−v+(√(v^2 +2g_l L)))/g_l ) u_1 ^2 =u^2 −2g_r R(1−cos ϕ) N=mg_r cos ϕ+((mu_1 ^2 )/R)=mg_r cos ϕ+((mu^2 −2mg_r R(1−cos ϕ))/R) N=mg_r (3 cos ϕ−2)+((mu^2 )/R)≥0 (u^2 /R)≥5g_r =5g cos θ ⇒u≥(√(5gR cos θ)) u_1 =((Rdϕ)/dt)=(√(u^2 −2g_r R(1−cos ϕ))) (√(R/(2g_r )))× (dϕ/( (√((u^2 /(2g_r R))−1+cos ϕ))))=dt (√((2R)/g_r ))∫_0 ^π (dϕ/( (√((u^2 /(2g_r R))−1+cos ϕ))))=t (√((2R)/g_r ))∫_0 ^π (dϕ/( (√((u^2 /(2g_r R))−1+cos ϕ))))=((−v+(√(v^2 +2g_l L)))/g_l ) ⇒(√((2R)/(g cos θ)))∫_0 ^π (dϕ/( (√((u^2 /(2gR cos θ))−1+cos ϕ))))=((−v+(√(v^2 +2gL sin θ)))/(g sin θ)) ⇒((4gR sin θ)/( u))F((π/2)∣((4gR cos θ)/( u^2 )))=(√(v^2 +2gL sin θ))−v

$${g}_{{l}} ={g}\:\mathrm{sin}\:\theta \\ $$$${g}_{{r}} ={g}\:\mathrm{cos}\:\theta \\ $$$${L}={vt}+\frac{{g}_{{l}} {t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${t}=\frac{−{v}+\sqrt{{v}^{\mathrm{2}} +\mathrm{2}{g}_{{l}} {L}}}{{g}_{{l}} } \\ $$$${u}_{\mathrm{1}} ^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}{g}_{{r}} {R}\left(\mathrm{1}−\mathrm{cos}\:\varphi\right) \\ $$$${N}={mg}_{{r}} \mathrm{cos}\:\varphi+\frac{{mu}_{\mathrm{1}} ^{\mathrm{2}} }{{R}}={mg}_{{r}} \mathrm{cos}\:\varphi+\frac{{mu}^{\mathrm{2}} −\mathrm{2}{mg}_{{r}} {R}\left(\mathrm{1}−\mathrm{cos}\:\varphi\right)}{{R}} \\ $$$${N}={mg}_{{r}} \left(\mathrm{3}\:\mathrm{cos}\:\varphi−\mathrm{2}\right)+\frac{{mu}^{\mathrm{2}} }{{R}}\geqslant\mathrm{0} \\ $$$$\frac{{u}^{\mathrm{2}} }{{R}}\geqslant\mathrm{5}{g}_{{r}} =\mathrm{5}{g}\:\mathrm{cos}\:\theta\:\Rightarrow{u}\geqslant\sqrt{\mathrm{5}{gR}\:\mathrm{cos}\:\theta} \\ $$$${u}_{\mathrm{1}} =\frac{{Rd}\varphi}{{dt}}=\sqrt{{u}^{\mathrm{2}} −\mathrm{2}{g}_{{r}} {R}\left(\mathrm{1}−\mathrm{cos}\:\varphi\right)} \\ $$$$\sqrt{\frac{{R}}{\mathrm{2}{g}_{{r}} }}×\:\frac{{d}\varphi}{\:\sqrt{\frac{{u}^{\mathrm{2}} }{\mathrm{2}{g}_{{r}} {R}}−\mathrm{1}+\mathrm{cos}\:\varphi}}={dt} \\ $$$$\sqrt{\frac{\mathrm{2}{R}}{{g}_{{r}} }}\int_{\mathrm{0}} ^{\pi} \frac{{d}\varphi}{\:\sqrt{\frac{{u}^{\mathrm{2}} }{\mathrm{2}{g}_{{r}} {R}}−\mathrm{1}+\mathrm{cos}\:\varphi}}={t} \\ $$$$\sqrt{\frac{\mathrm{2}{R}}{{g}_{{r}} }}\int_{\mathrm{0}} ^{\pi} \frac{{d}\varphi}{\:\sqrt{\frac{{u}^{\mathrm{2}} }{\mathrm{2}{g}_{{r}} {R}}−\mathrm{1}+\mathrm{cos}\:\varphi}}=\frac{−{v}+\sqrt{{v}^{\mathrm{2}} +\mathrm{2}{g}_{{l}} {L}}}{{g}_{{l}} } \\ $$$$\Rightarrow\sqrt{\frac{\mathrm{2}{R}}{{g}\:\mathrm{cos}\:\theta}}\int_{\mathrm{0}} ^{\pi} \frac{{d}\varphi}{\:\sqrt{\frac{{u}^{\mathrm{2}} }{\mathrm{2}{gR}\:\mathrm{cos}\:\theta}−\mathrm{1}+\mathrm{cos}\:\varphi}}=\frac{−{v}+\sqrt{{v}^{\mathrm{2}} +\mathrm{2}{gL}\:\mathrm{sin}\:\theta}}{{g}\:\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\frac{\mathrm{4}{gR}\:\mathrm{sin}\:\theta}{\:{u}}{F}\left(\frac{\pi}{\mathrm{2}}\mid\frac{\mathrm{4}{gR}\:\mathrm{cos}\:\theta}{\:{u}^{\mathrm{2}} }\right)=\sqrt{{v}^{\mathrm{2}} +\mathrm{2}{gL}\:\mathrm{sin}\:\theta}−{v} \\ $$

Commented by ajfour last updated on 09/Feb/23

I get ∫_0 ^( 2π) ((√(Rtan θ))/( (√(((u_1 ^2 /(2g_R R))−1)+cos ϕ))))dϕ =∫_0 ^( L) (dx/( (√(x+(v_0 ^2 /(2g_L )))))) but i want to deal this using Torque-Angular Momentum ways too!

$${I}\:{get}\: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{\sqrt{{R}\mathrm{tan}\:\theta}}{\:\sqrt{\left(\frac{{u}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}{g}_{{R}} {R}}−\mathrm{1}\right)+\mathrm{cos}\:\varphi}}{d}\varphi \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\:{L}} \frac{{dx}}{\:\sqrt{{x}+\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}{g}_{{L}} }}} \\ $$$${but}\:{i}\:{want}\:{to}\:{deal}\:{this}\:{using} \\ $$$${Torque}-{Angular}\:{Momentum}\:{ways} \\ $$$${too}! \\ $$

Commented by ajfour last updated on 09/Feb/23

Thank you Sir. What is this F(theta) Fourier Integral?

Commented by mr W last updated on 10/Feb/23

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