Question Number 213920 by ajfour last updated on 21/Nov/24
Commented by ajfour last updated on 21/Nov/24
$${Can}\:{we}\:{find}\:{at}\:{what}\:{speed}\:{does}\:{point} \\ $$$${P}\:{approach}\:{the}\:{ground}\:{just}\:{before} \\ $$$${hitting}\:{the}\:{ground}\:{if}\:{released}\:{at}\:{say} \\ $$$${the}\:{lower}\:{edge}\:{at}\:\mathrm{45}°\:{to}\:{horizontal}. \\ $$$${The}\:{radius}\:{of}\:{semi}-{disc}\:{is}\:{r},\:{mass}\:{m}. \\ $$
Answered by mr W last updated on 22/Nov/24
Commented by mr W last updated on 23/Nov/24
![e=((4r)/(3π))=kr with k=(4/(3π)) I_0 =((1/2)−((16)/(9π^2 )))mr^2 =ξmr^2 with ξ=(1/2)−((16)/(9π^2 )) the center point G of disc moves only vertically, because no force is acting in horizontal direction. y_G =r sin θ+e cos θ=r(sin θ+k cos θ) v_(Gy) =−(dy_G /dt)=r(cos θ−k sin θ)ω with ω=−(dθ/dt) at t=0: θ_0 =(π/4) y_(G0) =r(sin θ_0 +k cos θ_0 ) mg(y_(G0) −r sin θ−e cos θ)=((I_0 ω^2 )/2)+((mv_(Gy) ^2 )/2) 2mgr(sin θ_0 +k cos θ_0 −sin θ−k cos θ) =mξr^2 ω^2 +mr^2 (cos θ−k sin θ)^2 ω^2 2g(sin θ_0 +k cos θ_0 −sin θ−k cos θ) =r[ξ+(cos θ−k sin θ)^2 ]ω^2 ⇒ω=(√(g/r))(√((2(sin θ_0 +k cos θ_0 −sin θ−k cos θ))/(ξ+(cos θ−k sin θ)^2 ))) y_P =2r sin θ v_(Py) =−(dy_P /dt)=−2r cos θ (dθ/dt)=2r cos θ ω ⇒v_(Py) =2 cos θ(√((2gr(sin θ_0 +k cos θ_0 −sin θ−k cos θ))/(ξ+(cos θ−k sin θ)^2 ))) at θ=0: v_(Py) =(√(([(√2)−((4(2−(√2)))/(3π))]gr)/((3/8)−(4/(9π^2 ))))) ≈1.87948239(√(gr)) +++++++++++++ (dθ/dt)=−(√((2g(sin θ_0 +k cos θ_0 −sin θ−k cos θ))/(r[ξ+(cos θ−k sin θ)^2 ]))) dt=−(√((r[ξ+(cos θ−k sin θ)^2 ])/(2g(sin θ_0 +k cos θ_0 −sin θ−k cos θ)))) dθ ∫_0 ^T dt=(√(r/g))∫_0 ^(π/4) (√((ξ+(cos θ−k sin θ)^2 )/(2(sin θ_0 +k cos θ_0 −sin θ−k cos θ)))) dθ T=(√(r/g))∫_0 ^(π/4) (√(((1/2)−((16)/(9π^2 ))+(cos θ−((4 sin θ)/(3π)))^2 )/( (√2)+((4(√2))/( 3π))−2 sin θ−((8 cos θ)/(3π))))) dθ ≈1.47971689(√(r/g)) +++++++++++++ let f(θ)=(√((2(sin θ_0 +k cos θ_0 −sin θ−k cos θ))/(ξ+(cos θ−k sin θ)^2 ))) ω=(√(g/r))f(θ) α=(dω/dt)=ω(dω/dθ)=(g/r)f(θ)((df(θ))/dθ) N(r cos θ−e sin θ)=I_0 α Nr(cos θ−k sin θ)=mξr^2 ω(dω/dθ) ⇒(N/(mg))=((((1/2)−((16)/(9π^2 )))f(θ)((df(θ))/dθ))/(cos θ−k sin θ))](https://www.tinkutara.com/question/Q213959.png)
$${e}=\frac{\mathrm{4}{r}}{\mathrm{3}\pi}={kr}\:{with}\:{k}=\frac{\mathrm{4}}{\mathrm{3}\pi} \\ $$$${I}_{\mathrm{0}} =\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{16}}{\mathrm{9}\pi^{\mathrm{2}} }\right){mr}^{\mathrm{2}} =\xi{mr}^{\mathrm{2}} \:{with}\:\xi=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{16}}{\mathrm{9}\pi^{\mathrm{2}} } \\ $$$${the}\:{center}\:{point}\:{G}\:{of}\:{disc}\:{moves}\: \\ $$$${only}\:{vertically},\:{because}\:{no}\:{force} \\ $$$${is}\:{acting}\:{in}\:{horizontal}\:{direction}. \\ $$$${y}_{{G}} ={r}\:\mathrm{sin}\:\theta+{e}\:\mathrm{cos}\:\theta={r}\left(\mathrm{sin}\:\theta+{k}\:\mathrm{cos}\:\theta\right) \\ $$$${v}_{{Gy}} =−\frac{{dy}_{{G}} }{{dt}}={r}\left(\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)\omega \\ $$$${with}\:\omega=−\frac{{d}\theta}{{dt}} \\ $$$${at}\:{t}=\mathrm{0}:\:\theta_{\mathrm{0}} =\frac{\pi}{\mathrm{4}} \\ $$$${y}_{{G}\mathrm{0}} ={r}\left(\mathrm{sin}\:\theta_{\mathrm{0}} +{k}\:\mathrm{cos}\:\theta_{\mathrm{0}} \right) \\ $$$${mg}\left({y}_{{G}\mathrm{0}} −{r}\:\mathrm{sin}\:\theta−{e}\:\mathrm{cos}\:\theta\right)=\frac{{I}_{\mathrm{0}} \omega^{\mathrm{2}} }{\mathrm{2}}+\frac{{mv}_{{Gy}} ^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{2}{mgr}\left(\mathrm{sin}\:\theta_{\mathrm{0}} +{k}\:\mathrm{cos}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta−{k}\:\mathrm{cos}\:\theta\right) \\ $$$$\:\:={m}\xi{r}^{\mathrm{2}} \omega^{\mathrm{2}} +{mr}^{\mathrm{2}} \left(\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \omega^{\mathrm{2}} \\ $$$$\mathrm{2}{g}\left(\mathrm{sin}\:\theta_{\mathrm{0}} +{k}\:\mathrm{cos}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta−{k}\:\mathrm{cos}\:\theta\right) \\ $$$$\:\:={r}\left[\xi+\left(\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \right]\omega^{\mathrm{2}} \\ $$$$\Rightarrow\omega=\sqrt{\frac{{g}}{{r}}}\sqrt{\frac{\mathrm{2}\left(\mathrm{sin}\:\theta_{\mathrm{0}} +{k}\:\mathrm{cos}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta−{k}\:\mathrm{cos}\:\theta\right)}{\xi+\left(\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }} \\ $$$${y}_{{P}} =\mathrm{2}{r}\:\mathrm{sin}\:\theta \\ $$$${v}_{{Py}} =−\frac{{dy}_{{P}} }{{dt}}=−\mathrm{2}{r}\:\mathrm{cos}\:\theta\:\frac{{d}\theta}{{dt}}=\mathrm{2}{r}\:\mathrm{cos}\:\theta\:\omega \\ $$$$\Rightarrow{v}_{{Py}} =\mathrm{2}\:\mathrm{cos}\:\theta\sqrt{\frac{\mathrm{2}{gr}\left(\mathrm{sin}\:\theta_{\mathrm{0}} +{k}\:\mathrm{cos}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta−{k}\:\mathrm{cos}\:\theta\right)}{\xi+\left(\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }} \\ $$$${at}\:\theta=\mathrm{0}: \\ $$$${v}_{{Py}} =\sqrt{\frac{\left[\sqrt{\mathrm{2}}−\frac{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}{\mathrm{3}\pi}\right]{gr}}{\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{4}}{\mathrm{9}\pi^{\mathrm{2}} }}} \\ $$$$\:\:\:\:\:\:\approx\mathrm{1}.\mathrm{87948239}\sqrt{{gr}} \\ $$$$+++++++++++++ \\ $$$$\frac{{d}\theta}{{dt}}=−\sqrt{\frac{\mathrm{2}{g}\left(\mathrm{sin}\:\theta_{\mathrm{0}} +{k}\:\mathrm{cos}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta−{k}\:\mathrm{cos}\:\theta\right)}{{r}\left[\xi+\left(\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \right]}} \\ $$$${dt}=−\sqrt{\frac{{r}\left[\xi+\left(\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \right]}{\mathrm{2}{g}\left(\mathrm{sin}\:\theta_{\mathrm{0}} +{k}\:\mathrm{cos}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta−{k}\:\mathrm{cos}\:\theta\right)}}\:{d}\theta \\ $$$$\int_{\mathrm{0}} ^{{T}} {dt}=\sqrt{\frac{{r}}{{g}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\frac{\xi+\left(\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{sin}\:\theta_{\mathrm{0}} +{k}\:\mathrm{cos}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta−{k}\:\mathrm{cos}\:\theta\right)}}\:{d}\theta \\ $$$${T}=\sqrt{\frac{{r}}{{g}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\frac{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{16}}{\mathrm{9}\pi^{\mathrm{2}} }+\left(\mathrm{cos}\:\theta−\frac{\mathrm{4}\:\mathrm{sin}\:\theta}{\mathrm{3}\pi}\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{2}}+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\:\mathrm{3}\pi}−\mathrm{2}\:\mathrm{sin}\:\theta−\frac{\mathrm{8}\:\mathrm{cos}\:\theta}{\mathrm{3}\pi}}}\:{d}\theta \\ $$$$\:\:\:\approx\mathrm{1}.\mathrm{47971689}\sqrt{\frac{{r}}{{g}}} \\ $$$$+++++++++++++ \\ $$$${let}\:{f}\left(\theta\right)=\sqrt{\frac{\mathrm{2}\left(\mathrm{sin}\:\theta_{\mathrm{0}} +{k}\:\mathrm{cos}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta−{k}\:\mathrm{cos}\:\theta\right)}{\xi+\left(\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }} \\ $$$$\omega=\sqrt{\frac{{g}}{{r}}}{f}\left(\theta\right) \\ $$$$\alpha=\frac{{d}\omega}{{dt}}=\omega\frac{{d}\omega}{{d}\theta}=\frac{{g}}{{r}}{f}\left(\theta\right)\frac{{df}\left(\theta\right)}{{d}\theta} \\ $$$${N}\left({r}\:\mathrm{cos}\:\theta−{e}\:\mathrm{sin}\:\theta\right)={I}_{\mathrm{0}} \alpha \\ $$$${Nr}\left(\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)={m}\xi{r}^{\mathrm{2}} \omega\frac{{d}\omega}{{d}\theta} \\ $$$$\Rightarrow\frac{{N}}{{mg}}=\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{16}}{\mathrm{9}\pi^{\mathrm{2}} }\right){f}\left(\theta\right)\frac{{df}\left(\theta\right)}{{d}\theta}}{\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta} \\ $$
Commented by ajfour last updated on 23/Nov/24
$${v}_{{Py}} ^{\mathrm{2}} =\left(\mathrm{2}{gr}\right)\left\{\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({k}+\mathrm{1}\right)−{k}}{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{k}^{\mathrm{2}} \right)−{k}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{k}\right)^{\mathrm{2}} }\right\} \\ $$$${v}_{{Py}} ^{\mathrm{2}} =\sqrt{\mathrm{2}}{gr}\left\{\frac{\mathrm{1}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){k}}{\mathrm{1}−{k}}\right\} \\ $$$$\:\:\:\:\:\:\:\approx\:\mathrm{1}.\mathrm{423056}\:\sqrt{{gr}} \\ $$
Commented by mr W last updated on 23/Nov/24
$${did}\:{you}\:{take}\:{the}\:{lower}\:{point}\:{as}\:{a} \\ $$$${hinge}?\:{i}\:{thought}\:{it}\:{were}\:{a}\:{roller}. \\ $$
Commented by ajfour last updated on 23/Nov/24
$${yeah}\:{roller}\:{Sir},\:{i}\:{have}\:{used} \\ $$$${K}.{E}.\:=\:\frac{\mathrm{1}}{\mathrm{2}}{I}_{{cm}} \omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{cm}} ^{\mathrm{2}} \\ $$
Commented by ajfour last updated on 23/Nov/24
Commented by ajfour last updated on 23/Nov/24
$${Excellent}\:{sir},\:{thank}\:{you}\:{v}\:{very}\:{much}. \\ $$
Commented by mr W last updated on 23/Nov/24
Commented by mr W last updated on 23/Nov/24
![please recheck! at t=0: θ=45° P.E.=mg R sin (θ+β)=mg(r sin θ+h cos θ) at t=T: P.E.=mgh v_G =ωr=ωR cos β K.E.=((I_0 ω^2 )/2)+((mv_G ^2 )/2)=((mω^2 R^2 )/2)((I_0 /(mR^2 ))+cos^2 β) mgR sin (θ+β)−mgh=((mω^2 R^2 )/2)((I_0 /(mR^2 ))+cos^2 β) 2g[R sin (θ+β)−h]=ω^2 ((I_0 /m)+r^2 ) 2g[r sin θ+h cos θ−h]=ω^2 ((1/2)−((16)/(9π^2 ))+1)r^2 2g(sin θ+k cos θ−k)=ω^2 ((1/2)−((16)/(9π^2 ))+1)r ⇒ω^2 =((2g(sin θ+k cos θ−k))/(r((3/2)−((16)/(9π^2 )))))=((g((√2)+(√2)k−2k))/(r((3/2)−((16)/(9π^2 ))))) v_P =2rω=2(√((((√2)+(√2)k−2k)gr)/((3/2)−((16)/(9π^2 )))))](https://www.tinkutara.com/question/Q213974.png)
$${please}\:{recheck}! \\ $$$${at}\:{t}=\mathrm{0}: \\ $$$$\theta=\mathrm{45}° \\ $$$${P}.{E}.={mg}\:{R}\:\mathrm{sin}\:\left(\theta+\beta\right)={mg}\left({r}\:\mathrm{sin}\:\theta+{h}\:\mathrm{cos}\:\theta\right) \\ $$$${at}\:{t}={T}: \\ $$$${P}.{E}.={mgh} \\ $$$${v}_{{G}} =\omega{r}=\omega{R}\:\mathrm{cos}\:\beta \\ $$$${K}.{E}.=\frac{{I}_{\mathrm{0}} \omega^{\mathrm{2}} }{\mathrm{2}}+\frac{{mv}_{{G}} ^{\mathrm{2}} }{\mathrm{2}}=\frac{{m}\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{{I}_{\mathrm{0}} }{{mR}^{\mathrm{2}} }+\mathrm{cos}^{\mathrm{2}} \:\beta\right) \\ $$$${mgR}\:\mathrm{sin}\:\left(\theta+\beta\right)−{mgh}=\frac{{m}\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{{I}_{\mathrm{0}} }{{mR}^{\mathrm{2}} }+\mathrm{cos}^{\mathrm{2}} \:\beta\right) \\ $$$$\mathrm{2}{g}\left[{R}\:\mathrm{sin}\:\left(\theta+\beta\right)−{h}\right]=\omega^{\mathrm{2}} \left(\frac{{I}_{\mathrm{0}} }{{m}}+{r}^{\mathrm{2}} \right) \\ $$$$\mathrm{2}{g}\left[{r}\:\mathrm{sin}\:\theta+{h}\:\mathrm{cos}\:\theta−{h}\right]=\omega^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{16}}{\mathrm{9}\pi^{\mathrm{2}} }+\mathrm{1}\right){r}^{\mathrm{2}} \\ $$$$\mathrm{2}{g}\left(\mathrm{sin}\:\theta+{k}\:\mathrm{cos}\:\theta−{k}\right)=\omega^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{16}}{\mathrm{9}\pi^{\mathrm{2}} }+\mathrm{1}\right){r} \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{2}{g}\left(\mathrm{sin}\:\theta+{k}\:\mathrm{cos}\:\theta−{k}\right)}{{r}\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{16}}{\mathrm{9}\pi^{\mathrm{2}} }\right)}=\frac{{g}\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{k}−\mathrm{2}{k}\right)}{{r}\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{16}}{\mathrm{9}\pi^{\mathrm{2}} }\right)} \\ $$$${v}_{{P}} =\mathrm{2}{r}\omega=\mathrm{2}\sqrt{\frac{\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{k}−\mathrm{2}{k}\right){gr}}{\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{16}}{\mathrm{9}\pi^{\mathrm{2}} }}} \\ $$