Question Number 108588 by ajfour last updated on 17/Aug/20
Commented by ajfour last updated on 21/Aug/20
$${There}\:{is}\:{no}\:{friction}\:{between}\:{block} \\ $$$${and}\:{sphere},\:{but}\:{enough}\:{friction} \\ $$$${between}\:{ground}\:{and}\:{sphere},\:{such} \\ $$$${that}\:{sphere}\:{rolls}\:{as}\:{block}\:{slides} \\ $$$${backwards}.\:{Find}\:{maximum} \\ $$$${velocity}\:{acquired}\:{by}\:{sphere}\:\left({hollow}\right). \\ $$
Answered by ajfour last updated on 21/Aug/20
Commented by ajfour last updated on 21/Aug/20
![(Nsin θ)R(1+cos θ)=Iα ...(i) A=αR mgcos θ−N−mαRsin θ=((mv^2 )/R) ..(ii) mgsin θ+mαRcos θ=((mvdv)/(Rdθ)) ...(iii) (v/R)=(dθ/dt) ...(iv) ⇒ mgsin θ−((Iα)/(Rsin θ(1+cos θ))) −mαRsin θ=mR((dθ/dt))^2 from (iii) Rα=(1/(cos θ))(((vdv)/(Rdθ))−gsin θ) and let (I/(mR^2 ))=k , then gsin θ−((k/(sin θ(1+cos θ)))+sin θ)αR =R((dθ/dt))^2 ⇒ (d^2 θ/dt^2 )−((g/R))sin θ=(([((g/R))sin θ−((dθ/dt))^2 ]cos θ)/(sin θ+(k/(sin θ(1+cos θ))))) ......................................................](https://www.tinkutara.com/question/Q109134.png)
$$\left({N}\mathrm{sin}\:\theta\right){R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)={I}\alpha\:\:\:…\left({i}\right) \\ $$$${A}=\alpha{R} \\ $$$${mg}\mathrm{cos}\:\theta−{N}−{m}\alpha{R}\mathrm{sin}\:\theta=\frac{{mv}^{\mathrm{2}} }{{R}}\:\:..\left({ii}\right) \\ $$$${mg}\mathrm{sin}\:\theta+{m}\alpha{R}\mathrm{cos}\:\theta=\frac{{mvdv}}{{Rd}\theta}\:\:\:…\left({iii}\right) \\ $$$$\frac{{v}}{{R}}=\frac{{d}\theta}{{dt}}\:\:\:…\left({iv}\right) \\ $$$$\Rightarrow\:\:\: \\ $$$${mg}\mathrm{sin}\:\theta−\frac{{I}\alpha}{{R}\mathrm{sin}\:\theta\left(\mathrm{1}+\mathrm{cos}\:\theta\right)} \\ $$$$\:\:−{m}\alpha{R}\mathrm{sin}\:\theta={mR}\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} \\ $$$${from}\:\left({iii}\right) \\ $$$$\:\:\:{R}\alpha=\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\left(\frac{{vdv}}{{Rd}\theta}−{g}\mathrm{sin}\:\theta\right)\:\:\: \\ $$$${and}\:\:{let}\:\:\frac{{I}}{{mR}^{\mathrm{2}} }={k}\:\:\:\:,\:{then} \\ $$$${g}\mathrm{sin}\:\theta−\left(\frac{{k}}{\mathrm{sin}\:\theta\left(\mathrm{1}+\mathrm{cos}\:\theta\right)}+\mathrm{sin}\:\theta\right)\alpha{R} \\ $$$$\:\:\:\:\:\:\:={R}\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} \:\:\:\:\Rightarrow \\ $$$$\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }−\left(\frac{{g}}{{R}}\right)\mathrm{sin}\:\theta=\frac{\left[\left(\frac{{g}}{{R}}\right)\mathrm{sin}\:\theta−\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} \right]\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta+\frac{{k}}{\mathrm{sin}\:\theta\left(\mathrm{1}+\mathrm{cos}\:\theta\right)}} \\ $$$$……………………………………………… \\ $$
Answered by mr W last updated on 22/Aug/20
