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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 (hollow).

$${There}\:{is}\:{no}\:{friction}\:{between}\:{block} \\ $$$${and}\:{sphere},\:{but}\:{enough}\:{friction} \\ $$$${between}\:{ground}\:{and}\:{sphere},\:{such} \\ $$$${that}\:{sphere}\:{rolls}\:{as}\:{block}\:{slides} \\ $$$${backwards}.\:{Find}\:{maximum} \\ $$$${velocity}\:{acquired}\:{by}\:{sphere}\:\left({hollow}\right). \\ $$

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)))))  ......................................................

$$\left({N}\mathrm{sin}\:\theta\right){R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)={I}\alpha\:\:\:...\left({i}\right) \\ $$$${A}=\alpha{R} \\ $$$${mg}\mathrm{cos}\:\theta−{N}−{m}\alpha{R}\mathrm{sin}\:\theta=\frac{{mv}^{\mathrm{2}} }{{R}}\:\:..\left({ii}\right) \\ $$$${mg}\mathrm{sin}\:\theta+{m}\alpha{R}\mathrm{cos}\:\theta=\frac{{mvdv}}{{Rd}\theta}\:\:\:...\left({iii}\right) \\ $$$$\frac{{v}}{{R}}=\frac{{d}\theta}{{dt}}\:\:\:...\left({iv}\right) \\ $$$$\Rightarrow\:\:\: \\ $$$${mg}\mathrm{sin}\:\theta−\frac{{I}\alpha}{{R}\mathrm{sin}\:\theta\left(\mathrm{1}+\mathrm{cos}\:\theta\right)} \\ $$$$\:\:−{m}\alpha{R}\mathrm{sin}\:\theta={mR}\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} \\ $$$${from}\:\left({iii}\right) \\ $$$$\:\:\:{R}\alpha=\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\left(\frac{{vdv}}{{Rd}\theta}−{g}\mathrm{sin}\:\theta\right)\:\:\: \\ $$$${and}\:\:{let}\:\:\frac{{I}}{{mR}^{\mathrm{2}} }={k}\:\:\:\:,\:{then} \\ $$$${g}\mathrm{sin}\:\theta−\left(\frac{{k}}{\mathrm{sin}\:\theta\left(\mathrm{1}+\mathrm{cos}\:\theta\right)}+\mathrm{sin}\:\theta\right)\alpha{R} \\ $$$$\:\:\:\:\:\:\:={R}\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} \:\:\:\:\Rightarrow \\ $$$$\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }−\left(\frac{{g}}{{R}}\right)\mathrm{sin}\:\theta=\frac{\left[\left(\frac{{g}}{{R}}\right)\mathrm{sin}\:\theta−\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} \right]\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta+\frac{{k}}{\mathrm{sin}\:\theta\left(\mathrm{1}+\mathrm{cos}\:\theta\right)}} \\ $$$$...................................................... \\ $$

Answered by mr W last updated on 22/Aug/20

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