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Question-71421




Question Number 71421 by ajfour last updated on 15/Oct/19
Commented by ajfour last updated on 15/Oct/19
Find speed of sphere as it rolls  down the wedge, and is about to  touch the ground. Also find the  least coefficient of friction,  so that sphere comes rolling all  the way; wedge isn′t fixed,  assume that ground is smooth.
$${Find}\:{speed}\:{of}\:{sphere}\:{as}\:{it}\:{rolls} \\ $$$${down}\:{the}\:{wedge},\:{and}\:{is}\:{about}\:{to} \\ $$$${touch}\:{the}\:{ground}.\:{Also}\:{find}\:{the} \\ $$$${least}\:{coefficient}\:{of}\:{friction}, \\ $$$${so}\:{that}\:{sphere}\:{comes}\:{rolling}\:{all} \\ $$$${the}\:{way};\:{wedge}\:{isn}'{t}\:{fixed}, \\ $$$${assume}\:{that}\:{ground}\:{is}\:{smooth}. \\ $$
Answered by mr W last updated on 16/Oct/19
Commented by mr W last updated on 19/Oct/19
v=(ds/dt)=ωr  A=rα  α=(dω/dt)=(dω/ds)×(ds/dt)=(v/r)×(dv/ds)  I=((2mr^2 )/5)  Iα=rf  α=((rf)/I)=((5f)/(2mr))    Ma=−f cos θ+N sin θ  a=((−f cos θ+N sin θ)/M)    f=mg sin θ+ma cos θ−mrα  (7/2)f=mg sin θ+ma cos θ  N=mg cos θ−ma sin θ  (7/2)f=mg sin θ+(m/M)(−f cos θ+N sin θ) cos θ  ⇒((7/2)+(m/M) cos^2  θ)f−(m/M) sin θ cos θ N=mg sin θ  N=mg cos θ−(m/M)(−f cos θ+N sin θ) sin θ  ⇒−(m/M) sin θ cos θ f+(1+(m/M) sin^2  θ)N=mg cos θ  ⇒f=((mg sin θ (1+(m/M) sin^2  θ)−mg cos θ (−(m/M) sin θ cos θ))/(((7/2)+(m/M) cos^2  θ)(1+(m/M) sin^2  θ)−(−(m/M) sin θ cos θ)^2 ))  ⇒f=((mg sin θ (1+(m/M)))/((7/2)+(m/M)((5/2) sin^2  θ+1)))    ⇒N=((mg cos θ ((7/2)−(m/M) cos^2  θ)−mg sin θ (−(m/M) sin θ cos θ))/(((7/2)+(m/M) cos^2  θ)(1+(m/M) sin^2  θ)−(−(m/M) sin θ cos θ)^2 ))  ⇒N=((mg cos θ((7/2)+(m/M)))/((7/2)+(m/M)((5/2) sin^2  θ+1)))  f≤μN  ((mg sin θ (1+(m/M)))/((7/2)+(m/M)((5/2) sin^2  θ+1)))≤μ((mg cos θ((7/2)+(m/M)))/((7/2)+(m/M)((5/2) sin^2  θ+1)))  ⇒μ≥((1+(m/M))/((7/2)+(m/M)))×tan θ  ⇒α=((5f)/(2mr))=((5g)/(2r))×((sin θ(1+(m/M)))/((7/2)+(m/M)((5/2) sin^2  θ+1)))  ⇒(v/r)×(dv/ds)=((5g)/(2r))×((sin θ(1+(m/M)))/((7/2)+(m/M)((5/2) sin^2  θ+1)))  ⇒vdv=(((5/2)g sin θ(1+(m/M)))/((7/2)+(m/M)((5/2) sin^2  θ+1)))ds=λds  with λ=(((5/2)g sin θ(1+(m/M)))/((7/2)+(m/M)((5/2) sin^2  θ+1)))  =((5g sin θ)/(((5M+2M+2m)/(M+m))+((5m(1−cos^2  θ))/(M+m))))  =((g sin θ)/((7/5)−((m cos^2  θ)/(M+m))))  ⇒(v^2 /2)=λs  ⇒v=(√(2λs))=(√((2gs sin θ)/((7/5)−((m cos^2  θ)/(M+m)))))  when the ball is about to touch the  ground, s=(h/(sin θ))−(r/(tan (θ/2)))  ⇒v=(√((2g(h−2r cos^2  (θ/2)) )/((7/5)−((m cos^2  θ)/(M+m)))))
$${v}=\frac{{ds}}{{dt}}=\omega{r} \\ $$$${A}={r}\alpha \\ $$$$\alpha=\frac{{d}\omega}{{dt}}=\frac{{d}\omega}{{ds}}×\frac{{ds}}{{dt}}=\frac{{v}}{{r}}×\frac{{dv}}{{ds}} \\ $$$${I}=\frac{\mathrm{2}{mr}^{\mathrm{2}} }{\mathrm{5}} \\ $$$${I}\alpha={rf} \\ $$$$\alpha=\frac{{rf}}{{I}}=\frac{\mathrm{5}{f}}{\mathrm{2}{mr}} \\ $$$$ \\ $$$${Ma}=−{f}\:\mathrm{cos}\:\theta+{N}\:\mathrm{sin}\:\theta \\ $$$${a}=\frac{−{f}\:\mathrm{cos}\:\theta+{N}\:\mathrm{sin}\:\theta}{{M}} \\ $$$$ \\ $$$${f}={mg}\:\mathrm{sin}\:\theta+{ma}\:\mathrm{cos}\:\theta−{mr}\alpha \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}{f}={mg}\:\mathrm{sin}\:\theta+{ma}\:\mathrm{cos}\:\theta \\ $$$${N}={mg}\:\mathrm{cos}\:\theta−{ma}\:\mathrm{sin}\:\theta \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}{f}={mg}\:\mathrm{sin}\:\theta+\frac{{m}}{{M}}\left(−{f}\:\mathrm{cos}\:\theta+{N}\:\mathrm{sin}\:\theta\right)\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\left(\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\:\mathrm{cos}^{\mathrm{2}} \:\theta\right){f}−\frac{{m}}{{M}}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:{N}={mg}\:\mathrm{sin}\:\theta \\ $$$${N}={mg}\:\mathrm{cos}\:\theta−\frac{{m}}{{M}}\left(−{f}\:\mathrm{cos}\:\theta+{N}\:\mathrm{sin}\:\theta\right)\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow−\frac{{m}}{{M}}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:{f}+\left(\mathrm{1}+\frac{{m}}{{M}}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right){N}={mg}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{f}=\frac{{mg}\:\mathrm{sin}\:\theta\:\left(\mathrm{1}+\frac{{m}}{{M}}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right)−{mg}\:\mathrm{cos}\:\theta\:\left(−\frac{{m}}{{M}}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\right)}{\left(\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)\left(\mathrm{1}+\frac{{m}}{{M}}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right)−\left(−\frac{{m}}{{M}}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{f}=\frac{{mg}\:\mathrm{sin}\:\theta\:\left(\mathrm{1}+\frac{{m}}{{M}}\right)}{\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\left(\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{1}\right)} \\ $$$$ \\ $$$$\Rightarrow{N}=\frac{{mg}\:\mathrm{cos}\:\theta\:\left(\frac{\mathrm{7}}{\mathrm{2}}−\frac{{m}}{{M}}\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)−{mg}\:\mathrm{sin}\:\theta\:\left(−\frac{{m}}{{M}}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\right)}{\left(\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)\left(\mathrm{1}+\frac{{m}}{{M}}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right)−\left(−\frac{{m}}{{M}}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{N}=\frac{{mg}\:\mathrm{cos}\:\theta\left(\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\right)}{\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\left(\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{1}\right)} \\ $$$${f}\leqslant\mu{N} \\ $$$$\frac{{mg}\:\mathrm{sin}\:\theta\:\left(\mathrm{1}+\frac{{m}}{{M}}\right)}{\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\left(\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{1}\right)}\leqslant\mu\frac{{mg}\:\mathrm{cos}\:\theta\left(\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\right)}{\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\left(\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{1}\right)} \\ $$$$\Rightarrow\mu\geqslant\frac{\mathrm{1}+\frac{{m}}{{M}}}{\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}}×\mathrm{tan}\:\theta \\ $$$$\Rightarrow\alpha=\frac{\mathrm{5}{f}}{\mathrm{2}{mr}}=\frac{\mathrm{5}{g}}{\mathrm{2}{r}}×\frac{\mathrm{sin}\:\theta\left(\mathrm{1}+\frac{{m}}{{M}}\right)}{\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\left(\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{{v}}{{r}}×\frac{{dv}}{{ds}}=\frac{\mathrm{5}{g}}{\mathrm{2}{r}}×\frac{\mathrm{sin}\:\theta\left(\mathrm{1}+\frac{{m}}{{M}}\right)}{\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\left(\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{1}\right)} \\ $$$$\Rightarrow{vdv}=\frac{\frac{\mathrm{5}}{\mathrm{2}}{g}\:\mathrm{sin}\:\theta\left(\mathrm{1}+\frac{{m}}{{M}}\right)}{\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\left(\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{1}\right)}{ds}=\lambda{ds} \\ $$$${with}\:\lambda=\frac{\frac{\mathrm{5}}{\mathrm{2}}{g}\:\mathrm{sin}\:\theta\left(\mathrm{1}+\frac{{m}}{{M}}\right)}{\frac{\mathrm{7}}{\mathrm{2}}+\frac{{m}}{{M}}\left(\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{5}{g}\:\mathrm{sin}\:\theta}{\frac{\mathrm{5}{M}+\mathrm{2}{M}+\mathrm{2}{m}}{{M}+{m}}+\frac{\mathrm{5}{m}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\theta\right)}{{M}+{m}}} \\ $$$$=\frac{{g}\:\mathrm{sin}\:\theta}{\frac{\mathrm{7}}{\mathrm{5}}−\frac{{m}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{{M}+{m}}} \\ $$$$\Rightarrow\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\lambda{s} \\ $$$$\Rightarrow{v}=\sqrt{\mathrm{2}\lambda{s}}=\sqrt{\frac{\mathrm{2}{gs}\:\mathrm{sin}\:\theta}{\frac{\mathrm{7}}{\mathrm{5}}−\frac{{m}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{{M}+{m}}}} \\ $$$${when}\:{the}\:{ball}\:{is}\:{about}\:{to}\:{touch}\:{the} \\ $$$${ground},\:{s}=\frac{{h}}{\mathrm{sin}\:\theta}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$$\Rightarrow{v}=\sqrt{\frac{\mathrm{2}{g}\left({h}−\mathrm{2}{r}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)\:}{\frac{\mathrm{7}}{\mathrm{5}}−\frac{{m}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{{M}+{m}}}} \\ $$
Commented by ajfour last updated on 18/Oct/19
Commented by ajfour last updated on 18/Oct/19
    s=(h/(sin θ))−rtan (θ/2)
$$\:\:\:\:{s}=\frac{{h}}{\mathrm{sin}\:\theta}−{r}\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 18/Oct/19
mgssin θ=((MV^( 2) )/2)+(1/2)(((2mr^2 )/5))(v^2 /r^2 )             +(m/2)[(vcos θ−V)^2 +v^2 sin^2 θ]   &     mvcos θ=(M+m)V  ⇒ V=((mvcos θ)/(M+m))  &    vcos θ−V = ((Mvcos θ)/(M+m))  substituting in first eq.    mgssin θ=mv^2 ((1/5)+((sin^2 θ)/2))        +(M/2)(((mvcos θ)/(M+m)))^2 +(m/2)(((Mvcos θ)/(M+m)))^2   ⇒ mgssin θ=mv^2 ((1/5)+((sin^2 θ)/2))                                +((Mmv^2 cos^2 θ)/(2(M+m)))  ⇒  v^2 =((gssin θ)/((1/5)+((sin^2 θ)/2)+((Mcos^2 θ)/(2(M+m)))))     v = (√((gssin θ)/((1/5)+((sin^2 θ)/2)+((Mcos^2 θ)/(2(M+m))))))         v=(√((2gssin θ)/((7/5)−((mcos^2 θ)/(M+m)))))   And  with s=(h/(sin θ))−rtan (θ/2)    v=(√((g(h−rtan (θ/2)sin θ))/((1/5)+((sin^2 θ)/2)+((Mcos^2 θ)/(2(M+m)))))) .  vcos θ−V = ((Mvcos θ)/(M+m))    ⇒ v_(actual) ^2 =(((Mvcos θ)/(M+m)))^2 +(vsin θ)^2       v_(actual) =v(√(((M/(M+m)))^2 cos^2 θ+sin^2 θ))    this is speed of sphere    relative to wedge..
$${mgs}\mathrm{sin}\:\theta=\frac{{MV}^{\:\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}{mr}^{\mathrm{2}} }{\mathrm{5}}\right)\frac{{v}^{\mathrm{2}} }{{r}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\frac{{m}}{\mathrm{2}}\left[\left({v}\mathrm{cos}\:\theta−{V}\right)^{\mathrm{2}} +{v}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta\right] \\ $$$$\:\&\:\:\:\:\:{mv}\mathrm{cos}\:\theta=\left({M}+{m}\right){V} \\ $$$$\Rightarrow\:{V}=\frac{{mv}\mathrm{cos}\:\theta}{{M}+{m}} \\ $$$$\&\:\:\:\:{v}\mathrm{cos}\:\theta−{V}\:=\:\frac{{Mv}\mathrm{cos}\:\theta}{{M}+{m}} \\ $$$${substituting}\:{in}\:{first}\:{eq}. \\ $$$$\:\:{mgs}\mathrm{sin}\:\theta={mv}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:+\frac{{M}}{\mathrm{2}}\left(\frac{{mv}\mathrm{cos}\:\theta}{{M}+{m}}\right)^{\mathrm{2}} +\frac{{m}}{\mathrm{2}}\left(\frac{{Mv}\mathrm{cos}\:\theta}{{M}+{m}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{mgs}\mathrm{sin}\:\theta={mv}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{Mmv}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{2}\left({M}+{m}\right)} \\ $$$$\Rightarrow\:\:{v}^{\mathrm{2}} =\frac{{gs}\mathrm{sin}\:\theta}{\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}}+\frac{{M}\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{2}\left({M}+{m}\right)}} \\ $$$$\:\:\:{v}\:=\:\sqrt{\frac{{gs}\mathrm{sin}\:\theta}{\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}}+\frac{{M}\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{2}\left({M}+{m}\right)}}}\: \\ $$$$\:\:\:\:\:\:{v}=\sqrt{\frac{\mathrm{2}{gs}\mathrm{sin}\:\theta}{\frac{\mathrm{7}}{\mathrm{5}}−\frac{{m}\mathrm{cos}\:^{\mathrm{2}} \theta}{{M}+{m}}}}\: \\ $$$${And}\:\:{with}\:{s}=\frac{{h}}{\mathrm{sin}\:\theta}−{r}\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\:\:{v}=\sqrt{\frac{{g}\left({h}−{r}\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\mathrm{sin}\:\theta\right)}{\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}}+\frac{{M}\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{2}\left({M}+{m}\right)}}}\:. \\ $$$${v}\mathrm{cos}\:\theta−{V}\:=\:\frac{{Mv}\mathrm{cos}\:\theta}{{M}+{m}} \\ $$$$\:\:\Rightarrow\:{v}_{{actual}} ^{\mathrm{2}} =\left(\frac{{Mv}\mathrm{cos}\:\theta}{{M}+{m}}\right)^{\mathrm{2}} +\left({v}\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$\:\:\:\:{v}_{{actual}} ={v}\sqrt{\left(\frac{{M}}{{M}+{m}}\right)^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$\:\:{this}\:{is}\:{speed}\:{of}\:{sphere} \\ $$$$\:\:{relative}\:{to}\:{wedge}.. \\ $$$$ \\ $$
Commented by mr W last updated on 17/Oct/19
your method is more effective, very  nice! but i didn′t find where my error  is.
$${your}\:{method}\:{is}\:{more}\:{effective},\:{very} \\ $$$${nice}!\:{but}\:{i}\:{didn}'{t}\:{find}\:{where}\:{my}\:{error} \\ $$$${is}. \\ $$
Commented by ajfour last updated on 18/Oct/19
Nsin θ−fcos θ=Ma         ...(i)  N+masin θ=mgcos θ      ...(ii)  mgsin θ+macos θ=f+mαr    ...(iii)  rf=(2/5)mr^2 α                ....(iv)  ⇒ masin^2 θ+Ma+fcos θ             =mgsin θcos θ  ⇒ (M+msin^2 θ)(((f−mgsin θ+mαr)/(mcos θ)))         +fcos θ=mgsin θcos θ  f=((mgsin θcos θ+(((M+msin^2 θ)(gsin θ−αr))/(cos θ)))/(cos θ+((M+msin^2 θ)/(mcos θ))))     =((m[(M+m)gsin θ−αr(M+msin^2 θ)])/(M+m))    = (2/5)mrα        [from (iv)]   ⇒  αr[M+msin^2 θ+(2/5)(M+m)]            = (M+m)gsin θ  ⇒  αr = ((gsin θ)/((7/5)−((mcos^2 θ)/(M+m))))       ((vdv)/ds)=αr  ⇒ v=(√(2αrs))  ⇒  v=(√((2gssin θ)/((7/5)−((mcos^2 θ)/(M+m))))) .
$${N}\mathrm{sin}\:\theta−{f}\mathrm{cos}\:\theta={Ma}\:\:\:\:\:\:\:\:\:…\left({i}\right) \\ $$$${N}+{ma}\mathrm{sin}\:\theta={mg}\mathrm{cos}\:\theta\:\:\:\:\:\:…\left({ii}\right) \\ $$$${mg}\mathrm{sin}\:\theta+{ma}\mathrm{cos}\:\theta={f}+{m}\alpha{r}\:\:\:\:…\left({iii}\right) \\ $$$${rf}=\frac{\mathrm{2}}{\mathrm{5}}{mr}^{\mathrm{2}} \alpha\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({iv}\right) \\ $$$$\Rightarrow\:{ma}\mathrm{sin}\:^{\mathrm{2}} \theta+{Ma}+{f}\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={mg}\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\left({M}+{m}\mathrm{sin}\:^{\mathrm{2}} \theta\right)\left(\frac{{f}−{mg}\mathrm{sin}\:\theta+{m}\alpha{r}}{{m}\mathrm{cos}\:\theta}\right) \\ $$$$\:\:\:\:\:\:\:+{f}\mathrm{cos}\:\theta={mg}\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$${f}=\frac{{mg}\mathrm{sin}\:\theta\mathrm{cos}\:\theta+\frac{\left({M}+{m}\mathrm{sin}\:^{\mathrm{2}} \theta\right)\left({g}\mathrm{sin}\:\theta−\alpha{r}\right)}{\mathrm{cos}\:\theta}}{\mathrm{cos}\:\theta+\frac{{M}+{m}\mathrm{sin}\:^{\mathrm{2}} \theta}{{m}\mathrm{cos}\:\theta}} \\ $$$$\:\:\:=\frac{{m}\left[\left({M}+{m}\right){g}\mathrm{sin}\:\theta−\alpha{r}\left({M}+{m}\mathrm{sin}\:^{\mathrm{2}} \theta\right)\right]}{{M}+{m}} \\ $$$$\:\:=\:\frac{\mathrm{2}}{\mathrm{5}}{mr}\alpha\:\:\:\:\:\:\:\:\left[{from}\:\left({iv}\right)\right]\:\:\:\Rightarrow \\ $$$$\alpha{r}\left[{M}+{m}\mathrm{sin}\:^{\mathrm{2}} \theta+\frac{\mathrm{2}}{\mathrm{5}}\left({M}+{m}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\left({M}+{m}\right){g}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\alpha{r}\:=\:\frac{{g}\mathrm{sin}\:\theta}{\frac{\mathrm{7}}{\mathrm{5}}−\frac{{m}\mathrm{cos}\:^{\mathrm{2}} \theta}{{M}+{m}}}\:\: \\ $$$$\:\:\:\frac{{vdv}}{{ds}}=\alpha{r}\:\:\Rightarrow\:{v}=\sqrt{\mathrm{2}\alpha{rs}} \\ $$$$\Rightarrow\:\:{v}=\sqrt{\frac{\mathrm{2}{gs}\mathrm{sin}\:\theta}{\frac{\mathrm{7}}{\mathrm{5}}−\frac{{m}\mathrm{cos}\:^{\mathrm{2}} \theta}{{M}+{m}}}}\:. \\ $$

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