Question Number 96817 by peter frank last updated on 05/Jun/20
Commented by mr W last updated on 06/Jun/20
$${both}\:{results}\:{in}\:{question}\:{are}\:{wrong}! \\ $$$${please}\:{check}\:{if}\:{they}\:{are}\:{your}\:{typos}\:{or} \\ $$$${question}\:{in}\:{book}\:{is}\:{really}\:{wrong}. \\ $$
Commented by peter frank last updated on 07/Jun/20
$$\mathrm{yes}\:\mathrm{your}\:\mathrm{right}\:\mathrm{i}\:\mathrm{see}\: \\ $$
Answered by mr W last updated on 06/Jun/20
Commented by mr W last updated on 06/Jun/20
![(1/2)mv^2 =(1/2)mu_0 ^2 +mgr(1−cos α) ⇒v^2 =u_0 ^2 +2gr(1−cos α) mg cos α−N=((mv^2 )/r) ⇒N=m(g cos α−(v^2 /r))=m[g cos α−(u_0 ^2 /r)−2g(1−cos α)] ⇒N=m[(3 cos α−2)g−(u_0 ^2 /r)] when N=0, i.e. contact looses: (3 cos α−2)g−(u_0 ^2 /r)=0 ⇒cos α=(1/3)(2+(u_0 ^2 /(rg))) v^2 =u_0 ^2 +2gr(1−(2/3)−(u_0 ^2 /(3rg))) v^2 =u_0 ^2 +(2/3)(gr−u_0 ^2 ) v^2 =((u_0 ^2 +2gr)/3) ⇒v=(√((u_0 ^2 +2gr)/3))](Q97146.png)
$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mu}_{\mathrm{0}} ^{\mathrm{2}} +{mgr}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right) \\ $$$$\Rightarrow{v}^{\mathrm{2}} ={u}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}{gr}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right) \\ $$$${mg}\:\mathrm{cos}\:\alpha−{N}=\frac{{mv}^{\mathrm{2}} }{{r}} \\ $$$$\Rightarrow{N}={m}\left({g}\:\mathrm{cos}\:\alpha−\frac{{v}^{\mathrm{2}} }{{r}}\right)={m}\left[{g}\:\mathrm{cos}\:\alpha−\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} }{{r}}−\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)\right] \\ $$$$\Rightarrow{N}={m}\left[\left(\mathrm{3}\:\mathrm{cos}\:\alpha−\mathrm{2}\right){g}−\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} }{{r}}\right] \\ $$$$ \\ $$$${when}\:{N}=\mathrm{0},\:{i}.{e}.\:{contact}\:{looses}: \\ $$$$\left(\mathrm{3}\:\mathrm{cos}\:\alpha−\mathrm{2}\right){g}−\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} }{{r}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{2}+\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} }{{rg}}\right) \\ $$$${v}^{\mathrm{2}} ={u}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}{gr}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}−\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{3}{rg}}\right) \\ $$$${v}^{\mathrm{2}} ={u}_{\mathrm{0}} ^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}\left({gr}−{u}_{\mathrm{0}} ^{\mathrm{2}} \right) \\ $$$${v}^{\mathrm{2}} =\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}{gr}}{\mathrm{3}} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}{gr}}{\mathrm{3}}} \\ $$
Commented by peter frank last updated on 07/Jun/20
$$\mathrm{thank}\:\mathrm{you}\: \\ $$
Commented by peter frank last updated on 07/Jun/20
$$\mathrm{my}\:\mathrm{be}\:\mathrm{when}\:\mathrm{N}=\mathrm{0}\:\mathrm{the} \\ $$$$\mathrm{initil}\:\mathrm{velocity}\:\mathrm{u}_{\mathrm{o}} \:\mathrm{also}\:\mathrm{0} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}{gr}}{\mathrm{3}}} \\ $$$$\mathrm{u}=\mathrm{0} \\ $$$$\Rightarrow{v}=\sqrt{\frac{\mathrm{2}{gr}}{\mathrm{3}}} \\ $$$$ \\ $$