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Question-187640

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Question Number 187640 by ajfour last updated on 19/Feb/23
Commented by ajfour last updated on 19/Feb/23
If the hemisphere radius R. The ball radius b and mass m. Find Tension in string in terms of mg, θ, b, R.
$${If}\:{the}\:{hemisphere}\:{radius}\:{R}.\:{The} \\ $$$${ball}\:{radius}\:{b}\:{and}\:{mass}\:{m}.\:{Find}\: \\ $$$${Tension}\:{in}\:{string}\:{in}\:{terms}\:{of}\:{mg}, \\ $$$$\theta,\:{b},\:{R}. \\ $$
Answered by a.lgnaoui last updated on 19/Feb/23
tan θ=((ma_c )/(mg))=(v^2 /(Rg)) v=R(dθ/dt) T { ((Tx=Tcos θ=−mg)),((Ty=Tsin θ=ma_c )) :} T=m(√(g^2 +R((dθ/dt) )^2 )) (suite) η=(m/V)=(m/(4/3πb^3 )) T=(4/3)πb^3 η(√(g^2 +R((dθ/dt))^2 )) η=masse volumique
$$\mathrm{tan}\:\theta=\frac{{ma}_{{c}} }{{mg}}=\frac{{v}^{\mathrm{2}} }{{Rg}}\:\:\:\:\:{v}={R}\frac{{d}\theta}{{dt}} \\ $$$${T\begin{cases}{{Tx}={T}\mathrm{cos}\:\theta=−{mg}}\\{{Ty}={T}\mathrm{sin}\:\theta={ma}_{{c}} }\end{cases}} \\ $$$$\:\:\:\:\:\:\:\:{T}={m}\sqrt{{g}^{\mathrm{2}} +{R}\left(\frac{{d}\theta}{{dt}}\:\right)^{\mathrm{2}} } \\ $$$$\left({suite}\right) \\ $$$$\eta=\frac{{m}}{{V}}=\frac{{m}}{\mathrm{4}/\mathrm{3}\pi{b}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:{T}=\frac{\mathrm{4}}{\mathrm{3}}\pi{b}^{\mathrm{3}} \eta\sqrt{{g}^{\mathrm{2}} +{R}\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\eta={masse}\:{volumique}\: \\ $$
Answered by mr W last updated on 20/Feb/23
Commented by mr W last updated on 20/Feb/23
cos φ=(R/(R+b)) T cos φ=mg cos θ ⇒T=((mg cos θ)/(cos φ))=(1+(b/R))mg cos θ ✓
$$\mathrm{cos}\:\phi=\frac{{R}}{{R}+{b}} \\ $$$${T}\:\mathrm{cos}\:\phi={mg}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{T}=\frac{{mg}\:\mathrm{cos}\:\theta}{\mathrm{cos}\:\phi}=\left(\mathrm{1}+\frac{{b}}{{R}}\right){mg}\:\mathrm{cos}\:\theta\:\checkmark \\ $$
Commented by ajfour last updated on 20/Feb/23
very fine, thanks sir! What if string be attached to the surface of the ball (and could be heavy as well) ?
$${very}\:{fine},\:{thanks}\:{sir}! \\ $$$${What}\:{if}\:{string}\:{be}\:{attached}\:{to} \\ $$$${the}\:{surface}\:{of}\:{the}\:{ball}\:\left({and}\:{could}\right. \\ $$$$\left.{be}\:{heavy}\:{as}\:{well}\right)\:?\: \\ $$
Commented by mr W last updated on 20/Feb/23
if string is weightless, then it makes no difference whether the string is attached to the surface or to the center of the ball, the tension of string passes always the center of the ball where N and mg intersect.
$${if}\:{string}\:{is}\:{weightless},\:{then}\:{it}\:{makes} \\ $$$${no}\:{difference}\:{whether}\:{the}\:{string}\:{is} \\ $$$${attached}\:{to}\:{the}\:{surface}\:{or}\:{to}\:{the} \\ $$$${center}\:{of}\:{the}\:{ball},\:{the}\:{tension}\:{of} \\ $$$${string}\:{passes}\:{always}\:{the}\:{center}\:{of}\: \\ $$$${the}\:{ball}\:{where}\:{N}\:{and}\:{mg}\:{intersect}. \\ $$
Commented by ajfour last updated on 20/Feb/23
yes sir without friction and light string the solution is complete.
$${yes}\:{sir}\:{without}\:{friction}\:{and}\:{light}\: \\ $$$${string}\:{the}\:{solution}\:{is}\:{complete}. \\ $$
Commented by mr W last updated on 20/Feb/23
if the string has uniform mass ρ, then we have following situation: (BC=hanging string)
$${if}\:{the}\:{string}\:{has}\:{uniform}\:{mass}\:\rho, \\ $$$${then}\:{we}\:{have}\:{following}\:{situation}: \\ $$$$\left({BC}={hanging}\:{string}\right) \\ $$
Commented by mr W last updated on 20/Feb/23
Commented by mr W last updated on 20/Feb/23
Commented by mr W last updated on 20/Feb/23
Tsin φ=mg cos θ ⇒T=((mg cos θ)/(sin φ)) T_0 =T cos (φ−θ) a=(T_0 /(ρg))=((mg cos θ cos (φ−θ))/(ρg sin φ)) =(m/(2ρ))(((1+cos 2θ)/(tan φ))+sin 2θ) in xy−system with origin at center of hemisphere: x_B =R cos (ϕ+θ) y_B =R sin (ϕ+θ) x_C =(R+b) cos θ−b cos (φ−θ) y_C =(R+b) sin θ+b sin (φ−θ) ⇒x_C −x_B =(R+b) cos θ−b cos (φ−θ)−R cos (ϕ+θ) ⇒y_B −y_C =R sin (ϕ+θ)−(R+b) sin θ−b sin (φ−θ) hanging string BC is a segment of catenary with following equation in xy−system as shown: y=a cosh (x/a) at point C: tan (φ−θ)=sinh (x_C /a) (x_C /a)=sinh^(−1) (tan (φ−θ) y_C =a cosh (x_C /a) ⇒y_C =a (√(1+tan^2 (φ−θ))) at point B: (1/(tan (ϕ+θ)))=sinh (x_B /a) ⇒(x_B /a)=sinh^(−1) (1/(tan (ϕ+θ))) y_B =a cosh (x_B /a) ⇒y_B =a (√(1+(1/(tan^2 (ϕ+θ))))) x_B −x_C =a[sinh^(−1) (1/(tan (ϕ+θ)))−sinh^(−1) (tan (φ−θ))] ⇒(1+(b/R)) cos θ−(b/R) cos (φ−θ)−cos (ϕ+θ)=(m/(2Rρ))(((1+cos 2θ)/(tan φ))+sin 2θ)[sinh^(−1) (1/(tan (ϕ+θ)))−sinh^(−1) (tan (φ−θ))] ...(i) y_B −y_C =a[(√(1+(1/(tan^2 (ϕ+θ)))))−(√(1+tan^2 (φ−θ)))] ⇒sin (ϕ+θ)−(1+(b/R)) sin θ−(b/R) sin (φ−θ)=(m/(2Rρ))(((1+cos 2θ)/(tan φ))+sin 2θ)[(√(1+(1/(tan^2 (ϕ+θ)))))−(√(1+tan^2 (φ−θ)))] ...(ii) we can solve (i) and (ii) for ϕ and φ.
$${T}\mathrm{sin}\:\phi={mg}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{T}=\frac{{mg}\:\mathrm{cos}\:\theta}{\mathrm{sin}\:\phi} \\ $$$${T}_{\mathrm{0}} ={T}\:\mathrm{cos}\:\left(\phi−\theta\right) \\ $$$${a}=\frac{{T}_{\mathrm{0}} }{\rho{g}}=\frac{{mg}\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\left(\phi−\theta\right)}{\rho{g}\:\mathrm{sin}\:\phi} \\ $$$$\:\:\:=\frac{{m}}{\mathrm{2}\rho}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{tan}\:\phi}+\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$${in}\:{xy}−{system}\:{with}\:{origin}\:{at}\:{center} \\ $$$${of}\:{hemisphere}: \\ $$$${x}_{{B}} ={R}\:\mathrm{cos}\:\left(\varphi+\theta\right) \\ $$$${y}_{{B}} ={R}\:\mathrm{sin}\:\left(\varphi+\theta\right) \\ $$$${x}_{{C}} =\left({R}+{b}\right)\:\mathrm{cos}\:\theta−{b}\:\mathrm{cos}\:\left(\phi−\theta\right) \\ $$$${y}_{{C}} =\left({R}+{b}\right)\:\mathrm{sin}\:\theta+{b}\:\mathrm{sin}\:\left(\phi−\theta\right) \\ $$$$\Rightarrow{x}_{{C}} −{x}_{{B}} =\left({R}+{b}\right)\:\mathrm{cos}\:\theta−{b}\:\mathrm{cos}\:\left(\phi−\theta\right)−{R}\:\mathrm{cos}\:\left(\varphi+\theta\right) \\ $$$$\Rightarrow{y}_{{B}} −{y}_{{C}} ={R}\:\mathrm{sin}\:\left(\varphi+\theta\right)−\left({R}+{b}\right)\:\mathrm{sin}\:\theta−{b}\:\mathrm{sin}\:\left(\phi−\theta\right) \\ $$$$ \\ $$$${hanging}\:{string}\:{BC}\:{is}\:{a}\:{segment}\:{of} \\ $$$${catenary}\:{with}\:{following}\:{equation}\: \\ $$$${in}\:{xy}−{system}\:{as}\:{shown}: \\ $$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${at}\:{point}\:{C}: \\ $$$$\mathrm{tan}\:\left(\phi−\theta\right)=\mathrm{sinh}\:\frac{{x}_{{C}} }{{a}} \\ $$$$\frac{{x}_{{C}} }{{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\left(\phi−\theta\right)\right. \\ $$$${y}_{{C}} ={a}\:\mathrm{cosh}\:\frac{{x}_{{C}} }{{a}} \\ $$$$\Rightarrow{y}_{{C}} ={a}\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\phi−\theta\right)} \\ $$$${at}\:{point}\:{B}: \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\left(\varphi+\theta\right)}=\mathrm{sinh}\:\frac{{x}_{{B}} }{{a}} \\ $$$$\Rightarrow\frac{{x}_{{B}} }{{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{tan}\:\left(\varphi+\theta\right)} \\ $$$${y}_{{B}} ={a}\:\mathrm{cosh}\:\frac{{x}_{{B}} }{{a}} \\ $$$$\Rightarrow{y}_{{B}} ={a}\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\varphi+\theta\right)}} \\ $$$${x}_{{B}} −{x}_{{C}} ={a}\left[\mathrm{sinh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{tan}\:\left(\varphi+\theta\right)}−\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\left(\phi−\theta\right)\right)\right] \\ $$$$\Rightarrow\left(\mathrm{1}+\frac{{b}}{{R}}\right)\:\mathrm{cos}\:\theta−\frac{{b}}{{R}}\:\mathrm{cos}\:\left(\phi−\theta\right)−\mathrm{cos}\:\left(\varphi+\theta\right)=\frac{{m}}{\mathrm{2}{R}\rho}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{tan}\:\phi}+\mathrm{sin}\:\mathrm{2}\theta\right)\left[\mathrm{sinh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{tan}\:\left(\varphi+\theta\right)}−\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\left(\phi−\theta\right)\right)\right]\:\:\:…\left({i}\right) \\ $$$${y}_{{B}} −{y}_{{C}} ={a}\left[\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\varphi+\theta\right)}}−\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\phi−\theta\right)}\right] \\ $$$$\Rightarrow\mathrm{sin}\:\left(\varphi+\theta\right)−\left(\mathrm{1}+\frac{{b}}{{R}}\right)\:\mathrm{sin}\:\theta−\frac{{b}}{{R}}\:\mathrm{sin}\:\left(\phi−\theta\right)=\frac{{m}}{\mathrm{2}{R}\rho}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{tan}\:\phi}+\mathrm{sin}\:\mathrm{2}\theta\right)\left[\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\varphi+\theta\right)}}−\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\phi−\theta\right)}\right]\:\:\:…\left({ii}\right) \\ $$$${we}\:{can}\:{solve}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{for}\:\varphi\:{and}\:\phi. \\ $$

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