Question Number 226386 by fantastic2 last updated on 26/Nov/25
Commented by fantastic2 last updated on 27/Nov/25
$${l}=\mathrm{5}\pi{R} \\ $$$${a}\:{force}\:{F}\:{is}\:{given}\:{horizontally} \\ $$$${for}\:\mathrm{1}{sec}\:{such}\:{that}\:{the}\:{speed}\:{of}\:{the}\:{ball} \\ $$$${becomes}\:\mathrm{0}\:{just}\:{before}\:{touching}\:{the}\:{cylinder} \\ $$$${find}\:{the}\:{time}\:{taken}\:{to}\:{touch}\:{and}\:{the}\:{force} \\ $$
Commented by mr W last updated on 28/Nov/25
$${what}\:{do}\:{you}\:{mean}\:{with}\:“{just}\:{before} \\ $$$${touching}\:{the}\:{cylinder}''?\:{is}\:{it}\:{the} \\ $$$${instant}\:{when}\:{the}\:{string}\:{is}\:{getting} \\ $$$${completely}\:{wrapped}\:{around}\:{the}\: \\ $$$${cylinder}? \\ $$
Commented by fantastic2 last updated on 27/Nov/25
$${yes}\: \\ $$
Commented by mr W last updated on 28/Nov/25
$${that}\:{means}\:{the}\:{initial}\:{velocity}\:{of} \\ $$$${the}\:{mass}\:{must}\:{be}\:{large}\:{enoughso} \\ $$$${that}\:{the}\:{string}\:{keeps}\:{being}\:{taut}\:{all} \\ $$$${the}\:{time}.\:{then}\:{the}\:{velocity}\:{of}\:{the}\: \\ $$$${mass}\:{can}\:{never}\:{become}\:{zero}. \\ $$
Answered by mr W last updated on 28/Nov/25
Commented by mr W last updated on 29/Nov/25
![mV=Ft, t=1 sec. ⇒V=((Ft)/m) phase 2: b=L−Rθ ((mV^2 )/2)=((mu^2 )/2)+mg(L+R−R cos θ+b sin θ) u^2 =V^2 −2g[L+R−R(cos θ+θ sin θ)+L sin θ] mg sin θ+N=((mu^2 )/b) let λ=(L/R), =(V^( 2) /(gR)) (N/(mg))=(u^2 /(gb))−sin θ (N/(mg))=((V^2 −2g[L+R−R(cos θ+θ sin θ)+L sin θ])/(g(L−Rθ)))−sin θ (N/(mg))=(( −2(λ+1−cos θ−θ sin θ+λ sin θ))/(λ−θ))−sin θ (N/(mg))=(( −2(λ+1−cos θ))/(λ−θ))−3 sin θ such that the string can get wrapped around the cylinder completely, it must be kept taut all the time, i.e. for 0≤θ<λ: N≥0. for N_(min) : ((−2 sin θ)/(λ−θ))+(( −2(λ+1−cos θ))/((λ−θ)^2 ))−3 cos θ=0 ⇒(λ−θ)[2 sin θ+3(λ−θ)cos θ]+2(λ+1−cos θ)= for N_(min) =0: (λ−θ)[2 sin θ+3(λ−θ)cos θ]+2(λ+1−cos θ)=2(λ+1−cos θ)+3(λ−θ)sin θ ⇒θ_m +((tan θ_m )/3)=λ ⇒ =(V^2 /(gR))≥2(λ+1−cos θ_m )+3(λ−θ_m ) sin θ_m for λ=5π we get θ_m ≈1.54725≈88.65° approximately N_(min) is at θ_m ≈(π/2), ≥≈2(λ+1)+3(λ−(π/2))=5λ+2−((3π)/2) (u^2 /(gR))= −2[λ+1−cos θ+(λ−θ) sin θ] ⇒u=(√(gR{ −2[λ+1−cos θ+(λ−θ) sin θ]})) at θ=λ: u∣_(θ=λ) =(√(gR[ −2(λ+1−cos λ)])) this can never be zero. it′s clear, the mass can never reach a height such that the initial kinetical energy ((mV^2 )/2) completely transfered into potential energy. u=bω=(L−Rθ)(dθ/dt)=R(λ−θ)(dθ/dt) (√(gR{ −2[λ+1−cos θ+(λ−θ) sin θ]}))=R(λ−θ)(dθ/dt) dt=(√(R/g))(((λ−θ)dθ)/( (√( −2[λ+1−cos θ+(λ−θ) sin θ])))) ⇒T_2 =(√(R/g))∫_0 ^λ (((λ−θ)dθ)/( (√( −2[λ+1−cos θ+(λ−θ) sin θ])))) phase 1: u^2 +2gL(1−cos φ)=V^2 u=Lω=L(dφ/dt)=(√(V^2 −2gL(1−cos φ))) (dφ/dt)=(√((g/L)((V^2 /(gL))−2+2 cos φ))) dt=(√(L/g))(dφ/( (√(( /λ)−2+2 cos φ)))) ⇒T_1 =(√(L/g))∫_0 ^(π/2) (dφ/( (√(( /λ)−2+2 cos φ)))) total time needed: T=T_1 +T_2](https://www.tinkutara.com/question/Q226437.png)
$${mV}={Ft},\:{t}=\mathrm{1}\:{sec}. \\ $$$$\Rightarrow{V}=\frac{{Ft}}{{m}} \\ $$$$\underline{{phase}\:\mathrm{2}:} \\ $$$${b}={L}−{R}\theta \\ $$$$\frac{{mV}^{\mathrm{2}} }{\mathrm{2}}=\frac{{mu}^{\mathrm{2}} }{\mathrm{2}}+{mg}\left({L}+{R}−{R}\:\mathrm{cos}\:\theta+{b}\:\mathrm{sin}\:\theta\right) \\ $$$${u}^{\mathrm{2}} ={V}^{\mathrm{2}} −\mathrm{2}{g}\left[{L}+{R}−{R}\left(\mathrm{cos}\:\theta+\theta\:\mathrm{sin}\:\theta\right)+{L}\:\mathrm{sin}\:\theta\right] \\ $$$${mg}\:\mathrm{sin}\:\theta+{N}=\frac{{mu}^{\mathrm{2}} }{{b}} \\ $$$${let}\:\lambda=\frac{{L}}{{R}},\:\:=\frac{{V}^{\:\mathrm{2}} }{{gR}} \\ $$$$\frac{{N}}{{mg}}=\frac{{u}^{\mathrm{2}} }{{gb}}−\mathrm{sin}\:\theta \\ $$$$\frac{{N}}{{mg}}=\frac{{V}^{\mathrm{2}} −\mathrm{2}{g}\left[{L}+{R}−{R}\left(\mathrm{cos}\:\theta+\theta\:\mathrm{sin}\:\theta\right)+{L}\:\mathrm{sin}\:\theta\right]}{{g}\left({L}−{R}\theta\right)}−\mathrm{sin}\:\theta \\ $$$$\frac{{N}}{{mg}}=\frac{\:−\mathrm{2}\left(\lambda+\mathrm{1}−\mathrm{cos}\:\theta−\theta\:\mathrm{sin}\:\theta+\lambda\:\mathrm{sin}\:\theta\right)}{\lambda−\theta}−\mathrm{sin}\:\theta \\ $$$$\frac{{N}}{{mg}}=\frac{\:−\mathrm{2}\left(\lambda+\mathrm{1}−\mathrm{cos}\:\theta\right)}{\lambda−\theta}−\mathrm{3}\:\mathrm{sin}\:\theta \\ $$$${such}\:{that}\:{the}\:{string}\:{can}\:{get}\:{wrapped} \\ $$$${around}\:{the}\:{cylinder}\:{completely},\:{it}\: \\ $$$${must}\:{be}\:{kept}\:{taut}\:{all}\:{the}\:{time},\:{i}.{e}.\: \\ $$$${for}\:\mathrm{0}\leqslant\theta<\lambda:\:{N}\geqslant\mathrm{0}. \\ $$$${for}\:{N}_{{min}} : \\ $$$$\frac{−\mathrm{2}\:\mathrm{sin}\:\theta}{\lambda−\theta}+\frac{\:−\mathrm{2}\left(\lambda+\mathrm{1}−\mathrm{cos}\:\theta\right)}{\left(\lambda−\theta\right)^{\mathrm{2}} }−\mathrm{3}\:\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\left(\lambda−\theta\right)\left[\mathrm{2}\:\mathrm{sin}\:\theta+\mathrm{3}\left(\lambda−\theta\right)\mathrm{cos}\:\theta\right]+\mathrm{2}\left(\lambda+\mathrm{1}−\mathrm{cos}\:\theta\right)=\: \\ $$$${for}\:{N}_{{min}} =\mathrm{0}: \\ $$$$\left(\lambda−\theta\right)\left[\mathrm{2}\:\mathrm{sin}\:\theta+\mathrm{3}\left(\lambda−\theta\right)\mathrm{cos}\:\theta\right]+\mathrm{2}\left(\lambda+\mathrm{1}−\mathrm{cos}\:\theta\right)=\mathrm{2}\left(\lambda+\mathrm{1}−\mathrm{cos}\:\theta\right)+\mathrm{3}\left(\lambda−\theta\right)\mathrm{sin}\:\theta \\ $$$$\Rightarrow\theta_{{m}} +\frac{\mathrm{tan}\:\theta_{{m}} }{\mathrm{3}}=\lambda \\ $$$$\Rightarrow\:=\frac{{V}^{\mathrm{2}} }{{gR}}\geqslant\mathrm{2}\left(\lambda+\mathrm{1}−\mathrm{cos}\:\theta_{{m}} \right)+\mathrm{3}\left(\lambda−\theta_{{m}} \right)\:\mathrm{sin}\:\theta_{{m}} \\ $$$${for}\:\lambda=\mathrm{5}\pi\:{we}\:{get}\:\theta_{{m}} \approx\mathrm{1}.\mathrm{54725}\approx\mathrm{88}.\mathrm{65}° \\ $$$${approximately}\:{N}_{{min}} \:{is}\:{at}\:\theta_{{m}} \approx\frac{\pi}{\mathrm{2}}, \\ $$$$\:\geqslant\approx\mathrm{2}\left(\lambda+\mathrm{1}\right)+\mathrm{3}\left(\lambda−\frac{\pi}{\mathrm{2}}\right)=\mathrm{5}\lambda+\mathrm{2}−\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$ \\ $$$$\frac{{u}^{\mathrm{2}} }{{gR}}=\:−\mathrm{2}\left[\lambda+\mathrm{1}−\mathrm{cos}\:\theta+\left(\lambda−\theta\right)\:\mathrm{sin}\:\theta\right] \\ $$$$\Rightarrow{u}=\sqrt{{gR}\left\{\:−\mathrm{2}\left[\lambda+\mathrm{1}−\mathrm{cos}\:\theta+\left(\lambda−\theta\right)\:\mathrm{sin}\:\theta\right]\right\}} \\ $$$${at}\:\theta=\lambda: \\ $$$${u}\mid_{\theta=\lambda} =\sqrt{{gR}\left[\:−\mathrm{2}\left(\lambda+\mathrm{1}−\mathrm{cos}\:\lambda\right)\right]} \\ $$$${this}\:{can}\:{never}\:{be}\:{zero}. \\ $$$${it}'{s}\:{clear},\:{the}\:{mass}\:{can}\:{never}\:{reach} \\ $$$${a}\:{height}\:{such}\:{that}\:{the}\:{initial}\: \\ $$$${kinetical}\:{energy}\:\frac{{mV}^{\mathrm{2}} }{\mathrm{2}}\:{completely}\: \\ $$$${transfered}\:{into}\:{potential}\:{energy}. \\ $$$$ \\ $$$${u}={b}\omega=\left({L}−{R}\theta\right)\frac{{d}\theta}{{dt}}={R}\left(\lambda−\theta\right)\frac{{d}\theta}{{dt}} \\ $$$$\sqrt{{gR}\left\{\:−\mathrm{2}\left[\lambda+\mathrm{1}−\mathrm{cos}\:\theta+\left(\lambda−\theta\right)\:\mathrm{sin}\:\theta\right]\right\}}={R}\left(\lambda−\theta\right)\frac{{d}\theta}{{dt}} \\ $$$${dt}=\sqrt{\frac{{R}}{{g}}}\frac{\left(\lambda−\theta\right){d}\theta}{\:\sqrt{\:−\mathrm{2}\left[\lambda+\mathrm{1}−\mathrm{cos}\:\theta+\left(\lambda−\theta\right)\:\mathrm{sin}\:\theta\right]}} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\sqrt{\frac{{R}}{{g}}}\int_{\mathrm{0}} ^{\lambda} \frac{\left(\lambda−\theta\right){d}\theta}{\:\sqrt{\:−\mathrm{2}\left[\lambda+\mathrm{1}−\mathrm{cos}\:\theta+\left(\lambda−\theta\right)\:\mathrm{sin}\:\theta\right]}} \\ $$$$ \\ $$$$\underline{{phase}\:\mathrm{1}:} \\ $$$${u}^{\mathrm{2}} +\mathrm{2}{gL}\left(\mathrm{1}−\mathrm{cos}\:\phi\right)={V}^{\mathrm{2}} \\ $$$${u}={L}\omega={L}\frac{{d}\phi}{{dt}}=\sqrt{{V}^{\mathrm{2}} −\mathrm{2}{gL}\left(\mathrm{1}−\mathrm{cos}\:\phi\right)} \\ $$$$\frac{{d}\phi}{{dt}}=\sqrt{\frac{{g}}{{L}}\left(\frac{{V}^{\mathrm{2}} }{{gL}}−\mathrm{2}+\mathrm{2}\:\mathrm{cos}\:\phi\right)} \\ $$$${dt}=\sqrt{\frac{{L}}{{g}}}\frac{{d}\phi}{\:\sqrt{\frac{\:}{\lambda}−\mathrm{2}+\mathrm{2}\:\mathrm{cos}\:\phi}} \\ $$$$\Rightarrow{T}_{\mathrm{1}} =\sqrt{\frac{{L}}{{g}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\phi}{\:\sqrt{\frac{\:}{\lambda}−\mathrm{2}+\mathrm{2}\:\mathrm{cos}\:\phi}} \\ $$$${total}\:{time}\:{needed}: \\ $$$${T}={T}_{\mathrm{1}} +{T}_{\mathrm{2}} \\ $$
Commented by mr W last updated on 29/Nov/25
Commented by fantastic2 last updated on 28/Nov/25
$${sir}\:{please}\:{explain}\:{why}\:{you}\:{took} \\ $$$$\lambda\:{as}\:{upper}\:{limit}. \\ $$$${others}\:{are}\:{crystal}\:{clear}.{thank}\:{you}. \\ $$
Commented by mr W last updated on 28/Nov/25
$$\lambda=\frac{{L}}{{R}} \\ $$$$\theta=\lambda\:{means}\:{the}\:{instant}\:{as}\:{the} \\ $$$${string}\:{is}\:{completely}\:{wrapped}. \\ $$$${you}\:{can}\:{not}\:{go}\:{further}! \\ $$
Commented by fantastic2 last updated on 28/Nov/25
$${thanks}\:{sir}! \\ $$$${i}\:{appretiate}\:{your}\:{time} \\ $$