Menu Close

Question-225713

Tinku Tara 0 Comments
Pin




Question Number 225713 by mr W last updated on 07/Nov/25
Commented by mr W last updated on 07/Nov/25
Answered by mahdipoor last updated on 08/Nov/25
in θ (from vertical axis): dN+dW+T_θ −T_(θ+dθ) =adm ⇒ T_θ −(T_θ +(dT/dθ)dθ)=dT=−dT(−cosθ,sinθ) ⇒ dN(sinθ,cosθ)+gdm(0,−1)−dT(−cosθ,sinθ)= (rα(cosθ,−sinθ)+rω^2 (−sinθ,cosθ))dm at relesed time ⇒ ω=0 , α=0 ((−sinθdN)/(−cosθdN))=((cosθdT)/(−gdm−sinθdT)) ⇒(sin^2 θ+cos^2 θ)dT=−gsinθdm ⇒(dT/dm)=−g.sinθ⇒dm=λrdθ ⇒(dT/dθ)=−λgsinθ⇒T(θ)=(T_0 −gλ)+gλcosθ T(0)=0⇒T=gλ(cosθ−1) T_(max) =T(θ_(max) )=end of rope
$$\mathrm{in}\:\theta\:\left(\mathrm{from}\:\mathrm{vertical}\:\mathrm{axis}\right): \\ $$$$\mathrm{d}\boldsymbol{\mathrm{N}}+\mathrm{d}\boldsymbol{\mathrm{W}}+\boldsymbol{\mathrm{T}}_{\theta} −\boldsymbol{\mathrm{T}}_{\theta+\mathrm{d}\theta} =\boldsymbol{\mathrm{a}}\mathrm{dm} \\ $$$$\Rightarrow \\ $$$$\boldsymbol{\mathrm{T}}_{\theta} −\left(\boldsymbol{\mathrm{T}}_{\theta} +\frac{\mathrm{d}\boldsymbol{\mathrm{T}}}{\mathrm{d}\theta}\mathrm{d}\theta\right)=\mathrm{d}\boldsymbol{\mathrm{T}}=−\mathrm{dT}\left(−\mathrm{cos}\theta,\mathrm{sin}\theta\right) \\ $$$$\Rightarrow \\ $$$$\mathrm{dN}\left(\mathrm{sin}\theta,\mathrm{cos}\theta\right)+\mathrm{gdm}\left(\mathrm{0},−\mathrm{1}\right)−\mathrm{dT}\left(−\mathrm{cos}\theta,\mathrm{sin}\theta\right)= \\ $$$$\left(\mathrm{r}\alpha\left(\mathrm{cos}\theta,−\mathrm{sin}\theta\right)+\mathrm{r}\omega^{\mathrm{2}} \left(−\mathrm{sin}\theta,\mathrm{cos}\theta\right)\right)\mathrm{dm} \\ $$$$\mathrm{at}\:\mathrm{relesed}\:\mathrm{time}\:\Rightarrow\:\omega=\mathrm{0}\:,\:\alpha=\mathrm{0} \\ $$$$\frac{−\mathrm{sin}\theta\mathrm{dN}}{−\mathrm{cos}\theta\mathrm{dN}}=\frac{\mathrm{cos}\theta\mathrm{dT}}{−\mathrm{gdm}−\mathrm{sin}\theta\mathrm{dT}} \\ $$$$\Rightarrow\left(\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta\right)\mathrm{dT}=−\mathrm{gsin}\theta\mathrm{dm} \\ $$$$\Rightarrow\frac{\mathrm{dT}}{\mathrm{dm}}=−\mathrm{g}.\mathrm{sin}\theta\Rightarrow\mathrm{dm}=\lambda\mathrm{rd}\theta \\ $$$$\Rightarrow\frac{\mathrm{dT}}{\mathrm{d}\theta}=−\lambda\mathrm{gsin}\theta\Rightarrow\mathrm{T}\left(\theta\right)=\left(\mathrm{T}_{\mathrm{0}} −\mathrm{g}\lambda\right)+\mathrm{g}\lambda\mathrm{cos}\theta \\ $$$$\mathrm{T}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow\mathrm{T}=\mathrm{g}\lambda\left(\mathrm{cos}\theta−\mathrm{1}\right) \\ $$$$\mathrm{T}_{\mathrm{max}} \:=\mathrm{T}\left(\theta_{\mathrm{max}} \right)=\mathrm{end}\:\mathrm{of}\:\mathrm{rope} \\ $$
Commented by mahdipoor last updated on 08/Nov/25
i think my answer is not true ....
$$\mathrm{i}\:\mathrm{think}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{not}\:\mathrm{true}\:…. \\ $$
Answered by fantastic last updated on 08/Nov/25
Commented by fantastic last updated on 08/Nov/25
angle is measured from verical F_(net on l) =∫_0 ^φ dmgsin α[rφ=l⇒φ=(l/r)] let mas−length density=λ dm=λdl=λrdα F_(net on l) =λrg[−cos α]_0 ^φ =λrg(1−cos ((l/r))) a=(F_(net on l) /(λl))=((rg(1−cos ((l/r))))/l) in dl in the left end let us say T is acting ↖ in the right T+dT is acting↘ so T+dT+dmgsin α−T=dma (dT/dm)=(a−gsin α) at T max (dT/dm)=0 ⇒a=gsin α α=sin^(−1) (((r(1−cos ((l/r))))/l))
$${angle}\:{is}\:{measured}\:{from}\:{verical} \\ $$$${F}_{{net}\:{on}\:{l}} =\int_{\mathrm{0}} ^{\phi} {dmg}\mathrm{sin}\:\alpha\left[{r}\phi={l}\Rightarrow\phi=\frac{{l}}{{r}}\right] \\ $$$${let}\:{mas}−{length}\:{density}=\lambda \\ $$$${dm}=\lambda{dl}=\lambda{rd}\alpha \\ $$$${F}_{{net}\:{on}\:{l}} =\lambda{rg}\left[−\mathrm{cos}\:\alpha\right]_{\mathrm{0}} ^{\phi} =\lambda{rg}\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{{l}}{{r}}\right)\right) \\ $$$${a}=\frac{{F}_{{net}\:{on}\:{l}} }{\lambda{l}}=\frac{{rg}\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{{l}}{{r}}\right)\right)}{{l}} \\ $$$${in}\:{dl}\:{in}\:{the}\:{left}\:{end}\:{let}\:{us} \\ $$$${say}\:{T}\:{is}\:{acting}\:\nwarrow \\ $$$${in}\:{the}\:{right}\:{T}+{dT}\:{is}\:{acting}\searrow \\ $$$${so} \\ $$$${T}+{dT}+{dmg}\mathrm{sin}\:\alpha−{T}={dma} \\ $$$$\frac{{dT}}{{dm}}=\left({a}−{g}\mathrm{sin}\:\alpha\right) \\ $$$${at}\:{T}\:{max}\:\frac{{dT}}{{dm}}=\mathrm{0} \\ $$$$\Rightarrow{a}={g}\mathrm{sin}\:\alpha \\ $$$$\alpha=\mathrm{sin}^{−\mathrm{1}} \left(\frac{{r}\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{{l}}{{r}}\right)\right)}{{l}}\right) \\ $$
Commented by mr W last updated on 09/Nov/25
correct!
$${correct}! \\ $$
Answered by mr W last updated on 09/Nov/25
Commented by mr W last updated on 09/Nov/25
at first let′s look at the motion of the whole rope. the whole rope can be treated as a solid object in the shape of a circular arc segment which rotates about O. say mass of rope is M. ρ=(M/l) φ=(l/r) ≤(π/2) I=Mr^2 at t=0: ω=0, v=0 Iα=τ with τ=∫_0 ^φ r sin θ×ρgrdθ=ρgr^2 ∫_0 ^φ sin θ dθ =ρgr^2 (1−cos φ)=ρgr^2 (1−cos (l/r)) Mr^2 α=(M/l)gr^2 (1−cos (l/r)) ⇒α=(g/l)(1−cos (l/r)) ⇒a=αr=((gr)/l)(1−cos (l/r))
$${at}\:{first}\:{let}'{s}\:{look}\:{at}\:{the}\:{motion}\:{of}\: \\ $$$${the}\:{whole}\:{rope}. \\ $$$${the}\:{whole}\:{rope}\:{can}\:{be}\:{treated}\:{as}\:{a} \\ $$$${solid}\:{object}\:{in}\:{the}\:{shape}\:{of}\:{a}\:{circular} \\ $$$${arc}\:{segment}\:{which}\:{rotates}\:{about}\:{O}. \\ $$$${say}\:{mass}\:{of}\:{rope}\:{is}\:{M}. \\ $$$$\rho=\frac{{M}}{{l}} \\ $$$$\phi=\frac{{l}}{{r}}\:\leqslant\frac{\pi}{\mathrm{2}} \\ $$$${I}={Mr}^{\mathrm{2}} \\ $$$${at}\:{t}=\mathrm{0}:\:\omega=\mathrm{0},\:{v}=\mathrm{0} \\ $$$${I}\alpha=\tau \\ $$$${with} \\ $$$$\tau=\int_{\mathrm{0}} ^{\phi} {r}\:\mathrm{sin}\:\theta×\rho{grd}\theta=\rho{gr}^{\mathrm{2}} \int_{\mathrm{0}} ^{\phi} \mathrm{sin}\:\theta\:{d}\theta \\ $$$$\:\:\:=\rho{gr}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\phi\right)=\rho{gr}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\frac{{l}}{{r}}\right) \\ $$$${Mr}^{\mathrm{2}} \alpha=\frac{{M}}{{l}}{gr}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\frac{{l}}{{r}}\right) \\ $$$$\Rightarrow\alpha=\frac{{g}}{{l}}\left(\mathrm{1}−\mathrm{cos}\:\frac{{l}}{{r}}\right) \\ $$$$\Rightarrow{a}=\alpha{r}=\frac{{gr}}{{l}}\left(\mathrm{1}−\mathrm{cos}\:\frac{{l}}{{r}}\right) \\ $$
Commented by mr W last updated on 09/Nov/25
Commented by mr W last updated on 09/Nov/25
now let′s look at the motion of a small piece of rope. it is the same as the motion of a particle on a incline with angle θ. m=ρrdθ ma=dT+ρgrdθ×sin θ aρrdθ=dT+ρgrdθ×sin θ dT=ρr(a−g sin θ)dθ ∫_0 ^T dT=ρr∫_0 ^θ (a−g sin θ)dθ T=ρr[aθ−g(1−cos θ)] T=((Mgr)/l)[((r(1−cos φ))/l)θ−1+cos θ] ⇒T=((Mg)/φ)[(((1−cos φ))/φ)θ−1+cos θ] at θ=φ: T_φ =((Mg)/φ)[(((1−cos φ))/φ)×φ−1+cos φ] =0 ✓ for maximum T: (dT/dθ)=0 ((Mg)/φ)[((1−cos φ)/φ)−sin θ]=0 ⇒sin θ=((1−cos φ)/φ) i.e. θ=sin^(−1) ((1−cos φ)/φ) and T_(max) =((Mg)/φ)[((1−cos φ)/φ) sin^(−1) ((1−cos φ)/φ)−1+(√(1−(((1−cos φ)/φ))^2 ))] examples: (l/r)=(π/4) (45°): θ≈21.9°, T_(max) ≈0.09 Mg (l/r)=(π/3) (60°): θ≈28.5°, T_(max) ≈0.11 Mg (l/r)=(π/2) (90°): θ≈39.5°, T_(max) ≈0.13 Mg we see the maximum tension in rope occurs always a little bit above the middle of the rope.
$${now}\:{let}'{s}\:{look}\:{at}\:{the}\:{motion}\:{of}\:{a} \\ $$$${small}\:{piece}\:{of}\:{rope}.\:{it}\:{is}\:{the}\:{same} \\ $$$${as}\:{the}\:{motion}\:{of}\:{a}\:{particle}\:{on}\:{a} \\ $$$${incline}\:{with}\:{angle}\:\theta. \\ $$$${m}=\rho{rd}\theta \\ $$$${ma}={dT}+\rho{grd}\theta×\mathrm{sin}\:\theta \\ $$$${a}\rho{rd}\theta={dT}+\rho{grd}\theta×\mathrm{sin}\:\theta \\ $$$${dT}=\rho{r}\left({a}−{g}\:\mathrm{sin}\:\theta\right){d}\theta \\ $$$$\int_{\mathrm{0}} ^{{T}} {dT}=\rho{r}\int_{\mathrm{0}} ^{\theta} \left({a}−{g}\:\mathrm{sin}\:\theta\right){d}\theta \\ $$$${T}=\rho{r}\left[{a}\theta−{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\right] \\ $$$${T}=\frac{{Mgr}}{{l}}\left[\frac{{r}\left(\mathrm{1}−\mathrm{cos}\:\phi\right)}{{l}}\theta−\mathrm{1}+\mathrm{cos}\:\theta\right] \\ $$$$\Rightarrow{T}=\frac{{Mg}}{\phi}\left[\frac{\left(\mathrm{1}−\mathrm{cos}\:\phi\right)}{\phi}\theta−\mathrm{1}+\mathrm{cos}\:\theta\right] \\ $$$$ \\ $$$${at}\:\theta=\phi: \\ $$$${T}_{\phi} =\frac{{Mg}}{\phi}\left[\frac{\left(\mathrm{1}−\mathrm{cos}\:\phi\right)}{\phi}×\phi−\mathrm{1}+\mathrm{cos}\:\phi\right] \\ $$$$\:\:\:\:=\mathrm{0}\:\checkmark \\ $$$$ \\ $$$${for}\:{maximum}\:{T}: \\ $$$$\frac{{dT}}{{d}\theta}=\mathrm{0} \\ $$$$\frac{{Mg}}{\phi}\left[\frac{\mathrm{1}−\mathrm{cos}\:\phi}{\phi}−\mathrm{sin}\:\theta\right]=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{1}−\mathrm{cos}\:\phi}{\phi} \\ $$$${i}.{e}.\:\theta=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}−\mathrm{cos}\:\phi}{\phi} \\ $$$${and} \\ $$$${T}_{{max}} =\frac{{Mg}}{\phi}\left[\frac{\mathrm{1}−\mathrm{cos}\:\phi}{\phi}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}−\mathrm{cos}\:\phi}{\phi}−\mathrm{1}+\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{cos}\:\phi}{\phi}\right)^{\mathrm{2}} }\right] \\ $$$$ \\ $$$${examples}:\: \\ $$$$\frac{{l}}{{r}}=\frac{\pi}{\mathrm{4}}\:\:\left(\mathrm{45}°\right):\:\theta\approx\mathrm{21}.\mathrm{9}°,\:{T}_{{max}} \approx\mathrm{0}.\mathrm{09}\:{Mg} \\ $$$$\frac{{l}}{{r}}=\frac{\pi}{\mathrm{3}}\:\:\left(\mathrm{60}°\right):\:\theta\approx\mathrm{28}.\mathrm{5}°,\:{T}_{{max}} \approx\mathrm{0}.\mathrm{11}\:{Mg} \\ $$$$\frac{{l}}{{r}}=\frac{\pi}{\mathrm{2}}\:\:\left(\mathrm{90}°\right):\:\theta\approx\mathrm{39}.\mathrm{5}°,\:{T}_{{max}} \approx\mathrm{0}.\mathrm{13}\:{Mg} \\ $$$${we}\:{see}\:{the}\:{maximum}\:{tension}\:{in} \\ $$$${rope}\:{occurs}\:{always}\:{a}\:{little}\:{bit} \\ $$$${above}\:{the}\:{middle}\:{of}\:{the}\:{rope}. \\ $$
Commented by fantastic last updated on 09/Nov/25
great sir!
$${great}\:{sir}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *