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Question Number 161783 by ajfour last updated on 22/Dec/21
Commented by ajfour last updated on 22/Dec/21
Cylinder is free to purely roll, find radius, if point mass ball is to be received successfully.
$${Cylinder}\:{is}\:{free}\:{to}\:{purely}\:{roll},\:{find} \\ $$$${radius},\:{if}\:{point}\:{mass}\:{ball}\:{is}\:{to}\:{be} \\ $$$${received}\:{successfully}. \\ $$
Commented by ajfour last updated on 22/Dec/21
Left d job, shall hunt online..
Commented by mr W last updated on 22/Dec/21
welcome back sir!
$${welcome}\:{back}\:{sir}! \\ $$
Commented by ajfour last updated on 22/Dec/21
thank you sir, wont u try this question..
$${thank}\:{you}\:{sir},\:{wont}\:{u}\:{try}\:{this}\:{question}.. \\ $$
Commented by mr W last updated on 23/Dec/21
i′ll try how far i can go.
$${i}'{ll}\:{try}\:{how}\:{far}\:{i}\:{can}\:{go}. \\ $$
Answered by mr W last updated on 23/Dec/21
Commented by mr W last updated on 23/Dec/21
x_1 =s ϕ=(s/R) let ω=(dθ/dt) x_2 =s−R sin θ y_2 =R(1+cos θ) V=(ds/dt)=ω(ds/dθ) A=(dV/dt)=ω(dV/dθ) N sin θ R=(((MR^2 )/2)+MR^2 )((A/R)) ⇒N=((3MA)/(2 sin θ))=((3M)/(2 sin θ))×((ωdV)/dθ) N=0 ⇔ (dV/dθ)=0 v_x =(dx_2 /dt)=V−R cos θ ω v_y =(dy_2 /dt)=−R sin θ ω (m/2)(v_x ^2 +v_y ^2 )+(1/2)×((3MR^2 )/2)×((V/R))^2 =mgR(1−cos θ) v_x ^2 +v_y ^2 +((3MV^2 )/(2m))=2gR(1−cos θ) V^2 +R^2 ω^2 −2VRω cos θ+((3MV^2 )/(2m))=2gR(1−cos θ) ω^2 −((2g(1−cos θ))/R)−2ω cos θ((V/R))+(((3M)/(2m))+1)((V/R))^2 =0 ⇒(V/R)=((ω cos θ+(√(ω^2 (cos^2 θ−μ)+((2μg(1−cos θ))/R))))/μ) with μ=((3M)/(2m))+1 ((ωds)/(Rdθ))=((ω cos θ+(√(ω^2 (cos^2 θ−μ)+((2μg(1−cos θ))/R))))/μ) (ds/dθ)=R×((cos θ+(√(cos^2 θ−μ+((2μg(1−cos θ))/(Rω^2 )))))/μ) ⇒s=R∫_0 ^θ ((cos θ+(√(cos^2 θ−μ+((2μg(1−cos θ))/(Rω^2 )))))/μ)dθ a_x =(dv_x /dt)=((ωd)/dθ)(V−R cos θ ω)=ω((dV/dθ)−Rcos θ (dω/dθ)+R sin θ ω) a_y =(dv_y /dt)=((ωd)/dθ)(−R sin θ ω)=ω(−R sin θ (dω/dθ)−R cos θ ω) ma_x =−N sin θ mω((dV/dθ)−Rcos θ (dω/dθ)+R sin θ ω)=−((3M)/2)×((ωdV)/dθ) ⇒μ(d/dθ)((V/R))−cos θ (dω/dθ)+sin θ ω=0 ma_y =N cos θ−mg mω(−R sin θ (dω/dθ)−R cos θ ω)=((3M)/(2 sin θ))×((ωdV)/dθ) cos θ−mg ⇒((3M)/(2m))×(d/dθ)((V/R))=(g/(ωR))−(sin θ (dω/dθ)+cos θ ω)tan θ ((3M)/(2mμ))cos θ (dω/dθ)−((3M)/(2mμ))sin θ ω=(g/(ωR))−((sin^2 θ)/(cos θ)) (dω/dθ)−sin θ ω determinant ((((((3M)/(3M+2m))+tan^2 θ)cos θ (dω/dθ)−(((2m)/(3M+2m)))sin θ ω−(g/(ωR))=0))) solve this d.e. (hard to do!)for ω=f(θ) under ω=0 at θ=0 ....... once we have ω=f(θ), we also get (V/R). from (d/dθ)((V/R))=0 we get θ_1 at which the small ball loses contact to cylinder. we also get the v_x and v_y at this instant. upon now the small ball has the motion of projectile and we can find the position where it hits the ground.
$${x}_{\mathrm{1}} ={s} \\ $$$$\varphi=\frac{{s}}{{R}} \\ $$$${let}\:\omega=\frac{{d}\theta}{{dt}} \\ $$$${x}_{\mathrm{2}} ={s}−{R}\:\mathrm{sin}\:\theta \\ $$$${y}_{\mathrm{2}} ={R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$${V}=\frac{{ds}}{{dt}}=\omega\frac{{ds}}{{d}\theta} \\ $$$${A}=\frac{{dV}}{{dt}}=\omega\frac{{dV}}{{d}\theta} \\ $$$${N}\:\mathrm{sin}\:\theta\:{R}=\left(\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}+{MR}^{\mathrm{2}} \right)\left(\frac{{A}}{{R}}\right) \\ $$$$\Rightarrow{N}=\frac{\mathrm{3}{MA}}{\mathrm{2}\:\mathrm{sin}\:\theta}=\frac{\mathrm{3}{M}}{\mathrm{2}\:\mathrm{sin}\:\theta}×\frac{\omega{dV}}{{d}\theta} \\ $$$${N}=\mathrm{0}\:\Leftrightarrow\:\frac{{dV}}{{d}\theta}=\mathrm{0} \\ $$$${v}_{{x}} =\frac{{dx}_{\mathrm{2}} }{{dt}}={V}−{R}\:\mathrm{cos}\:\theta\:\omega \\ $$$${v}_{{y}} =\frac{{dy}_{\mathrm{2}} }{{dt}}=−{R}\:\mathrm{sin}\:\theta\:\omega \\ $$$$\frac{{m}}{\mathrm{2}}\left({v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}{MR}^{\mathrm{2}} }{\mathrm{2}}×\left(\frac{{V}}{{R}}\right)^{\mathrm{2}} ={mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} +\frac{\mathrm{3}{MV}^{\mathrm{2}} }{\mathrm{2}{m}}=\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${V}^{\mathrm{2}} +{R}^{\mathrm{2}} \omega^{\mathrm{2}} −\mathrm{2}{VR}\omega\:\mathrm{cos}\:\theta+\frac{\mathrm{3}{MV}^{\mathrm{2}} }{\mathrm{2}{m}}=\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\omega^{\mathrm{2}} −\frac{\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}}−\mathrm{2}\omega\:\mathrm{cos}\:\theta\left(\frac{{V}}{{R}}\right)+\left(\frac{\mathrm{3}{M}}{\mathrm{2}{m}}+\mathrm{1}\right)\left(\frac{{V}}{{R}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\frac{{V}}{{R}}=\frac{\omega\:\mathrm{cos}\:\theta+\sqrt{\omega^{\mathrm{2}} \left(\mathrm{cos}^{\mathrm{2}} \:\theta−\mu\right)+\frac{\mathrm{2}\mu{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}}}}{\mu}\: \\ $$$${with}\:\mu=\frac{\mathrm{3}{M}}{\mathrm{2}{m}}+\mathrm{1} \\ $$$$\frac{\omega{ds}}{{Rd}\theta}=\frac{\omega\:\mathrm{cos}\:\theta+\sqrt{\omega^{\mathrm{2}} \left(\mathrm{cos}^{\mathrm{2}} \:\theta−\mu\right)+\frac{\mathrm{2}\mu{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}}}}{\mu} \\ $$$$\frac{{ds}}{{d}\theta}={R}×\frac{\mathrm{cos}\:\theta+\sqrt{\mathrm{cos}^{\mathrm{2}} \:\theta−\mu+\frac{\mathrm{2}\mu{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}\omega^{\mathrm{2}} }}}{\mu} \\ $$$$\Rightarrow{s}={R}\int_{\mathrm{0}} ^{\theta} \frac{\mathrm{cos}\:\theta+\sqrt{\mathrm{cos}^{\mathrm{2}} \:\theta−\mu+\frac{\mathrm{2}\mu{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}\omega^{\mathrm{2}} }}}{\mu}{d}\theta \\ $$$${a}_{{x}} =\frac{{dv}_{{x}} }{{dt}}=\frac{\omega{d}}{{d}\theta}\left({V}−{R}\:\mathrm{cos}\:\theta\:\omega\right)=\omega\left(\frac{{dV}}{{d}\theta}−{R}\mathrm{cos}\:\theta\:\frac{{d}\omega}{{d}\theta}+{R}\:\mathrm{sin}\:\theta\:\omega\right) \\ $$$${a}_{{y}} =\frac{{dv}_{{y}} }{{dt}}=\frac{\omega{d}}{{d}\theta}\left(−{R}\:\mathrm{sin}\:\theta\:\omega\right)=\omega\left(−{R}\:\mathrm{sin}\:\theta\:\frac{{d}\omega}{{d}\theta}−{R}\:\mathrm{cos}\:\theta\:\omega\right) \\ $$$${ma}_{{x}} =−{N}\:\mathrm{sin}\:\theta \\ $$$${m}\omega\left(\frac{{dV}}{{d}\theta}−{R}\mathrm{cos}\:\theta\:\frac{{d}\omega}{{d}\theta}+{R}\:\mathrm{sin}\:\theta\:\omega\right)=−\frac{\mathrm{3}{M}}{\mathrm{2}}×\frac{\omega{dV}}{{d}\theta} \\ $$$$\Rightarrow\mu\frac{{d}}{{d}\theta}\left(\frac{{V}}{{R}}\right)−\mathrm{cos}\:\theta\:\frac{{d}\omega}{{d}\theta}+\mathrm{sin}\:\theta\:\omega=\mathrm{0} \\ $$$${ma}_{{y}} ={N}\:\mathrm{cos}\:\theta−{mg} \\ $$$${m}\omega\left(−{R}\:\mathrm{sin}\:\theta\:\frac{{d}\omega}{{d}\theta}−{R}\:\mathrm{cos}\:\theta\:\omega\right)=\frac{\mathrm{3}{M}}{\mathrm{2}\:\mathrm{sin}\:\theta}×\frac{\omega{dV}}{{d}\theta}\:\mathrm{cos}\:\theta−{mg} \\ $$$$\Rightarrow\frac{\mathrm{3}{M}}{\mathrm{2}{m}}×\frac{{d}}{{d}\theta}\left(\frac{{V}}{{R}}\right)=\frac{{g}}{\omega{R}}−\left(\mathrm{sin}\:\theta\:\frac{{d}\omega}{{d}\theta}+\mathrm{cos}\:\theta\:\omega\right)\mathrm{tan}\:\theta \\ $$$$\frac{\mathrm{3}{M}}{\mathrm{2}{m}\mu}\mathrm{cos}\:\theta\:\frac{{d}\omega}{{d}\theta}−\frac{\mathrm{3}{M}}{\mathrm{2}{m}\mu}\mathrm{sin}\:\theta\:\omega=\frac{{g}}{\omega{R}}−\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{cos}\:\theta}\:\frac{{d}\omega}{{d}\theta}−\mathrm{sin}\:\theta\:\omega \\ $$$$\begin{array}{|c|}{\left(\frac{\mathrm{3}{M}}{\mathrm{3}{M}+\mathrm{2}{m}}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)\mathrm{cos}\:\theta\:\frac{{d}\omega}{{d}\theta}−\left(\frac{\mathrm{2}{m}}{\mathrm{3}{M}+\mathrm{2}{m}}\right)\mathrm{sin}\:\theta\:\omega−\frac{{g}}{\omega{R}}=\mathrm{0}}\\\hline\end{array} \\ $$$${solve}\:{this}\:{d}.{e}.\:\left({hard}\:{to}\:{do}!\right){for}\:\omega={f}\left(\theta\right)\: \\ $$$${under}\:\omega=\mathrm{0}\:{at}\:\theta=\mathrm{0} \\ $$$$……. \\ $$$${once}\:{we}\:{have}\:\omega={f}\left(\theta\right),\:{we}\:{also}\:{get}\:\frac{{V}}{{R}}. \\ $$$${from}\:\frac{{d}}{{d}\theta}\left(\frac{{V}}{{R}}\right)=\mathrm{0}\:{we}\:{get}\:\theta_{\mathrm{1}} \:{at}\:{which}\:{the} \\ $$$${small}\:{ball}\:{loses}\:{contact}\:{to}\:{cylinder}. \\ $$$${we}\:{also}\:{get}\:{the}\:{v}_{{x}} \:{and}\:{v}_{{y}} \:{at}\:{this}\:{instant}. \\ $$$${upon}\:{now}\:{the}\:{small}\:{ball}\:{has}\:{the} \\ $$$${motion}\:{of}\:{projectile}\:{and}\:{we}\:{can}\:{find} \\ $$$${the}\:{position}\:{where}\:{it}\:{hits}\:{the}\:{ground}. \\ $$
Commented by mr W last updated on 23/Dec/21
that′s the point i can maximally come to.
$${that}'{s}\:{the}\:{point}\:{i}\:{can}\:{maximally}\:{come} \\ $$$${to}. \\ $$
Commented by Tawa11 last updated on 23/Dec/21
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by ajfour last updated on 23/Dec/21
mg(2R)=(1/2)(((mR^2 )/2))ω^2 +(1/2)m(v_x ^2 +v_y ^2 ) +mgR(1+cos θ) ...(i) (energy conservation) mv_x =mωR (linear momentum ..(ii) conservation) mx=ms ..(iii) (center of mass ) 2s=Rcos θ ..)iv) mv_x R(1+cos θ)+mv_y x =((mR^2 )/2)ω+mR^2 ω ...(v) (angular momentum conservation) now v_y t+(1/2)gt^2 =R(1+cos θ) ..(vi) v_x t=(b−x) ..(vii) ................................................ so v_x =ωR (b−((Rcos θ)/2))((v_y /(ωR)))+(1/2)(g/(ω^2 R^2 ))(b−((Rcos θ)/2))^2 =R(1+cos θ) say (b/R)=λ ; (v_y /(ωR))=μ 4(2λ−cos θ)μ+(g/(ω^2 R))(2λ−cos θ)^2 μ^2 =8(1+cos θ) .....(A) λ =? (so we need μ and cos θ) and from ..(v) 2(1+cos θ)+μcos θ=3 ⇒ cos θ=(1/(2+μ)) And from ..(i) ((8g)/R)=ω^2 +2ω^2 (1+μ^2 )+((4g)/R)(1+cos θ) ⇒ ω^2 =((4g)/R)(((1−cos θ)/(3+2μ^2 ))) ...(I) ⇒ ω^2 R=4g(((1−(1/(2+μ)))/(3+2μ^2 ))) also (2ωRcos θ)^2 +(μωRsin θ)^2 =gRcos θ ⇒ ω^2 = (g/R)(((cos θ)/(4cos^2 θ+μ^2 sin^2 θ))) ..(II) from ...(I) & (II) 4(1−cos θ)(4cos^2 θ+μ^2 sin^2 θ) =cos θ(3+2μ^2 ) but cos θ=(1/(2+μ)) hence we obtain 𝛍 from 4(1−(1/(2+μ))){((2/(2+μ)))^2 +(μ^2 /(1−(1/((2+μ)^2 ))))} =(((3+2μ^2 )/(2+μ))) and now eq..(A) gives us λ=(b/R)
$${mg}\left(\mathrm{2}{R}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\right)\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}\left({v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{mgR}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\:\:\:…\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\left({energy}\:{conservation}\right) \\ $$$${mv}_{{x}} ={m}\omega{R}\:\:\:\:\left({linear}\:{momentum}\:\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:..\left({ii}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{conservation}\right) \\ $$$${mx}={ms}\:\:\:..\left({iii}\right)\:\:\:\:\left({center}\:{of}\:{mass}\:\right) \\ $$$$\left.\mathrm{2}\left.{s}={R}\mathrm{cos}\:\theta\:\:..\right){iv}\right) \\ $$$${mv}_{{x}} {R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)+{mv}_{{y}} {x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\omega+{mR}^{\mathrm{2}} \omega\:\:\:…\left({v}\right) \\ $$$$\left({angular}\:{momentum}\:{conservation}\right) \\ $$$${now}\:\:\:{v}_{{y}} {t}+\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} ={R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\:..\left({vi}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{v}_{{x}} {t}=\left({b}−{x}\right)\:\:\:..\left({vii}\right) \\ $$$$\:\:\:………………………………………… \\ $$$${so}\:\:\:{v}_{{x}} =\omega{R} \\ $$$$\left({b}−\frac{{R}\mathrm{cos}\:\theta}{\mathrm{2}}\right)\left(\frac{{v}_{{y}} }{\omega{R}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\frac{{g}}{\omega^{\mathrm{2}} {R}^{\mathrm{2}} }\left({b}−\frac{{R}\mathrm{cos}\:\theta}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:={R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$${say}\:\:\frac{{b}}{{R}}=\lambda\:\:\:;\:\:\frac{{v}_{{y}} }{\omega{R}}=\mu \\ $$$$\mathrm{4}\left(\mathrm{2}\lambda−\mathrm{cos}\:\theta\right)\mu+\frac{{g}}{\omega^{\mathrm{2}} {R}}\left(\mathrm{2}\lambda−\mathrm{cos}\:\theta\right)^{\mathrm{2}} \mu^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{8}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\:\:\:\:\:…..\left({A}\right) \\ $$$$\:\:\lambda\:=?\:\:\:\left({so}\:{we}\:{need}\:\:\:\mu\:{and}\:\mathrm{cos}\:\theta\right) \\ $$$${and}\:{from}\:..\left({v}\right) \\ $$$$\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)+\mu\mathrm{cos}\:\theta=\mathrm{3} \\ $$$$\Rightarrow\:\:\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}+\mu} \\ $$$${And}\:{from}\:..\left({i}\right) \\ $$$$\frac{\mathrm{8}{g}}{{R}}=\omega^{\mathrm{2}} +\mathrm{2}\omega^{\mathrm{2}} \left(\mathrm{1}+\mu^{\mathrm{2}} \right)+\frac{\mathrm{4}{g}}{{R}}\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\:\:\omega^{\mathrm{2}} =\frac{\mathrm{4}{g}}{{R}}\left(\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{3}+\mathrm{2}\mu^{\mathrm{2}} }\right)\:\:\:…\left({I}\right) \\ $$$$\Rightarrow\:\: \\ $$$$\omega^{\mathrm{2}} {R}=\mathrm{4}{g}\left(\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}+\mu}}{\mathrm{3}+\mathrm{2}\mu^{\mathrm{2}} }\right) \\ $$$${also}\:\:\:\:\:\left(\mathrm{2}\omega{R}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left(\mu\omega{R}\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:={gR}\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\omega^{\mathrm{2}} =\:\frac{{g}}{{R}}\left(\frac{\mathrm{cos}\:\theta}{\mathrm{4cos}\:^{\mathrm{2}} \theta+\mu^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}\right)\:\:..\left({II}\right) \\ $$$${from}\:\:…\left({I}\right)\:\&\:\left({II}\right) \\ $$$$\mathrm{4}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left(\mathrm{4cos}\:^{\mathrm{2}} \theta+\mu^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta\right) \\ $$$$\:\:\:\:\:=\mathrm{cos}\:\theta\left(\mathrm{3}+\mathrm{2}\mu^{\mathrm{2}} \right) \\ $$$${but}\:\:\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}+\mu} \\ $$$${hence}\:{we}\:{obtain}\:\boldsymbol{\mu}\:{from} \\ $$$$\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}+\mu}\right)\left\{\left(\frac{\mathrm{2}}{\mathrm{2}+\mu}\right)^{\mathrm{2}} +\frac{\mu^{\mathrm{2}} }{\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{2}+\mu\right)^{\mathrm{2}} }}\right\} \\ $$$$\:\:\:\:\:\:=\left(\frac{\mathrm{3}+\mathrm{2}\mu^{\mathrm{2}} }{\mathrm{2}+\mu}\right) \\ $$$${and}\:{now}\:\:{eq}..\left({A}\right)\:\:\:\:{gives}\:{us}\:\lambda=\frac{{b}}{{R}} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 23/Dec/21
great! i need some time to follow it. i think with ω you don′t mean ω=(dθ/dt), right?
$${great}! \\ $$$${i}\:{need}\:{some}\:{time}\:{to}\:{follow}\:{it}. \\ $$$${i}\:{think}\:{with}\:\omega\:{you}\:{don}'{t}\:{mean} \\ $$$$\omega=\frac{{d}\theta}{{dt}},\:{right}? \\ $$
Commented by ajfour last updated on 23/Dec/21
yeah its not (dθ/dt) , its the angular velocity of rotation of the cylinder.
$${yeah}\:{its}\:{not}\:\frac{{d}\theta}{{dt}}\:,\:{its}\:{the}\:{angular} \\ $$$${velocity}\:{of}\:{rotation}\:{of}\:{the}\:{cylinder}. \\ $$

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