Question Number 219770 by ajfour last updated on 01/May/25
Commented by ajfour last updated on 01/May/25
$${A}\:{rolling}\:{cylinder}\:{of}\:{length}\:{L}\:{rolls} \\ $$$$\:{gathering}\:{moss}\:{at}\:{the}\:{rate}\:\frac{{dm}}{{dt}}=\rho_{\mathrm{0}} {Lv}.\: \\ $$$${Find}\:{radius}\:{r}\left({t}\right){and}\:{speed}\:{v}\left({t}\right)\:{of}\: \\ $$$${center}\:{of}\:{mass}. \\ $$
Commented by ajfour last updated on 01/May/25
Commented by mr W last updated on 02/May/25
![please recheck (dm/dt)=ρ_0 Lv both sides have different units. therefore i have added a constand λ in the unit [m]. or did you even mean: (dm/dt)=ρ_0 rLv ?](https://www.tinkutara.com/question/Q219791.png)
$${please}\:{recheck} \\ $$$$\frac{{dm}}{{dt}}=\rho_{\mathrm{0}} {Lv} \\ $$$${both}\:{sides}\:{have}\:{different}\:{units}. \\ $$$${therefore}\:{i}\:{have}\:{added}\:{a}\:{constand}\:\lambda \\ $$$${in}\:{the}\:{unit}\:\left[{m}\right]. \\ $$$${or}\:{did}\:{you}\:{even}\:{mean}: \\ $$$$\frac{{dm}}{{dt}}=\rho_{\mathrm{0}} {rLv}\:? \\ $$
Answered by mr W last updated on 02/May/25
![(dm/dt)=ρ_0 λLv (λ constant in [m]) m=ρ_0 πr^2 L (dm/dt)=2ρ_0 πLr(dr/dt) 2ρ_0 πLr(dr/dt)=ρ_0 λLv ⇒(dr/dt)=((λv)/(2πr)) I=((mr^2 )/2)=((ρ_0 πLr^4 )/2) ω=(v/r) (d/dt)(mv)=f (f=friction force) (d/dt)(Iω)=−fr (d/dt)(Iω)=−r(d/dt)(mv) (d/dt)(((ρ_0 πLr^3 v)/2))=−r(d/dt)(ρ_0 πr^2 Lv) (d/dt)(r^3 v)=−2r(d/dt)(r^2 v) r^3 (dv/dt)+3r^2 v(dr/dt)=−2r(r^2 (dv/dt)+2rv(dr/dt)) 3r(dv/dt)+7v(dr/dt)=0 3∫_v_0 ^v (dv/v)=−7∫_r_0 ^r (dr/r) 3ln (v/v_0 )=−7ln (r/r_0 ) ⇒(v/v_0 )=((r/r_0 ))^(−(7/3)) (d/dt)((r/r_0 ))=((λv_0 )/(2πr_0 ^2 ))((r/r_0 ))^(−((10)/3)) ∫_1 ^(r/r_0 ) ((r/r_0 ))^((10)/3) d((r/r_0 ))=((λv_0 )/(2πr_0 ^2 ))∫_0 ^t dt (3/(13))[((r/r_0 ))^((13)/3) −1]=((λv_0 t)/(2πr_0 ^2 )) ⇒(r/r_0 )=(1+((13λv_0 t)/(6πr_0 ^2 )))^(3/(13)) ⇒(v/v_0 )=(1+((13λv_0 t)/(6πr_0 ^2 )))^(−(7/(13)))](https://www.tinkutara.com/question/Q219790.png)
$$\frac{{dm}}{{dt}}=\rho_{\mathrm{0}} \lambda{Lv}\:\:\:\left(\lambda\:{constant}\:{in}\:\left[{m}\right]\right) \\ $$$${m}=\rho_{\mathrm{0}} \pi{r}^{\mathrm{2}} {L} \\ $$$$\frac{{dm}}{{dt}}=\mathrm{2}\rho_{\mathrm{0}} \pi{Lr}\frac{{dr}}{{dt}} \\ $$$$\mathrm{2}\rho_{\mathrm{0}} \pi{Lr}\frac{{dr}}{{dt}}=\rho_{\mathrm{0}} \lambda{Lv} \\ $$$$\Rightarrow\frac{{dr}}{{dt}}=\frac{\lambda{v}}{\mathrm{2}\pi{r}} \\ $$$${I}=\frac{{mr}^{\mathrm{2}} }{\mathrm{2}}=\frac{\rho_{\mathrm{0}} \pi{Lr}^{\mathrm{4}} }{\mathrm{2}} \\ $$$$\omega=\frac{{v}}{{r}} \\ $$$$\frac{{d}}{{dt}}\left({mv}\right)={f}\:\:\:\:\left({f}={friction}\:{force}\right) \\ $$$$\frac{{d}}{{dt}}\left({I}\omega\right)=−{fr} \\ $$$$\frac{{d}}{{dt}}\left({I}\omega\right)=−{r}\frac{{d}}{{dt}}\left({mv}\right) \\ $$$$\frac{{d}}{{dt}}\left(\frac{\rho_{\mathrm{0}} \pi{Lr}^{\mathrm{3}} {v}}{\mathrm{2}}\right)=−{r}\frac{{d}}{{dt}}\left(\rho_{\mathrm{0}} \pi{r}^{\mathrm{2}} {Lv}\right) \\ $$$$\frac{{d}}{{dt}}\left({r}^{\mathrm{3}} {v}\right)=−\mathrm{2}{r}\frac{{d}}{{dt}}\left({r}^{\mathrm{2}} {v}\right) \\ $$$${r}^{\mathrm{3}} \frac{{dv}}{{dt}}+\mathrm{3}{r}^{\mathrm{2}} {v}\frac{{dr}}{{dt}}=−\mathrm{2}{r}\left({r}^{\mathrm{2}} \frac{{dv}}{{dt}}+\mathrm{2}{rv}\frac{{dr}}{{dt}}\right) \\ $$$$\mathrm{3}{r}\frac{{dv}}{{dt}}+\mathrm{7}{v}\frac{{dr}}{{dt}}=\mathrm{0} \\ $$$$\mathrm{3}\int_{{v}_{\mathrm{0}} } ^{{v}} \frac{{dv}}{{v}}=−\mathrm{7}\int_{{r}_{\mathrm{0}} } ^{{r}} \frac{{dr}}{{r}} \\ $$$$\mathrm{3ln}\:\frac{{v}}{{v}_{\mathrm{0}} }=−\mathrm{7ln}\:\frac{{r}}{{r}_{\mathrm{0}} } \\ $$$$\Rightarrow\frac{{v}}{{v}_{\mathrm{0}} }=\left(\frac{{r}}{{r}_{\mathrm{0}} }\right)^{−\frac{\mathrm{7}}{\mathrm{3}}} \\ $$$$\frac{{d}}{{dt}}\left(\frac{{r}}{{r}_{\mathrm{0}} }\right)=\frac{\lambda{v}_{\mathrm{0}} }{\mathrm{2}\pi{r}_{\mathrm{0}} ^{\mathrm{2}} }\left(\frac{{r}}{{r}_{\mathrm{0}} }\right)^{−\frac{\mathrm{10}}{\mathrm{3}}} \\ $$$$\int_{\mathrm{1}} ^{\frac{{r}}{{r}_{\mathrm{0}} }} \left(\frac{{r}}{{r}_{\mathrm{0}} }\right)^{\frac{\mathrm{10}}{\mathrm{3}}} {d}\left(\frac{{r}}{{r}_{\mathrm{0}} }\right)=\frac{\lambda{v}_{\mathrm{0}} }{\mathrm{2}\pi{r}_{\mathrm{0}} ^{\mathrm{2}} }\int_{\mathrm{0}} ^{{t}} {dt} \\ $$$$\frac{\mathrm{3}}{\mathrm{13}}\left[\left(\frac{{r}}{{r}_{\mathrm{0}} }\right)^{\frac{\mathrm{13}}{\mathrm{3}}} −\mathrm{1}\right]=\frac{\lambda{v}_{\mathrm{0}} {t}}{\mathrm{2}\pi{r}_{\mathrm{0}} ^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{r}}{{r}_{\mathrm{0}} }=\left(\mathrm{1}+\frac{\mathrm{13}\lambda{v}_{\mathrm{0}} {t}}{\mathrm{6}\pi{r}_{\mathrm{0}} ^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{13}}} \\ $$$$\Rightarrow\frac{{v}}{{v}_{\mathrm{0}} }=\left(\mathrm{1}+\frac{\mathrm{13}\lambda{v}_{\mathrm{0}} {t}}{\mathrm{6}\pi{r}_{\mathrm{0}} ^{\mathrm{2}} }\right)^{−\frac{\mathrm{7}}{\mathrm{13}}} \\ $$