Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 215650 by mr W last updated on 13/Jan/25

Commented by mr W last updated on 14/Jan/25

$$arotatingdiscwithangularvelocity$$$$\mathit{\omega }isputonaninclinewithrough$$$$surfaceasshown.$$$$findthemaximumdistancethe$$$$disccanmoveupwardsalongthe$$$$incline.$$$$thefrictioncoefficientis\mathit{\mu }.$$$$$$$$(originalquestionseeyoutube$$$$channelfromajfoursir)$$

Commented by mr W last updated on 15/Jan/25

https://youtu.be/3wiUgK_riz8?si=HlRNIT6W5V9kkg1q

Answered by Wuji last updated on 13/Jan/25

$${\mathrm{E}}_{\mathrm{rot}}=\frac{1}{2}\mathrm{I}{\omega }^2$$$$\mathrm{for}\mathrm{solid}\mathrm{disc},\mathrm{the}\mathrm{momment}\mathrm{of}$$$$\mathrm{inertia}\mathrm{I}\mathrm{is}$$$$\mathrm{I}=\frac{1}{2}{\mathrm{mr}}^2$$$$\mathrm{gravutational}\mathrm{potential}\mathrm{energy}$$$$\mathrm{at}\mathrm{height}\mathrm{h}$$$${\mathrm{E}}_{\mathrm{grav}}=\mathrm{mgh}$$$$\mathrm{Workdone}\mathrm{aginst}\mathrm{friction}$$$${\mathrm{W}}_{\mathrm{f}}=\mu \mathrm{mgcos}\theta \mathrm{S}$$$$\mathrm{from}\mathrm{Energy}\mathrm{conservation}$$$${\mathrm{E}}_{\mathrm{rot}}={\mathrm{E}}_{\mathrm{grav}}+{\mathrm{W}}_{\mathrm{f}}$$$$\frac{1}{4}{\mathrm{mr}}^2{\omega }^2=\mathrm{mgh}+\mu \mathrm{mgcos}\theta \mathrm{S}$$$$\mathrm{S}=\frac{\frac{1}{4}{\mathrm{mr}}^2{\omega }^2-\mathrm{mgh}}{\mu \mathrm{mgcos}\theta }$$$$\mathrm{S}=\frac{{\mathrm{mr}}^2{\omega }^2-4\mathrm{mgh}}{4\mu \mathrm{mgcos}\theta }$$$$\mathrm{S}=\frac{\mathrm{m}({\mathrm{r}}^2{\omega }^2-4\mathrm{gh})}{\mathrm{m}\left(4\mu \mathrm{gcos}\theta \right)}$$$$\mathrm{S}=\frac{{\mathrm{r}}^2{\omega }^2-4\mathrm{gh}}{4\mu \mathrm{gcos}\theta }$$

Commented by mr W last updated on 13/Jan/25

$$notcorrect!onlyfrictionbyslipping$$$$causeslossofenergy.butnotthe$$$$wholedistancesisslipping.only$$$$apartofitisslipping,theother$$$$partisrolling.therollingpart$$$${doesn}^{\prime }tcauseenergyloss.$$

Answered by mr W last updated on 14/Jan/25

Commented by mr W last updated on 16/Jan/25

$$\mathit{c}\mathit{a}\mathit{s}\mathit{e}1:\mathit{t}\mathit{h}\mathit{e}\mathit{f}\mathit{r}\mathit{i}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{c}\mathit{o}\mathit{e}\mathit{f}\mathit{f}\mathit{i}\mathit{c}\mathit{i}\mathit{e}\mathit{n}\mathit{t}$$$$\mathit{i}\mathit{s}\mathit{l}\mathit{a}\mathit{r}\mathit{g}\mathit{e}:\mathit{\mu }>\mathbf{tan}\mathit{\theta }$$$$$$$$inthiscasetherotatingdiscwill$$$$moveupwardsalongtheincline$$$$afteritisreleased.$$$$forsoliddisc:I=\frac{{mr}^2}{2}$$$$\underset{―}{t=0\to t_1:rolling\mathrm{\&}slipping}$$$$frictionforcef=\mu mg\cos \theta (\nearrow )$$$$a=\frac{\mu mg\cos \theta -mg\sin \theta }{m}=g(\mu \cos \theta -\sin \theta )$$$$\alpha =\frac{fr}{I}=\frac{r\mu mg\cos \theta }{\frac{{mr}^2}{2}}=\frac{2\mu g\cos \theta }{r}$$$$v_1={\omega }_{1}r$$$$t_1=\frac{v_1}{a}=\frac{\omega -{\omega }_1}{\alpha }$$$$\frac{{\omega }_{1}r}{g(\mu \cos \theta -\sin \theta )}=\frac{(\omega -{\omega }_1)r}{2\mu g\cos \theta }$$$$\frac{{\omega }_1}{\mu -\tan \theta }=\frac{\omega -{\omega }_1}{2\mu }$$$$\Rightarrow {\omega }_1=\frac{(\mu -\tan \theta )\omega }{3\mu -\tan \theta }$$$$s_1=\frac{v_1^2}{2a}=\frac{{(\mu -\tan \theta )}^2{\omega }^2{r}^2}{2g(\mu \cos \theta -\sin \theta ){(3\mu -\tan \theta )}^2}$$$$\Rightarrow s_1=\frac{(\mu -\tan \theta ){\omega }^2{r}^2}{2g{(3\mu -\tan \theta )}^2\cos \theta }$$$$check:$$$$rotationofdisc{\phi }_1$$$${\phi }_1=\frac{{\omega }^2-{\omega }_1^2}{2\alpha }=\frac{r}{4\mu g\cos \theta }[1-\frac{{(\mu -\tan \theta )}^2}{{(3\mu -\tan \theta )}^2}]{\omega }^2$$$${\phi }_{1}r=\frac{(2\mu -\tan \theta ){\omega }^2{r}^2}{g{(3\mu -\tan \theta )}^2\cos \theta }>s_1$$$$\Rightarrow slippingindeed!$$$${\phi }_1=\frac{(2\mu -\tan \theta ){\omega }^{2}r}{g{(3\mu -\tan \theta )}^2\cos \theta }$$$$$$$$\underset{―}{t=t_1\to t_2:rollingonly}$$$$K.E.\Rightarrow P.E.$$$$\frac{1}{2}(\frac{{mr}^2}{2}+{mr}^2){\omega }_1^2=mg{s}_2\sin \theta$$$$s_2=\frac{3{mr}^2}{4mg\sin \theta }\times \frac{{(\mu -\tan \theta )}^2{\omega }^2}{{(3\mu -\tan \theta )}^2}$$$$\Rightarrow s_2=\frac{3{(\mu -\tan \theta )}^2{\omega }^2{r}^2}{4g{(3\mu -\tan \theta )}^2\sin \theta }$$$${\phi }_2=\frac{s_2}{r}=\frac{3{(\mu -\tan \theta )}^2{\omega }^{2}r}{4g{(3\mu -\tan \theta )}^2\sin \theta }$$$$\Rightarrow {\phi }_2=\frac{3{(\mu -\tan \theta )}^2{\omega }^{2}r}{4g{(3\mu -\tan \theta )}^2\sin \theta }$$$$$$$$totaldistances=s_1+s_2$$$$s=\frac{(\mu -\tan \theta ){\omega }^2{r}^2}{2g{(3\mu -\tan \theta )}^2\cos \theta }+\frac{3{(\mu -\tan \theta )}^2{\omega }^2{r}^2}{4g{(3\mu -\tan \theta )}^2\sin \theta }$$$$\Rightarrow s=\frac{(\mu -\tan \theta ){\omega }^2{r}^2}{4g(3\mu -\tan \theta )\sin \theta }$$$$or$$$$s\sin \theta =h=\frac{(\mu -\tan \theta ){\omega }^2{r}^2}{4g(3\mu -\tan \theta )}$$$$$$$${mgh}^{\prime }=\frac{1}{2}\times \frac{m{\omega }^2{r}^2}{2}$$$$h^{\prime }=\frac{{\omega }^2{r}^2}{4g}$$$$weseeh=\frac{(\mu -\tan \theta )h^{\prime }}{3\mu -\tan \theta }<h^{\prime }$$$$$$$$\phi ={\phi }_1+{\phi }_2=\frac{(2\mu -\tan \theta ){\omega }^{2}r}{g\cos \theta {(3\mu -\tan \theta )}^2}+\frac{3{(\mu -\tan \theta )}^2{\omega }^{2}r}{4g{(3\mu -\tan \theta )}^2\sin \theta }$$$$\Rightarrow \phi =\frac{(\mu +\tan \theta ){\omega }^{2}r}{4g(3\mu -\tan \theta )\sin \theta }$$

Commented by mr W last updated on 14/Jan/25

Commented by mr W last updated on 14/Jan/25

$$\mathit{c}\mathit{a}\mathit{s}\mathit{e}2:\mathit{t}\mathit{h}\mathit{e}\mathit{f}\mathit{r}\mathit{i}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{c}\mathit{o}\mathit{e}\mathit{f}\mathit{f}\mathit{i}\mathit{c}\mathit{i}\mathit{e}\mathit{n}\mathit{t}$$$$\mathit{i}\mathit{s}\mathit{s}\mathit{m}\mathit{a}\mathit{l}\mathit{l}:\mathit{\mu }<\mathbf{tan}\mathit{\theta }$$$$inthiscasetherotatingcannot$$$$moveupwardsalongtheincline.$$$$itwillmovedownwards.aftersame$$$$timethediscwillbeonlyrolling.$$$$f=\mu mg\cos \theta (\nearrow )$$$$a=\frac{mg\sin \theta -\mu mg\cos \theta }{m}=g(\sin \theta -\mu \cos \theta )$$$$\alpha =\frac{fr}{I}=\frac{\mu mgr\cos \theta }{\frac{{mr}^2}{2}}=\frac{2\mu g\cos \theta }{r}$$$$v_1={\omega }_{1}r$$$$t_1=\frac{v_1}{a}=\frac{{\omega }_1+\omega }{\alpha }$$$$\frac{{\omega }_{1}r}{g(\sin \theta -\mu \cos \theta )}=\frac{({\omega }_1+\omega )r}{2\mu g\cos \theta }$$$$\frac{{\omega }_1}{\tan \theta -\mu }=\frac{{\omega }_1+\omega }{2\mu }$$$$(3\mu -\tan \theta ){\omega }_1=(\tan \theta -\mu )\omega$$$$\Rightarrow {\omega }_1=\frac{(\tan \theta -\mu )\omega }{3\mu -\tan \theta }$$$$\frac{\tan \theta }{3}<\mu <\tan \theta$$$$s=\frac{v_1^2}{2a}=\frac{{(\tan \theta -\mu )}^2{\omega }^2{r}^2}{2g(\sin \theta -\mu \cos \theta ){(3\mu -\tan \theta )}^2}$$$$\Rightarrow s=\frac{(\tan \theta -\mu ){\omega }^2{r}^2}{2g{(3\mu -\tan \theta )}^2\cos \theta }$$$$$$$$if\mu ⩽\frac{\tan \theta }{3},thedisccanneverroll$$$$onlyontheincline,butalwaysroll$$$$andslip.$$

Commented by ajfour last updated on 15/Jan/25

https://youtu.be/-RRr7_nNNqI?si=mdeobU80KLO5c5m- A special case of a more general geometry question that mrW sir, MJS sir and myself had solved way back several yers.

Commented by mr W last updated on 15/Jan/25

��

Terms of Service

Privacy Policy

Contact: info@tinkutara.com