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Question Number 20891 by Tinkutara last updated on 06/Sep/17

$$\mathrm{The}\mathrm{Figure}\mathrm{shows}\mathrm{a}\mathrm{system}\mathrm{consisting}$$$$\mathrm{of}\left(i\right)\mathrm{a}\mathrm{ring}\mathrm{of}\mathrm{outer}\mathrm{radius}3R\mathrm{rolling}$$$$\mathrm{clockwise}\mathrm{without}\mathrm{slipping}\mathrm{on}\mathrm{a}$$$$\mathrm{horizontal}\mathrm{surface}\mathrm{with}\mathrm{angular}\mathrm{speed}$$$$\omega \mathrm{and}\left(ii\right)\mathrm{an}\mathrm{inner}\mathrm{disc}\mathrm{of}\mathrm{radius}2R$$$$\mathrm{rotating}\mathrm{anti}-\mathrm{clockwise}\mathrm{with}\mathrm{angular}$$$$\mathrm{speed}\omega /2.\mathrm{The}\mathrm{ring}\mathrm{and}\mathrm{disc}\mathrm{are}$$$$\mathrm{separated}\mathrm{by}\mathrm{frictionless}\mathrm{ball}\mathrm{bearing}.$$$$\mathrm{The}\mathrm{system}\mathrm{is}\mathrm{in}\mathrm{the}x-z\mathrm{plane}.\mathrm{The}$$$$\mathrm{point}P\mathrm{on}\mathrm{the}\mathrm{inner}\mathrm{disc}\mathrm{is}\mathrm{at}\mathrm{a}\mathrm{distance}$$$$R\mathrm{from}\mathrm{the}\mathrm{origin},\mathrm{where}OP\mathrm{makes}\mathrm{an}$$$$\mathrm{angle}30°\mathrm{with}\mathrm{the}\mathrm{horizontal}.\mathrm{Then}$$$$\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{the}\mathrm{horizontal}\mathrm{surface}$$$$\left(a\right)\mathrm{The}\mathrm{point}O\mathrm{has}\mathrm{a}\mathrm{linear}\mathrm{velocity}$$$$3R\omega \stackrel{\wedge }{i}$$$$\left(b\right)\mathrm{The}\mathrm{point}P\mathrm{has}\mathrm{a}\mathrm{linear}\mathrm{velocity}$$$$\frac{11}{4}R\omega \stackrel{\wedge }{i}+\frac{\sqrt{3}}{4}R\omega \stackrel{\wedge }{k}$$

Commented by Tinkutara last updated on 06/Sep/17

Commented by ajfour last updated on 06/Sep/17

Commented by Tinkutara last updated on 06/Sep/17

$$\mathrm{But}\mathrm{since}\mathrm{radius}\mathrm{at}O\mathrm{is}0,\mathrm{so}\mathrm{linear}$$$$\mathrm{velocity}=r\omega =0\left(\omega \right)=0?\mathrm{Why}\mathrm{not}?$$

Commented by ajfour last updated on 06/Sep/17

$$letvelocityofpointObe{v}_0.$$$$asthebottommostpointdoesnot$$$$slipitsvelocity=v_0-\omega \left(3R\right)=0$$$$\Rightarrow {\overrightarrow{v}}_0=3\omega R\hat{i}$$$$velocityofpointPbe{v}_P;then$$$${\overrightarrow{v}}_P={\overrightarrow{v}}_0+\frac{\omega R}{2}(-\sin 30°\hat{i}+\cos 30°\hat{k})$$$$=(3\omega R-\frac{\omega R}{4})\hat{i}+\frac{\omega R}{2}\times \frac{\sqrt{3}}{2}\hat{k}$$$$=\frac{11\omega R}{4}\hat{i}+\frac{\omega R\sqrt{3}}{4}\hat{k}.$$$$botharetrue.$$

Commented by ajfour last updated on 06/Sep/17

$$that?wouldbethecaseifyou$$$$hangthewheelinair,likeapulley.$$

Commented by ajfour last updated on 06/Sep/17

$$rollingisrotatingplussliding..$$

Commented by Tinkutara last updated on 06/Sep/17

$$\mathrm{Then}\mathrm{why}{v}_O\ne \left(2R\right)\left(\frac{\omega }{2}\right)=R\omega ?\mathrm{Why}$$$$\mathrm{outer}\mathrm{radius}\mathrm{is}\mathrm{taken}\mathrm{only}?$$

Commented by ajfour last updated on 06/Sep/17

$$owingtorotationthespeedis$$$$zero,butduetotranslationitis$$$$3\omega R.$$

Commented by Tinkutara last updated on 06/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Commented by ajfour last updated on 06/Sep/17

$$onlytheouterradiusisusedfor$$$$v_{O}becauseitistheouterring$$$$whichisincontactwithground.$$$$sowehavearelationonlyofthe$$$$angularvelocityofringand$$$$linearvelocityofitscenter.$$$$v_O=3\omega R.Thedisccamhaveany$$$$angularvelocity\left(here\omega /2\right)..$$

Commented by Tinkutara last updated on 06/Sep/17

$$\mathrm{Thanks}.\mathrm{Now}\mathrm{I}\mathrm{understand}.$$

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