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Chidren playinO plavg 0UE Ion E alconi Lity #130l Icse theit che parking cnepacner ReMe WWac Chrldten nhut [he nqur : belue Mne plargrouna acove the parkirc cre sCr...

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Chidren playinO plavg 0UE Ion E alconi Lity #130l Icse theit che parking cnepacner ReMe WWac Chrldten nhut [he nqur : belue Mne plargrouna acove the parkirc cre sCrmoi uilding 5 verzica Nin; rormhing 1.50 hiqh Diing around t= playground_ The alls Funcneo an anq 530u abov= che horlzonta Dolrit O from the base of tha puilcing The: hall taker neach rcni verlicall Aov Fhi: natijre this prc;olcm; MS munded Anreit Adiat: value: Ynuc [Aleulaticin: indudica anxicesuhrrited "chariiqt FInd the speed

Chidren playinO plavg 0UE Ion E alconi Lity #130l Icse theit che parking cnepacner ReMe WWac Chrldten nhut [he nqur : belue Mne plargrouna acove the parkirc cre sCrmoi uilding 5 verzica Nin; rormhing 1.50 hiqh Diing around t= playground_ The alls Funcneo an anq 530u abov= che horlzonta Dolrit O from the base of tha puilcing The: hall taker neach rcni verlicall Aov Fhi: natijre this prc;olcm; MS munded Anreit Adiat: value: Ynuc [Aleulaticin: indudica anxicesuhrrited "chariiqt FInd the speed (In Ms) shich the ball was lunched Ca Find thc vcrtici dishnac (in mit whah -hc ball ~Icare Chcei theonzontz dstance motonethe cllande Can You tind che tice tatesthe -rave the mnz eitic round? Can VoV then Telate this -he torz harlzonta cls-ance the travels anc then the hcizonta distanc trom the Y3i7 What In? the: [eAchur Alway'= Lnce; thc ball %ith nccc us? thc triqonomctic idcnzity 50" "(0; 44.59 above 01E horizantal nccd cuni ic '(0) ,} (4i #Na chi: minirnugri 4nule Jhyree$ hurizaniali ahicti #urich Ihc 34l Anid #GIi cIxar nlaygmiund railing? (icue - You Ma Wnzt wqui Joriontol distancc (in trom thc Y point cnc roct Yiher thc tnis C3sc: 2 #467 pproath Ycu nuui ideclical Id part {c), only na Urt: initial 4nulc Murd [4t 0i Need Help? Reld I



Answers

Let us fix the co-ordinate axis $x y z$ as shown in the fig. As the ball rolls without slipping along the rigid surface so, on the basis of the solution of problem $1.52$ : $\overrightarrow{v_{0}}=\overrightarrow{v_{c}}+\vec{\omega} \times \overrightarrow{r_{o c}}=0$ Thus $v_{c}=\omega R$ and $\vec{\omega} \uparrow \uparrow(-\vec{k})$ as $\left.\vec{v}_{c} \uparrow \uparrow \vec{i}\right\}$$\mathbf{3 1}$ and $w_{c}=\beta R$ and $\vec{\omega}_{c}+\vec{\beta} \times \vec{r}_{o c}=0$ At the position corresponding to that of Fig., in accordance with the problem, $w_{c}=w$, so $v_{c}=w t$ and $\quad \omega=\frac{v_{c}}{R}=\frac{w t}{R}$ and $\beta=\frac{w}{R}$ (using 1) (a) Let us fix the co-ordinate system with the frame attached with the rigid surface as shown in the Fig. As point $O$ is the instantaneous centre of rotation of the ball at the moment shown in Fig. so, $\overrightarrow{v_{0}}=0$ Now, $\overrightarrow{v_{A}}=\overrightarrow{v_{C}}+\vec{\omega} \times \overrightarrow{r_{A C}}$ $=v_{C} \vec{i}+\omega(-\vec{k}) \times R(\vec{j})=\left(v_{C}+\omega R\right) \vec{i}$ So, $\overrightarrow{v_{A}}=2 v_{C} \vec{i}=2 w t \vec{i}$ (using 1) Similalry $\overrightarrow{v_{B}}=\overrightarrow{v_{C}}+\vec{\omega} \times \overrightarrow{r_{B C}}=v_{C} \vec{i}+\omega(-\vec{k}) \times R(\overrightarrow{i)}$ $=v_{C} \vec{i}+\omega R(-\vec{j})=v_{C} \vec{i}+v_{C}(-\vec{j})$ So, $v_{B}=\sqrt{2} v_{c}=\sqrt{2} w t$ and $\vec{v}_{B}$ is at an angle $45^{\circ}$ from both $\vec{i}$ and $\vec{j}$ (Fig.) (b) $\vec{w}_{0}=\vec{w}_{C}+\omega^{2}\left(-\vec{r}_{o c}\right)+\vec{\beta} \times \vec{r}_{O C}$ $=\omega^{2}\left(-\vec{r}_{O C}\right)=\frac{\nu_{C}^{2}}{R}\left(-\hat{u}_{O C}\right)$ (using 1) where $\hat{u}_{o c}$ is the unit vector along $\vec{r}_{o c}$ so, $w_{0}=\frac{v_{0}^{2}}{R}=\frac{w^{2} t^{2}}{R}$ (using 2) and $\vec{w}_{0}$ is directed towards the centre of the ball Now $\vec{w}_{A}=\vec{w}_{C}+\omega^{2}\left(-\vec{r}_{A C}\right)+\vec{\beta} \times \vec{r}_{A C}$ $=w \vec{i}+\omega^{2} R(-\vec{j})+\beta(-\vec{k}) \times R \vec{j}$


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