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Question Number 30165 by rahul 19 last updated on 17/Feb/18

Commented by ajfour last updated on 19/Feb/18

$$letusfirstchooseafixedx-axis.$$$$\overline{\omega }=\omega \hat{k}andattimet\theta =\omega t$$$$\overline{r}=r(\cos \theta \hat{i}+\sin \theta \hat{j})$$$$=r(\cos \omega t\hat{i}+\sin \omega t\hat{j})=r{\hat{e}}_r$$$$\overline{v}=\frac{dr}{dt}{\hat{e}}_r+\omega r(-\sin \omega t\hat{i}+\cos \omega t\hat{j})$$$$=\frac{dr}{dt}{\hat{e}}_r+\omega r{\hat{e}}_{\theta }$$$$\overline{a}=\frac{d^{2}r}{{dt}^2}{\hat{e}}_r+\omega \frac{dr}{dt}{\hat{e}}_{\theta }+\omega \frac{dr}{dt}{\hat{e}}_{\theta }-{\omega }^{2}r{\hat{e}}_r$$$$asthereiscompleteabsenceof$$$$forcealongradialdirection,so$$$$radialcomponentofacceleration$$$$iszero.$$$$\Rightarrow \frac{d^{2}r}{{dt}^2}-{\omega }^{2}r=0$$$$let{v}_r=\frac{dr}{dt}$$$$\Rightarrow \frac{v_r{dv}_r}{dr}={\omega }^{2}r$$$${\int }_0^{v_r}{v}_{r}dr={\omega }^2{\int }_{R/2}^{r}rdr$$$$v_r^2={\omega }^2(r^2-\frac{R^2}{4})$$$$v_r=\frac{dr}{dt}=\omega \sqrt{r^2-{\left(\frac{R}{2}\right)}^2}$$$${\int }_{R/2}^r\frac{dr}{\sqrt{r^2-{\left(\frac{R}{2}\right)}^2}}=\omega {\int }_0^{t}dt$$$$\Rightarrow \ln (r+\sqrt{r^2-\frac{R^2}{4}}){\mid }_{R/2}^r=\omega t$$$$\Rightarrow r+\sqrt{r^2-\frac{R^2}{4}}=\frac{R}{2}{e}^{\omega t}...\left(i\right)$$$$rationalisingandtaking$$$$reciprocalwealsoget,$$$$r-\sqrt{r^2-\frac{R^2}{4}}=\frac{R}{2}{e}^{-\omega t}...\left(ii\right)$$$$hencebyadding\left(i\right)and\left(ii\right)$$$$\mathit{r}=\frac{\mathit{R}}{4}({\mathit{e}}^{\mathit{\omega }\mathit{t}}+{\mathit{e}}^{-\mathit{\omega }\mathit{t}}).$$$$Force\left(tangential\right)onblockby$$$$disc=\overline{F}=m\left(2\omega \frac{dr}{dt}\right){\hat{e}}_{\theta }$$$$\overline{F}=2m\omega v_r{\hat{e}}_{\theta }$$$$=\left(2m{\omega }^2\sqrt{r^2-\frac{R^2}{4}}\right){\hat{e}}_{\theta }$$$$\left(i\right)-\left(ii\right)gives$$$$\sqrt{r^2-\frac{R^2}{4}}=\frac{R}{4}(e^{\omega t}-e^{-\omega t})$$$$hence\overline{F}=\frac{m{\omega }^{2}R}{2}(e^{\omega t}-e^{-\omega t}){\hat{e}}_{\theta }$$$$andifxaxisischosentobe$$$$alongradialdirectionrotating$$$$alongwithdiscthen\hat{i}isalong{\hat{e}}_r$$$$and\hat{j}isalong{\hat{e}}_{\theta }.$$$$Then\overline{F}=\frac{m{\omega }^{2}R}{2}(e^{\omega t}-e^{-\omega t})\hat{j}$$$$discalsosupportsweightofblock$$$$hencewealsohaveaforce$$$$N_z=mg\hat{k}$$$$\overline{\mathit{N}}thereactionforceofdiscon$$$$blockis\overline{N}=\overline{F}+mg\hat{k}$$$$\overline{N}=\frac{m{\omega }^{2}R}{2}(e^{\omega t}-e^{\omega t})\hat{j}+mg\hat{k}.$$

Commented by ajfour last updated on 19/Feb/18

Commented by rahul 19 last updated on 19/Feb/18

$$\mathcal{W}\mathfrak{O}\mathcal{W}\dots !$$

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