Question Number 31187 by rahul 19 last updated on 03/Mar/18
Commented by ajfour last updated on 03/Mar/18
Commented by ajfour last updated on 03/Mar/18
$${v}=\omega\left({R}+{r}\right) \\ $$$${u}=\omega{R} \\ $$$$\Rightarrow\:\:\boldsymbol{{u}}=\frac{\boldsymbol{{vR}}}{\boldsymbol{{R}}+\boldsymbol{{r}}} \\ $$$${displacement}\:{of}\:{point}\:{A}\:{by}\:{the}\:{time} \\ $$$${spool}\:{makes}\:{one}\:{rotation} \\ $$$$\:\:\:\:\:\:{s}_{{A}} \:=\:{v}\:{T}\:=\:{v}\left(\frac{\mathrm{2}\pi}{\omega}\right)\:=\:{v}×\frac{\mathrm{2}\pi}{\left(\frac{{v}}{{R}+{r}}\right)} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{2}\boldsymbol{\pi}\left(\boldsymbol{{R}}+\boldsymbol{{r}}\right) \\ $$$${x}={ut}\:\:\:{and}\:\:\:{x}={R}\mathrm{cot}\:\left(\frac{\theta}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\:{R}\mathrm{cot}\:\left(\frac{\theta}{\mathrm{2}}\right)={ut} \\ $$$$\Rightarrow\:\:\:−\frac{{R}}{\mathrm{2}}\mathrm{cosec}\:^{\mathrm{2}} \left(\theta/\mathrm{2}\right)\frac{{d}\theta}{{dt}}\:={u}=\frac{{vR}}{{R}+{r}} \\ $$$$\Rightarrow\:\:\mid\frac{{d}\theta}{{dt}}\mid=\left(\frac{\mathrm{2}\boldsymbol{{v}}}{\boldsymbol{{R}}+\boldsymbol{{r}}}\right)\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right) \\ $$$${hence}\:\:\:\left(\mathrm{3}\right),\:{and}\:\left(\mathrm{4}\right)\:{are}\:{correct}. \\ $$
Commented by rahul 19 last updated on 03/Mar/18
$${but}\:{correct}\:{ans}.\:{is}\:{ACD}. \\ $$
Commented by ajfour last updated on 03/Mar/18
$$\left({A}\right)\:{isn}'{t}\:{correct}. \\ $$