Question Number 53841 by rahul 19 last updated on 26/Jan/19
Commented by rahul 19 last updated on 27/Jan/19
$$\left.{Mrw}\:{sir},\:\mathrm{1}\right){when}\:{do}\:{we}\:{apply}\:{formula} \\ $$$$\left.{K}\left..{E}=\:\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} \:?\:\:\:\:\:\mathrm{2}\right)\right)\:{what}\:{is}\:{the}\:{diff}. \\ $$$${between}\:{circular}\:\&\:{rotational}\:{motion}\: \\ $$$${for}\:{a}\:{point}\:\:{object}\:{having}\:{mass}\:{m}\:?? \\ $$
Commented by rahul 19 last updated on 26/Jan/19
$$\Delta{U}\:+\:\Delta{K}=\mathrm{0} \\ $$$$\:{Now}\:\Delta{U}=−{mg}\left(\mathrm{2}{R}\right)\: \\ $$$${and}\:\Delta{K}=\:\frac{\mathrm{1}}{\mathrm{2}}\:{I}\omega^{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{2}}\left({mR}^{\mathrm{2}} +\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{{v}}{{R}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\Delta{K}=\:\frac{\mathrm{3}}{\mathrm{4}}{mv}^{\mathrm{2}} . \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}\:{mv}^{\mathrm{2}} =\:\mathrm{2}{mgR} \\ $$$$\Rightarrow\:{v}=\:\sqrt{\frac{\mathrm{8}}{\mathrm{3}}{gR}}. \\ $$
Commented by rahul 19 last updated on 26/Jan/19
$${My}\:{doubt}:\:{I}\:'{ve}\:{studied}\:{that}\:{we}\:{can} \\ $$$${apply}\:{the}\:{formula}\::\:\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} \:{only}\:{when} \\ $$$${the}\:{system}\:{performs}\:{pure}\:{rotation}. \\ $$$${But}\:{here}\:{bead}\:{is}\:{performing}\:{circular} \\ $$$${motion}\:\left({only}\:{disc}\:{is}\:{performing}\right. \\ $$$$\left.{pure}\:{rotation}\right),\:{then}\:{why}\:{the}\:{above} \\ $$$${method}\:{is}\:{giving}\:{correct}\:{answer}\:? \\ $$
Commented by mr W last updated on 26/Jan/19
$${each}\:{single}\:{point}\:{on}\:{the}\:{disc}\:{performs} \\ $$$${circular}\:{motion}.\:{rotation}\:{of}\:{an}\:{object}= \\ $$$${circular}\:{motion}\:{of}\:{all}\:{its}\:{points}\:{about} \\ $$$${the}\:{same}\:{axis}! \\ $$$${bead}\:{is}\:{here}\:{a}\:{normal}\:{point}\:{on}\:{the}\:{disc} \\ $$$${like}\:{every}\:{other}\:{point}\:{on}\:{it},\:{they}\:{all} \\ $$$${make}\:{circular}\:{motion}\:{about}\:{the}\:{same} \\ $$$${point}.\:{the}\:{difference}\:{is}\:{that}\:{the}\:{disc} \\ $$$${with}\:{the}\:{bead}\:{has}\:{an}\:{other}\:{I}\:{than}\:{the} \\ $$$${disc}\:{without}\:{bead}.\:{but}\:{you}\:{have}\:{taken} \\ $$$${this}\:{into}\:{consideration},\:{therefore} \\ $$$${you}\:{got}\:{the}\:{right}\:{result}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jan/19
$${moment}\:{of}\:{inertia}={M}.{I}\:{of}\:{disc}+{M}.{I}\:{of}\:{welded}\:{bead} \\ $$$$\:\:=\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}+{mR}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by mr W last updated on 26/Jan/19
$${rotation}\:{of}\:{a}\:{body}\:=\Sigma\:{circular}\:{motion}\:{of} \\ $$$${all}\:{its}\:{masses} \\ $$$${K}.{E}.=\int\frac{\mathrm{1}}{\mathrm{2}}{dmv}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\int{r}^{\mathrm{2}} \omega^{\mathrm{2}} {dm}=\frac{\mathrm{1}}{\mathrm{2}}\left(\int{r}^{\mathrm{2}} {dm}\right)\omega^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} \\ $$
Commented by rahul 19 last updated on 27/Jan/19
$${Thank}\:{you}\:{Sir}! \\ $$$${Q}.\mathrm{2}\:?? \\ $$
Commented by mr W last updated on 27/Jan/19
$${again}:\:{circular}\:{motion}={rotation} \\ $$$${a}\:{point}\:{mass}\:{is}\:{also}\:{an}\:{object}. \\ $$$${KE}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mr}^{\mathrm{2}} \omega^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} \\ $$$${I}={mr}^{\mathrm{2}} ={IOM}\:{of}\:{the}\:{mass} \\ $$$${the}\:{physics}\:{for}\:{circular}\:{motion}\:{and} \\ $$$${rotation}\:{is}\:{the}\:{same}.\:{this}\:{is}\:{my} \\ $$$${understanding}. \\ $$