Question Number 224807 by fantastic last updated on 05/Oct/25
Commented by fantastic last updated on 05/Oct/25
![It took me 10 to 15 minutes to solve this question.It was not too hard but the question is very interesting Here is the question: A bottle of syllendrical shape of r=5cm is rotating around its axis at ω rad/s. the height difference between the center of the bottle and water edge sticked with the bottle is h h=?[in terms of g,ω,r] g=1000cm/s^2](https://www.tinkutara.com/question/Q224808.png)
$${It}\:{took}\:{me}\:\mathrm{10}\:{to}\:\mathrm{15}\:{minutes}\:{to}\:{solve} \\ $$$${this}\:{question}.{It}\:{was}\:{not}\:{too}\:{hard} \\ $$$${but}\:{the}\:{question}\:{is}\:{very}\:{interesting} \\ $$$${Here}\:{is}\:{the}\:{question}: \\ $$$${A}\:{bottle}\:{of}\:{syllendrical}\:{shape} \\ $$$${of}\:{r}=\mathrm{5}{cm}\:{is}\:{rotating}\:{around}\:{its} \\ $$$${axis}\:{at}\:\omega\:{rad}/{s}. \\ $$$${the}\:{height}\:{difference}\:{between}\:{the}\:{center} \\ $$$${of}\:{the}\:{bottle}\:{and}\:\:{water}\:{edge}\:{sticked} \\ $$$${with}\:{the}\:{bottle}\:{is}\:{h} \\ $$$${h}=?\left[{in}\:{terms}\:{of}\:{g},\omega,{r}\right] \\ $$$${g}=\mathrm{1000}{cm}/{s}^{\mathrm{2}} \\ $$
Answered by mr W last updated on 05/Oct/25
Commented by mr W last updated on 05/Oct/25
$$\mathrm{tan}\:\theta={y}' \\ $$$${mg}\:\mathrm{tan}\:\theta={mx}\omega^{\mathrm{2}} \\ $$$${gy}'={x}\omega^{\mathrm{2}} \\ $$$$\Rightarrow{y}=\frac{\omega^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{g}}+{C}\:{with}\:{C}=\mathrm{0} \\ $$$$\Rightarrow\:{h}=\frac{\omega^{\mathrm{2}} {r}^{\mathrm{2}} }{\mathrm{2}{g}}\:\checkmark \\ $$
Commented by fantastic last updated on 05/Oct/25
$$\:\underline{\underbrace{\lesseqgtr}} \\ $$