Question Number 31149 by NECx last updated on 03/Mar/18
$${here}\:{is}\:{a}\:{question}\:{really}\:{troubling} \\ $$$${me}. \\ $$$$ \\ $$$${A}\:{cylindrical}\:{tube}\:{rolling}\:{down}\:{a} \\ $$$${slope}\:{of}\:{inclination}\:\theta\:{moves}\:{a} \\ $$$${distance}\:{L}\:{in}\:{the}\:{time}\:{T}.\:{The} \\ $$$${equation}\:{relating}\:{these}\:{quantities}\:{is} \\ $$$$ \\ $$$$\:\:\:{L}\left(\mathrm{3}+\frac{{a}^{\mathrm{2}} }{{P}}\right)={QT}^{\mathrm{2}} \mathrm{sin}\:\theta\:{where}\:{a}\:{is} \\ $$$${the}\:{internal}\:{radius}\:{of}\:{the}\:{tube}\:{and} \\ $$$${P}\:\:{and}\:{Q}\:{are}\:{constants}.{What}\:{are} \\ $$$${the}\:{units}\:{of}\:{P}\:{and}\:{Q}? \\ $$
Commented by NECx last updated on 03/Mar/18
$${here}\:{is}\:{what}\:{I}\:{did}\:..... \\ $$$$ \\ $$$$\mathrm{2}{gh}={v}^{\mathrm{2}} \left(\mathrm{1}+{K}^{\mathrm{2}} /{r}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$\mathrm{2}{gLsin}\theta={v}^{\mathrm{2}} \left(\mathrm{1}+{k}^{\mathrm{2}} /{r}^{\mathrm{2}} \right) \\ $$$$ \\ $$$${v}^{\mathrm{2}} \left(\mathrm{1}+{k}^{\mathrm{2}} /{r}^{\mathrm{2}} \right)=\mathrm{2}{gL}\mathrm{sin}\:\theta \\ $$$$ \\ $$$$\left(\frac{{L}^{\mathrm{2}} }{{T}^{\mathrm{2}} }\right)\left(\mathrm{1}+{k}^{\mathrm{2}} /{r}^{\mathrm{2}} \right)=\mathrm{2}{gLsin}\theta \\ $$$$ \\ $$$${L}\left(\mathrm{1}+{k}^{\mathrm{2}} /{r}^{\mathrm{2}} \right)=\mathrm{2}{gT}^{\mathrm{2}} {sin}\theta.....\left(\mathrm{1}\right) \\ $$$${L}\left(\mathrm{3}+{a}^{\mathrm{2}} /{P}\right)={QT}^{\mathrm{2}} \mathrm{sin}\:\theta.......\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\therefore\:{P}={L}^{\mathrm{2}} \:{and}\:{Q}={g}={LT}^{−\mathrm{2}} \\ $$$$ \\ $$$${I}\:{really}\:{have}\:{a}\:{doubt}\:{on}\:{this}\:{please} \\ $$$${check}\:{and}\:{correct}.\:{Thanks} \\ $$
Answered by mrW2 last updated on 03/Mar/18
![If you just need to know the units of P and Q, you don′t need to know all the formula, even you don′t need to understand physics. (a^2 /P) has the same unit as 3, so P has the same unit as a^2 , i.e. [L^2 ]. The unit of left side is [L], the right side must also have this unit. sin θ has no unit. T has the unit [T]. ⇒[Q]×[T^2 ]=[L] ⇒[Q]=[(L/T^2 )] i.e. the unit of Q is [(L/T^2 )], e.g. m/s^2 .](Q31155.png)
$${If}\:{you}\:{just}\:{need}\:{to}\:{know}\:{the}\:{units}\:{of} \\ $$$${P}\:{and}\:{Q},\:{you}\:{don}'{t}\:{need}\:{to}\: \\ $$$${know}\:{all}\:{the}\:{formula},\:{even}\:{you}\:{don}'{t} \\ $$$${need}\:{to}\:{understand}\:{physics}. \\ $$$$\frac{{a}^{\mathrm{2}} }{{P}}\:{has}\:{the}\:{same}\:{unit}\:{as}\:\mathrm{3},\:{so}\:{P}\:{has} \\ $$$${the}\:{same}\:{unit}\:{as}\:{a}^{\mathrm{2}} ,\:{i}.{e}.\:\left[{L}^{\mathrm{2}} \right]. \\ $$$${The}\:{unit}\:{of}\:{left}\:{side}\:{is}\:\left[{L}\right],\:{the}\:{right} \\ $$$${side}\:{must}\:{also}\:{have}\:{this}\:{unit}. \\ $$$${sin}\:\theta\:{has}\:{no}\:{unit}. \\ $$$${T}\:{has}\:{the}\:{unit}\:\left[{T}\right]. \\ $$$$\Rightarrow\left[{Q}\right]×\left[{T}^{\mathrm{2}} \right]=\left[{L}\right] \\ $$$$\Rightarrow\left[{Q}\right]=\left[\frac{{L}}{{T}^{\mathrm{2}} }\right] \\ $$$${i}.{e}.\:{the}\:{unit}\:{of}\:{Q}\:{is}\:\left[\frac{{L}}{{T}^{\mathrm{2}} }\right],\:{e}.{g}.\:{m}/{s}^{\mathrm{2}} . \\ $$
Commented by NECx last updated on 03/Mar/18
$${oh}....{Thanks} \\ $$