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

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