Experiment-shows-that-the-viscous-force-F-on-a-spherical-body-of-radius-r-moving-with-angular-velocity-n-is-F-Kr-a-n-b-c-Where-K-is-a-dimentionless-constant-using-the-method-of-dimentional-

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Question Number 9658 by tawakalitu last updated on 23/Dec/16

$$\mathrm{Experiment}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{viscous}\mathrm{force},\mathrm{F}$$$$\mathrm{on}\mathrm{a}\mathrm{spherical}\mathrm{body}\mathrm{of}\mathrm{radius},\mathrm{r}\mathrm{moving}\mathrm{with}$$$$\mathrm{angular}\mathrm{velocity},\mathrm{n}\mathrm{is},\mathrm{F}={\mathrm{Kr}}^{\mathrm{a}}{\mathrm{n}}^{\mathrm{b}}{\omega }^{\mathrm{c}}.$$$$\mathrm{Where}\mathrm{K}\mathrm{is}\mathrm{a}\mathrm{dimentionless}\mathrm{constant}.$$$$\mathrm{using}\mathrm{the}\mathrm{method}\mathrm{of}\mathrm{dimentional}\mathrm{analysis},$$$$\mathrm{determine}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}$$

Commented by ridwan balatif last updated on 23/Dec/16

$$\mathrm{what}\mathrm{is}\omega \mathrm{in}\mathrm{that}\mathrm{question}?$$$$\mathrm{is}\mathrm{that}\mathrm{a}\mathrm{mass}?$$

Commented by ridwan balatif last updated on 23/Dec/16

$$\mathrm{r}\mathrm{is}\mathrm{radius}\left(\mathrm{m}\right)$$$$\mathrm{n}\mathrm{is}\mathrm{angular}\mathrm{velocity}\left(\mathrm{rad}/\mathrm{s}\right)$$$$\mathrm{what}\mathrm{is}\omega ?$$

Commented by sandy_suhendra last updated on 23/Dec/16

$$\mathrm{I}\mathrm{think}\mathrm{that}\mathrm{is}\mathrm{F}={\mathrm{Kr}}^{\mathrm{a}}{\eta }^{\mathrm{b}}{\mathrm{v}}^{\mathrm{c}}$$$$\eta =\mathrm{coefficient}\mathrm{of}\mathrm{viscosity}\left(\mathrm{Pa}\mathrm{s}\right)$$$$\mathrm{r}=\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{ball}\left(\mathrm{m}\right)$$$$\mathrm{v}=\mathrm{velocity}\mathrm{of}\mathrm{the}\mathrm{ball}\left(\mathrm{m}/\mathrm{s}\right)$$$$\mathrm{and}\left[\mathrm{F}\right]=\left[\mathrm{M}\right]\left[\mathrm{L}\right]{\left[\mathrm{T}\right]}^{-2}$$$$\left[\mathrm{r}\right]=\left[\mathrm{L}\right]$$$$\left[\eta \right]=\left[\mathrm{M}\right]{\left[\mathrm{L}\right]}^{-1}{\left[\mathrm{T}\right]}^{-1}$$$$\left[\mathrm{v}\right]=\left[\mathrm{L}\right]{\left[\mathrm{T}\right]}^{-1}$$$$$$$$\mathrm{F}={\mathrm{Kr}}^{\mathrm{a}}{\eta }^{\mathrm{b}}{\mathrm{v}}^{\mathrm{c}}$$$$\left[\mathrm{M}\right]\left[\mathrm{L}\right]{\left[\mathrm{T}\right]}^{-2}={\left[\mathrm{L}\right]}^{\mathrm{a}}\left\{{\left[\mathrm{M}\right]}^{\mathrm{b}}{\left[\mathrm{L}\right]}^{-\mathrm{b}}{\left[\mathrm{T}\right]}^{-\mathrm{b}}\right\}\left\{{\left[\mathrm{L}\right]}^{\mathrm{c}}{\left[\mathrm{T}\right]}^{-\mathrm{c}}\right\}$$$$\left[\mathrm{M}\right]\left[\mathrm{L}\right]{\left[\mathrm{T}\right]}^{-2}={\left[\mathrm{M}\right]}^{\mathrm{b}}{\left[\mathrm{L}\right]}^{\mathrm{a}-\mathrm{b}+\mathrm{c}}{\left[\mathrm{T}\right]}^{-\mathrm{b}-\mathrm{c}}$$$$\mathrm{so}\mathrm{b}=1$$$$-\mathrm{b}-\mathrm{c}=-2\Rightarrow \mathrm{c}=1$$$$\mathrm{a}-\mathrm{b}+\mathrm{c}=1\Rightarrow \mathrm{a}=1$$

Commented by ridwan balatif last updated on 23/Dec/16

$$\mathrm{now}\mathrm{i}\mathrm{understand}$$

Commented by tawakalitu last updated on 23/Dec/16

$$\mathrm{i}\mathrm{really}\mathrm{appreciate}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Answered by ridwan balatif last updated on 23/Dec/16

$$\mathrm{dimentional}$$$$\mathrm{F}={\mathrm{kgms}}^{-2}$$$$=\left[\mathrm{M}\right]\left[\mathrm{L}\right]{\left[\mathrm{T}\right]}^{-2}$$$$\mathrm{r}=\mathrm{m}$$$$=\left[\mathrm{L}\right]$$$$\eta =\mathrm{Pa}.\mathrm{s}={\mathrm{kgm}}^{-1}{\mathrm{s}}^{-1}$$$$=\left[\mathrm{M}\right]{\left[\mathrm{L}\right]}^{-1}{\left[\mathrm{T}\right]}^{-1}$$$$\omega =\mathrm{rad}/\mathrm{s}$$$$={\left[\mathrm{T}\right]}^{-1}$$$$\mathrm{F}=\mathrm{k}.{\mathrm{r}}^{\mathrm{a}}.\eta {.}^{\mathrm{b}}{\omega }^{\mathrm{c}}$$$$\left[\mathrm{M}\right]\left[\mathrm{L}\right]{\left[\mathrm{T}\right]}^{-2}={\left(\left[\mathrm{L}\right]\right)}^{\mathrm{a}}{\left(\left[\mathrm{M}\right]{\left[\mathrm{L}\right]}^{-1}{\left[\mathrm{T}\right]}^{-1}\right)}^{\mathrm{b}}{\left({\left[\mathrm{T}\right]}^{-1}\right)}^{\mathrm{c}}$$$$\left[\mathrm{M}\right]\left[\mathrm{L}\right]{\left[\mathrm{T}\right]}^{-2}={\left[\mathrm{M}\right]}^{\mathrm{b}}{\left[\mathrm{L}\right]}^{\mathrm{a}-\mathrm{b}}{\left[\mathrm{T}\right]}^{-\mathrm{b}-\mathrm{c}}$$$$\mathrm{from}\mathrm{dimentional}\left[\mathrm{M}\right]\mathrm{we}\mathrm{get}\mathrm{b}=1$$$$\mathrm{from}\mathrm{dimentional}\left[\mathrm{L}\right]\mathrm{we}\mathrm{get}\mathrm{a}-\mathrm{b}=1$$$$\mathrm{a}-1=1$$$$\mathrm{a}=2$$$$\mathrm{from}\mathrm{dimentional}\left[\mathrm{T}\right]\mathrm{we}\mathrm{get}-\mathrm{b}-\mathrm{c}=-2$$$$-1-\mathrm{c}=-2$$$$\mathrm{c}=1$$$$\mathrm{so},\mathrm{a}=2\mathrm{b}=1\mathrm{c}=1$$$$\mathrm{F}={\mathrm{kr}}^2\eta \omega$$

Commented by sandy_suhendra last updated on 23/Dec/16

$$\mathrm{I}\mathrm{think}\mathrm{that}\mathrm{is}\mathrm{not}\omega \left(\mathrm{rad}/\mathrm{s}\right)\mathrm{but}\mathrm{v}=\mathrm{linear}\mathrm{velocity}\left(\mathrm{m}/\mathrm{s}\right)$$$$\mathrm{because}\mathrm{the}\mathrm{formula}\mathrm{must}\mathrm{be}\mathrm{F}=6\pi \mathrm{r}\eta \mathrm{v}$$$$\mathrm{and}\mathrm{K}=6\pi \left[{\mathrm{Stokes}}^{\prime }\mathrm{Law}\right]$$

Commented by ridwan balatif last updated on 23/Dec/16

$$\mathrm{yeah},\mathrm{i}\mathrm{remember}\mathrm{now}$$$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by tawakalitu last updated on 23/Dec/16

$$\mathrm{i}\mathrm{really}\mathrm{appreciate},\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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