Question Number 212941 by MrGaster last updated on 27/Oct/24
$$\:\:\:\:\:\:\:\:{Mathematival}\:{proof}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{F}={m}\omega^{\mathrm{2}} {r} \\ $$
Commented by MrGaster last updated on 27/Oct/24
What it means here is "using pure mathematics to prove" "
Answered by ajfour last updated on 27/Oct/24
Commented by ajfour last updated on 27/Oct/24
![Consider particle describing uniform circular motion. ⇒ equal change in angle in equal intervals of time. so ∣v_1 ∣=∣v_2 ∣=u=((s(arc lenth))/(Δt)) ω=((Δθ)/(Δt))=(((rΔθ))/(rΔt))=((s(arc length ))/(rΔt))=(u/r) Δt=((Δθ)/ω) F_(avg) Δt (Impulse)=Δp=mv_2 −mv_1 =m[2usin (((Δθ)/2))] F_(avg) =((m[2usin (((Δθ)/2))])/(Δt)) F_(inst) =lim_(Δθ→0) {m[2usin (((Δθ)/2))]×(1/(Δt))} F_(inst) =lim_(Δθ→0) {m[2usin (((Δθ)/2))]×(ω/(Δθ))} =mωulim_(Δθ→0) (((2sin (((Δθ)/2)))/(Δθ))) =mω^2 rlim_(((Δθ)/2) → 0) (((sin (((Δθ)/2)))/((((Δθ)/2))))) F_(centripetal) =mω^2 r×1 =mω^2 r](https://www.tinkutara.com/question/Q212947.png)
$${Consider}\:{particle}\:{describing}\:{uniform} \\ $$$${circular}\:{motion}.\:\Rightarrow\:{equal}\:{change} \\ $$$${in}\:{angle}\:{in}\:{equal}\:{intervals}\:{of}\:{time}. \\ $$$${so}\:\mid{v}_{\mathrm{1}} \mid=\mid{v}_{\mathrm{2}} \mid={u}=\frac{{s}\left({arc}\:{lenth}\right)}{\Delta{t}} \\ $$$$\omega=\frac{\Delta\theta}{\Delta{t}}=\frac{\left({r}\Delta\theta\right)}{{r}\Delta{t}}=\frac{{s}\left({arc}\:{length}\:\right)}{{r}\Delta{t}}=\frac{{u}}{{r}} \\ $$$$\Delta{t}=\frac{\Delta\theta}{\omega} \\ $$$${F}_{{avg}} \Delta{t}\:\left({Impulse}\right)=\Delta{p}={mv}_{\mathrm{2}} −{mv}_{\mathrm{1}} \\ $$$$={m}\left[\mathrm{2}{u}\mathrm{sin}\:\left(\frac{\Delta\theta}{\mathrm{2}}\right)\right] \\ $$$${F}_{{avg}} =\frac{{m}\left[\mathrm{2}{u}\mathrm{sin}\:\left(\frac{\Delta\theta}{\mathrm{2}}\right)\right]}{\Delta{t}} \\ $$$${F}_{{inst}} =\underset{\Delta\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{{m}\left[\mathrm{2}{u}\mathrm{sin}\:\left(\frac{\Delta\theta}{\mathrm{2}}\right)\right]×\frac{\mathrm{1}}{\Delta{t}}\right\} \\ $$$${F}_{{inst}} =\underset{\Delta\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{{m}\left[\mathrm{2}{u}\mathrm{sin}\:\left(\frac{\Delta\theta}{\mathrm{2}}\right)\right]×\frac{\omega}{\Delta\theta}\right\} \\ $$$$\:\:\:\:\:\:\:\:={m}\omega{u}\underset{\Delta\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2sin}\:\left(\frac{\Delta\theta}{\mathrm{2}}\right)}{\Delta\theta}\right) \\ $$$$\:\:\:\:\:\:\:={m}\omega^{\mathrm{2}} {r}\underset{\frac{\Delta\theta}{\mathrm{2}}\:\rightarrow\:\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\left(\frac{\Delta\theta}{\mathrm{2}}\right)}{\left(\frac{\Delta\theta}{\mathrm{2}}\right)}\right) \\ $$$$\:\:{F}_{{centripetal}} ={m}\omega^{\mathrm{2}} {r}×\mathrm{1}\:\:={m}\omega^{\mathrm{2}} {r}\: \\ $$$$ \\ $$
Commented by MrGaster last updated on 28/Oct/24
Thank you, sir.