Question-215032

Question Number 215032 by Tawa11 last updated on 26/Dec/24

Commented by Tawa11 last updated on 26/Dec/24

$$\mathrm{I}\:\mathrm{got}\:\:\mathrm{10}.\mathrm{3}\:\mathrm{m}/\mathrm{s} \\ $$

Answered by mr W last updated on 26/Dec/24

U^2 −u^2 =2gh tan θ=((√(U^2 −u^2 ))/u)=((2h)/d) ((√(2gh))/u)=((2h)/d) ⇒u=d(√(g/(2h)))=8(√((20)/(2×3)))≈10.3 m/s or t=(d/u) h=((gt^2 )/2)=((gd^2 )/(2u^2 )) ⇒u=d(√(g/(2h)))

$${U}^{\mathrm{2}} −{u}^{\mathrm{2}} =\mathrm{2}{gh} \\ $$$$\mathrm{tan}\:\theta=\frac{\sqrt{{U}^{\mathrm{2}} −{u}^{\mathrm{2}} }}{{u}}=\frac{\mathrm{2}{h}}{{d}} \\ $$$$\frac{\sqrt{\mathrm{2}{gh}}}{{u}}=\frac{\mathrm{2}{h}}{{d}} \\ $$$$\Rightarrow{u}={d}\sqrt{\frac{{g}}{\mathrm{2}{h}}}=\mathrm{8}\sqrt{\frac{\mathrm{20}}{\mathrm{2}×\mathrm{3}}}\approx\mathrm{10}.\mathrm{3}\:{m}/{s} \\ $$$${or} \\ $$$${t}=\frac{{d}}{{u}} \\ $$$${h}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=\frac{{gd}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} } \\ $$$$\Rightarrow{u}={d}\sqrt{\frac{{g}}{\mathrm{2}{h}}} \\ $$

Commented by ajfour last updated on 26/Dec/24

For both of you to watch this 13 min lecture of mine solving a rotation with slipping question, physics. Thank you. https://youtu.be/Hg4sQD7xK9g?si=OMljkOWfCw2sIKV4

Commented by mr W last updated on 27/Dec/24

$${agree} \\ $$

Commented by mr W last updated on 27/Dec/24

rotational momentum: ((mR^2 )/2)(ω_0 −ω_T )=μmgRT ⇒ω_0 −ω_T =((2μgT)/R) ...(i) translational momentum: m(Rω_T −0)=μmgT ⇒ω_T =((μgT)/R) ...(ii) ⇒ω_T =(ω_0 /3) ⇒v_T =((ω_0 R)/3) ⇒T=((ω_0 R)/(3μg)) ⇒s=((v_T T)/2)=((ω_0 ^2 R^2 )/(18μg))

$${rotational}\:{momentum}: \\ $$$$\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\left(\omega_{\mathrm{0}} −\omega_{{T}} \right)=\mu{mgRT} \\ $$$$\Rightarrow\omega_{\mathrm{0}} −\omega_{{T}} =\frac{\mathrm{2}\mu{gT}}{{R}}\:\:\:…\left({i}\right) \\ $$$${translational}\:{momentum}: \\ $$$${m}\left({R}\omega_{{T}} −\mathrm{0}\right)=\mu{mgT} \\ $$$$\Rightarrow\omega_{{T}} =\frac{\mu{gT}}{{R}}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\omega_{{T}} =\frac{\omega_{\mathrm{0}} }{\mathrm{3}}\:\Rightarrow{v}_{{T}} =\frac{\omega_{\mathrm{0}} {R}}{\mathrm{3}}\: \\ $$$$\Rightarrow{T}=\frac{\omega_{\mathrm{0}} {R}}{\mathrm{3}\mu{g}}\:\Rightarrow{s}=\frac{{v}_{{T}} {T}}{\mathrm{2}}=\frac{\omega_{\mathrm{0}} ^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{18}\mu{g}} \\ $$

Commented by ajfour last updated on 27/Dec/24

Thank you sir. Brilliant alternative way. wow!

Commented by Tawa11 last updated on 27/Dec/24

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Commented by ajfour last updated on 27/Dec/24

https://youtu.be/Ofwdgcvu2pg?si=fthaRVxGBuyalxko To determine radius of the circle shown.

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