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

Commented by Tawa11 last updated on 26/Dec/24

$$\mathrm{I}\mathrm{got}10.3\mathrm{m}/\mathrm{s}$$

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

$$U^2-u^2=2gh$$$$\tan \theta =\frac{\sqrt{U^2-u^2}}{u}=\frac{2h}{d}$$$$\frac{\sqrt{2gh}}{u}=\frac{2h}{d}$$$$\Rightarrow u=d\sqrt{\frac{g}{2h}}=8\sqrt{\frac{20}{2\times 3}}\approx 10.3m/s$$$$or$$$$t=\frac{d}{u}$$$$h=\frac{{gt}^2}{2}=\frac{{gd}^2}{2u^2}$$$$\Rightarrow u=d\sqrt{\frac{g}{2h}}$$

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

$$rotationalmomentum:$$$$\frac{{mR}^2}{2}({\omega }_0-{\omega }_T)=\mu mgRT$$$$\Rightarrow {\omega }_0-{\omega }_T=\frac{2\mu gT}{R}...\left(i\right)$$$$translationalmomentum:$$$$m(R{\omega }_T-0)=\mu mgT$$$$\Rightarrow {\omega }_T=\frac{\mu gT}{R}...\left(ii\right)$$$$\Rightarrow {\omega }_T=\frac{{\omega }_0}{3}\Rightarrow v_T=\frac{{\omega }_{0}R}{3}$$$$\Rightarrow T=\frac{{\omega }_{0}R}{3\mu g}\Rightarrow s=\frac{v_{T}T}{2}=\frac{{\omega }_0^2{R}^2}{18\mu g}$$

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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