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Question Number 215593 by mr W last updated on 11/Jan/25

Commented by mr W last updated on 11/Jan/25

$$asolidballwithradius\mathit{r}andmass$$$$\mathit{m}isgivenaspeed\mathit{u}onarough$$$$horizontalsurface.thefriction$$$$coefficientis\mathit{\mu }.$$$$findafterwhatadistanceandwith$$$$whataspeedtheballonlyrollsover$$$$thesurface.$$$$(originalquestionseeyoutube$$$$channelfromajfoursir)$$

Commented by mr W last updated on 11/Jan/25

https://youtu.be/Hg4sQD7xK9g?si=iWhsrRN19lc2lkzb

Answered by mahdipoor last updated on 11/Jan/25

$$M_{cm}=\mu mgr=I\stackrel{..}{\theta }(cw=+)$$$$m{\stackrel{..}{x}}_{cm}=-\mu mg$$$$\Rightarrow \stackrel{.}{\theta }=\frac{\mu mgr}{I}t+{\stackrel{.}{\theta }}_0=\frac{\mu mgr}{I}t$$$$\Rightarrow {\stackrel{.}{x}}_{cm}=-\mu gt+{\stackrel{.}{x}}_0=-\mu gt+u$$$$onlyroll\equiv {\stackrel{.}{x}}_{sur}={\stackrel{.}{x}}_{cm}-r\stackrel{.}{\theta }=0\Rightarrow$$$$t_a=\frac{u}{\mu g(\frac{{mr}^2}{I}+1)}=\frac{I}{\mu g({mr}^2+I)}u$$$$L_a=x_{cm}\left(t_a\right)=-\mu g\frac{t_a^2}{2}+{ut}_a=\frac{I(2{mr}^2+I)}{2\mu g{({mr}^2+I)}^2}{u}^2$$$$u_a={\stackrel{.}{x}}_{cm}\left(t_a\right)=\left(\frac{{mr}^2}{{mr}^2+I}\right)u$$

Commented by mr W last updated on 11/Jan/25

$$thanks!$$

Answered by mr W last updated on 11/Jan/25

Commented by mr W last updated on 11/Jan/25

$$whenonlyrolling:v=\omega r$$$$f=\mu mg$$$$a=-\frac{f}{m}=-\frac{\mu mg}{m}=-\mu g$$$$\alpha =\frac{fr}{I}=\frac{\mu mgr}{\frac{2{mr}^2}{5}}=\frac{5\mu g}{2r}$$$$t_1=\frac{\omega }{\alpha }=\frac{v-u}{a}$$$$\frac{2r\omega }{5\mu g}=-\frac{\omega r-u}{\mu g}$$$$\Rightarrow \omega r=v=\frac{5u}{7}✓$$$$s=\frac{v^2-u^2}{2a}=\frac{u^2-{\left(\frac{5u}{7}\right)}^2}{2\mu g}=\frac{12u^2}{\mu g}✓$$

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