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Question Number 215395 by mr W last updated on 05/Jan/25

Commented by mr W last updated on 06/Jan/25

$$aballwithmassmfallsfromthe$$$$highthandhitstheendoftherod$$$$juttingoutofftheedgeofatable.$$$$thelengthoftheuniformrodwith$$$$massMisL=a+b(a>b).$$$$$$$$1)findthemaximumhightwhich$$$$theballwillreachaftercollision.$$$$$$$$2)findthemaximumhightwhich$$$$therodwillreachaftercollision.$$$$(supposedthatthetableisremoved$$$$awayafterthecollisionsothatthe$$$$rodcanmovefreely).$$

Commented by ajfour last updated on 07/Jan/25

https://youtu.be/2O_ZoICD2Ss?si=r85hss2CP8u60lv6 Motion of charge in variable magnetic field

Commented by mr W last updated on 07/Jan/25

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Answered by mr W last updated on 07/Jan/25

Commented by mr W last updated on 10/Jan/25

Commented by mr W last updated on 10/Jan/25

$$f=\mu N=\mu mg\cos \theta$$$$v=\omega r$$$$I_p=\frac{2{mr}^2}{5}+{mr}^2=\frac{7{mr}^2}{5}$$$$mg(\sin \theta +\mu \cos \theta )s_1=\frac{{mu}^2}{2}-\frac{1}{2}\times \frac{7{mr}^2{\omega }^2}{5}$$$$\Rightarrow g(\sin \theta +\mu \cos \theta )s_1=\frac{u^2}{2}-\frac{7r^2{\omega }^2}{10}$$$$a=(\sin \theta +\mu \cos \theta )g$$$$\alpha =\frac{fr}{I}=\frac{\mu mgr\cos \theta }{\frac{2{mr}^2}{5}}=\frac{5\mu g\cos \theta }{2r}$$$$\frac{u-v}{a}=\frac{u-r\omega }{(\sin \theta +\mu \cos \theta )g}=\frac{\omega }{\alpha }=\frac{2r\omega }{5\mu g\cos \theta }$$$$\frac{u-r\omega }{\tan \theta +\mu }=\frac{2r\omega }{5\mu }$$$$\Rightarrow r\omega =\frac{5\mu u}{7\mu +2\tan \theta }$$$$g(\sin \theta +\mu \cos \theta )s_1=\frac{u^2}{2}-\frac{7}{10}\times {\left(\frac{5\mu u}{7\mu +2\tan \theta }\right)}^2$$$$s_1=\frac{[1-\frac{35{\mu }^2}{{(7\mu +2\tan \theta )}^2}]u^2}{2g(\sin \theta +\mu \cos \theta )}$$$${mgs}_2\sin \theta =\frac{1}{2}\times \frac{7{mr}^2{\omega }^2}{5}$$$$s_2=\frac{7r^2{\omega }^2}{10g\sin \theta }=\frac{7}{10g\sin \theta }\times {\left(\frac{5\mu u}{7\mu +2\tan \theta }\right)}^2$$$$\Rightarrow s_2=\frac{35{\mu }^2{u}^2}{2g\sin \theta {(7\mu +2\tan \theta )}^2}$$$$s=s_1+s_2$$$$\Rightarrow s=\frac{[1+\frac{35{\mu }^2}{{(7\mu +2\tan \theta )}^2}]u^2}{2g(\sin \theta +\mu \cos \theta )}+\frac{35{\mu }^2{u}^2}{2g\sin \theta {(7\mu +2\tan \theta )}^2}$$

Commented by mr W last updated on 08/Jan/25

$$u=\sqrt{2gh}$$$$mv=J_1-mu$$$$V=\frac{(a-b)\omega }{2}$$$$MV=J_2-J_1$$$$\frac{M(a-b)\omega }{2}=J_2-m(v+u)$$$$J_2=\frac{M(a-b)\omega }{2}+m(v+u)$$$$\frac{M{(a+b)}^2\omega }{12}=J_{1}b-\frac{(a-b)J_2}{2}$$$$\frac{M{(a+b)}^2\omega }{12}=m(v+u)b-\frac{(a-b)}{2}[\frac{M(a-b)\omega }{2}+m(v+u)]$$$$\frac{(a^2+b^2-ab)M\omega }{3}=\frac{m(3b-a)(v+u)}{2}$$$$\Rightarrow v=\frac{2M(a^2+b^2-ab)\omega }{3m(3b-a)}-u$$$$\frac{m(u^2-v^2)}{2}=\frac{{MV}^2}{2}+\frac{1}{2}\times \frac{M{(a+b)}^2{\omega }^2}{12}$$$$\frac{m(u^2-v^2)}{M}=\frac{{(a-b)}^2{\omega }^2}{4}+\frac{{(a+b)}^2{\omega }^2}{12}$$$$v^2=u^2-\frac{M(a^2+b^2-ab){\omega }^2}{3m}$$$$with\mu =\frac{m}{M}$$$${[\frac{2(a^2+b^2-ab)\omega }{3\mu (3b-a)}-u]}^2=u^2-\frac{(a^2+b^2-ab){\omega }^2}{3\mu }$$$$\Rightarrow \omega =\frac{12\mu (3b-a)u}{4(a^2+b^2-ab)+3\mu {(3b-a)}^2}$$$$x_A=\frac{(a+b)\cos \left(\omega t\right)}{2}$$$$y_A=Vt-\frac{{gt}^2}{2}+\frac{(a+b)\sin \left(\omega t\right)}{2}$$$$y_A=\frac{(a-b)\omega t}{2}-\frac{{gt}^2}{2}+\frac{(a+b)\sin \left(\omega t\right)}{2}$$$$v=[\frac{8}{4+\frac{3\mu {(3b-a)}^2}{a^2+b^2-ab}}-1]u$$$$v=0when\mu =\frac{4(a^2+b^2-ab)}{3{(3b-a)}^2}$$$$forv>0,i.e.whentheballrebounds$$$$fromtherod,itcanreachamaximum$$$$hight{h}_{max}with$$$$\frac{h_{max}}{h}={\left(\frac{v}{u}\right)}^2={[\frac{8}{4+\frac{3\mu {(3b-a)}^2}{a^2+b^2-ab}}-1]}^2$$$$with\frac{b}{a}=\lambda$$$$\frac{h_{max}}{h}={[\frac{8}{4+\frac{3\mu {(3\lambda -1)}^2}{{\lambda }^2-\lambda +1}}-1]}^2$$

Commented by ajfour last updated on 08/Jan/25

https://youtu.be/3wiUgK_riz8?si=pKPRqyotoxf-pwhm How far up the incline does solid ball reach?

Commented by mr W last updated on 08/Jan/25

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Commented by mr W last updated on 09/Jan/25

Commented by mr W last updated on 09/Jan/25

Commented by mr W last updated on 09/Jan/25

Commented by mr W last updated on 09/Jan/25

Commented by ajfour last updated on 10/Jan/25

Thanks for this. by the way any error/blunder in my video lecture for this question sir, did you notice?

Commented by mr W last updated on 10/Jan/25

$$igotx=s_1=\frac{[1-\frac{35{\mu }^2}{{(7\mu +2\tan \theta )}^2}]u^2}{2g(\sin \theta +\mu \cos \theta )}$$$$whileyougot$$$$x=\frac{[1-\frac{25{\mu }^2}{{(7\mu +2\tan \theta )}^2}]u^2}{2g(\sin \theta +\mu \cos \theta )}$$

Commented by ajfour last updated on 11/Jan/25

$$\Rightarrow g(\sin \theta +\underset{―}{\mu \cos \theta })s_1=\frac{u^2}{2}-\frac{7r^2{\omega }^2}{10}$$$$Pointofapplicationoffrictionforcemoves$$$$through{s}_1-r\mathit{\varphi }soaboveeq.shouldbe:$$$$g\left(\sin \theta \right)s_1+\left(\mu g\cos \theta \right)(s_1-r\mathit{\varphi })=\frac{u^2}{2}-\frac{7r^2{\omega }^2}{10}$$$$imeaninclineseesspheresurface$$$$hasrubbedpastthislength=s_1-r\mathit{\varphi }$$$$\frac{v^2}{r^2}={\omega }^2=2\left(\frac{\mu mgr\cos \theta }{\frac{2}{5}{mr}^2}\right)\mathit{\varphi }$$$$\frac{v^2}{r^2}={\omega }^2=\left(\frac{5\mu g}{r}\cos \theta \right)\varphi$$$$\mu rg\mathit{\varphi }\cos \theta =\frac{v^2}{5}=\frac{{\omega }^2{r}^2}{5}hence$$$$s_1(\sin \theta +\mu \cos \theta )g-\frac{2{\omega }^2{r}^2}{10}=\frac{u^2}{2}-\frac{7{\omega }^2{r}^2}{10}$$$$s_1=\frac{u^2-{\omega }^2{r}^2}{2g(\sin \theta +\mu \cos \theta )}$$$$nowasr\omega =\frac{5\mu u}{7\mu +2\tan \theta }$$$$s_1=\frac{[1-\frac{25{\mu }^2}{{(7\mu +2\tan \theta )}^2}]u^2}{2g(\sin \theta +\mu \cos \theta )}★$$$$Alongthisapproachiwouldgetthis,sir!$$

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

$$youarerightsir,thanksalot!$$

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