Question Number 103143 by ajfour last updated on 13/Jul/20
Commented by ajfour last updated on 13/Jul/20
$${Q}.\mathrm{103072}\:\:\:\:\left({A}\:{revisit}\right) \\ $$
Answered by ajfour last updated on 13/Jul/20
![lets the frictional force be f(θ). work-energy eq_(−) mgrsin θ=(1/2)MV^2 +∫fdx +(1/2)m[(usin θ−V)^2 +u^2 cos^2 θ] Impulse-momentum eq._(−) −MV+m(usin θ−V)=∫fdt friction force, normal N, acc. A_(−) N+mAcos θ−mgsin θ=((mu^2 )/r) f=μ(Nsin θ+Mg) Ncos θ−f=MA (everything much interdependent so not easy to resolve even if we desire to know just final V, u ....)](Q103146.png)
$${lets}\:\:{the}\:{frictional}\:{force}\:{be}\:{f}\left(\theta\right). \\ $$$$\underset{−} {{work}-{energy}\:{eq}} \\ $$$${mgr}\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}{MV}^{\mathrm{2}} +\int{fdx} \\ $$$$\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}{m}\left[\left({u}\mathrm{sin}\:\theta−{V}\right)^{\mathrm{2}} +{u}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta\right] \\ $$$$\underset{−} {{Impulse}-{momentum}\:{eq}.} \\ $$$$−{MV}+{m}\left({u}\mathrm{sin}\:\theta−{V}\right)=\int{fdt} \\ $$$$ \\ $$$$\underset{−} {{friction}\:{force},\:{normal}\:{N},\:{acc}.\:{A}} \\ $$$${N}+{mA}\mathrm{cos}\:\theta−{mg}\mathrm{sin}\:\theta=\frac{{mu}^{\mathrm{2}} }{{r}} \\ $$$$\:\:{f}=\mu\left({N}\mathrm{sin}\:\theta+{Mg}\right) \\ $$$$\:\:\:{N}\mathrm{cos}\:\theta−{f}={MA} \\ $$$$\left({everything}\:{much}\:{interdependent}\right. \\ $$$$\:{so}\:{not}\:{easy}\:{to}\:{resolve}\:{even}\:{if}\:{we} \\ $$$$\left.{desire}\:{to}\:{know}\:{just}\:{final}\:{V},\:{u}\:....\right) \\ $$
Commented by Dwaipayan Shikari last updated on 13/Jul/20
$${Is}\:{the}\:{circular}\:{path}\:\left({from}\:{where}\:{the}\:{small}\:{box}\:{slides}\right) \\ $$$${frictionless}?\:{If}\:{friction}\:{is}\:{absent}\:{so}\:{there}\:{is}\:{no}\:{need} \\ $$$${to}\:{calculate}\:{i}\:{think}\:{sir}! \\ $$$${As}\: \\ $$$$\mu\rightarrow\mathrm{0}\:\:\:\:\:\:{Mv}+{mV}=\mathrm{0}\:\:\:\left\{\:\:{Sir}\:{can}\:{i}\:{apply}\:{momentum}\:{conservaion}\:{here}?\right. \\ $$$$\:\:\:\:\:\:\:\:{v}=−\frac{{mV}}{{M}}\:\:\left({v}={velocity}\:{of}\:{the}\:{wedge}\:\:,{V}={velocity}\:{of}\:{th}\:{block}\right. \\ $$$$\:\:\:\:{v}=−\frac{{m}\sqrt{\mathrm{2}{gR}}}{{M}}\:\:\:\:\:\:\:\:\:\left\{\:\frac{\mathrm{1}}{\mathrm{2}}{mV}^{\mathrm{2}} ={mgR}\Rightarrow{V}=\sqrt{\mathrm{2}{gR}}\right. \\ $$$${I}\:\:{am}\:{not}\:{sure} \\ $$$${If}\:{i}\:{am}\:{wrong}\:{then}\:{kindly}\:{rectify}\:{me} \\ $$$$ \\ $$
Commented by ajfour last updated on 13/Jul/20
$${we}\:{are}\:{not}\:{solving}\:{for}\:\mu=\mathrm{0} \\ $$$${thank}\:{you}\:{Sir}. \\ $$