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Question Number 44454 by ajfour last updated on 29/Sep/18
Commented by ajfour last updated on 29/Sep/18
A sphere in pure rolling motion on a rough ground, encounters a stair, collides completely inelastically with its corner. Just after it comes atop the stair, find its velocity in terms of u, h, and R.
$${A}\:{sphere}\:{in}\:{pure}\:{rolling}\:{motion} \\ $$$${on}\:{a}\:{rough}\:{ground},\:{encounters}\:{a}\:{stair}, \\ $$$${collides}\:{completely}\:{inelastically} \\ $$$${with}\:{its}\:{corner}.\:{Just}\:{after}\:{it} \\ $$$${comes}\:{atop}\:{the}\:{stair},\:{find}\:{its} \\ $$$${velocity}\:{in}\:{terms}\:{of}\:\boldsymbol{{u}},\:\boldsymbol{{h}},\:{and}\:\boldsymbol{{R}}. \\ $$
Commented by ajfour last updated on 30/Sep/18
Commented by ajfour last updated on 30/Sep/18
Initial angular momentum of sphere about corner P : L_0 = ((2/5)MR^2 )(u/R)+Mu(R−h) just after collision L_1 = (7/5)MR^2 ((v_1 /R)) about P we can conserve amgular momentum, so L_1 = L_0 ⇒ ((7v_1 R)/5) = ((2uR)/5)+u(R−h) ...(i) v_1 = ((2u)/7)+((5u)/7)−(5/7)((h/R))u ⇒ v_1 = (1−((5h)/(7R)))u = λu Now as it rises above, (1/2)((7/5)MR^2 )((v_1 /R))^2 = Mgh + (1/2)((7/5)MR^2 )((v_2 /R))^2 using (i): (1/2)((7/5))λ^2 u^2 = gh+ (1/2)((7/5))v_2 ^2 ⇒ v_2 ^2 = λ^2 u^2 − ((10)/7)gh v_2 ^2 = (1−((5h)/(7R)))^2 u^2 −((10)/7)gh ⇒ v_2 = (√((1−((5h)/(7R)))^2 u^2 −((10)/7)gh)) .
$${Initial}\:{angular}\:{momentum} \\ $$$${of}\:{sphere}\:{about}\:{corner}\:{P}\:: \\ $$$$\:{L}_{\mathrm{0}} \:=\:\left(\frac{\mathrm{2}}{\mathrm{5}}{MR}^{\mathrm{2}} \right)\frac{{u}}{{R}}+{Mu}\left({R}−{h}\right) \\ $$$${just}\:{after}\:{collision} \\ $$$$\:\:{L}_{\mathrm{1}} \:=\:\frac{\mathrm{7}}{\mathrm{5}}{MR}^{\mathrm{2}} \left(\frac{{v}_{\mathrm{1}} }{{R}}\right) \\ $$$${about}\:{P}\:{we}\:{can}\:{conserve}\:{amgular} \\ $$$${momentum},\:{so}\:\:\:{L}_{\mathrm{1}} \:=\:{L}_{\mathrm{0}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{7}{v}_{\mathrm{1}} {R}}{\mathrm{5}}\:=\:\frac{\mathrm{2}{uR}}{\mathrm{5}}+{u}\left({R}−{h}\right)\:…\left({i}\right) \\ $$$$\:\:{v}_{\mathrm{1}} \:=\:\frac{\mathrm{2}{u}}{\mathrm{7}}+\frac{\mathrm{5}{u}}{\mathrm{7}}−\frac{\mathrm{5}}{\mathrm{7}}\left(\frac{{h}}{{R}}\right){u} \\ $$$$\Rightarrow\:\:\:\:{v}_{\mathrm{1}} =\:\left(\mathrm{1}−\frac{\mathrm{5}{h}}{\mathrm{7}{R}}\right){u}\:=\:\lambda{u} \\ $$$${Now}\:{as}\:{it}\:{rises}\:{above}, \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{7}}{\mathrm{5}}{MR}^{\mathrm{2}} \right)\left(\frac{{v}_{\mathrm{1}} }{{R}}\right)^{\mathrm{2}} \:=\:{Mgh} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{7}}{\mathrm{5}}{MR}^{\mathrm{2}} \right)\left(\frac{{v}_{\mathrm{2}} }{{R}}\right)^{\mathrm{2}} \\ $$$${using}\:\left({i}\right): \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{7}}{\mathrm{5}}\right)\lambda^{\mathrm{2}} {u}^{\mathrm{2}} \:=\:{gh}+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{7}}{\mathrm{5}}\right){v}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{v}_{\mathrm{2}} ^{\mathrm{2}} \:=\:\lambda^{\mathrm{2}} {u}^{\mathrm{2}} \:−\:\frac{\mathrm{10}}{\mathrm{7}}{gh} \\ $$$$\:\:\:\:{v}_{\mathrm{2}} ^{\mathrm{2}} \:=\:\left(\mathrm{1}−\frac{\mathrm{5}{h}}{\mathrm{7}{R}}\right)^{\mathrm{2}} {u}^{\mathrm{2}} −\frac{\mathrm{10}}{\mathrm{7}}{gh} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{v}}_{\mathrm{2}} \:=\:\sqrt{\left(\mathrm{1}−\frac{\mathrm{5}\boldsymbol{{h}}}{\mathrm{7}\boldsymbol{{R}}}\right)^{\mathrm{2}} \boldsymbol{{u}}^{\mathrm{2}} −\frac{\mathrm{10}}{\mathrm{7}}\boldsymbol{{gh}}}\:\:. \\ $$
Commented by MrW3 last updated on 30/Sep/18
correct and very nice sir!
$${correct}\:{and}\:{very}\:{nice}\:{sir}! \\ $$

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