(* Content-type: application/mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 6.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 145, 7] NotebookDataLength[ 113406, 3147] NotebookOptionsPosition[ 104603, 2857] NotebookOutlinePosition[ 105005, 2874] CellTagsIndexPosition[ 104962, 2871] WindowFrame->Normal ContainsDynamic->False*) (* Beginning of Notebook Content *) Notebook[{ Cell[CellGroupData[{ Cell["Phys 261 Hwk #8", "Subsubtitle", CellChangeTimes->{{3.436179744821*^9, 3.436179758259*^9}, { 3.4362193637917285`*^9, 3.4362193665137286`*^9}}], Cell[CellGroupData[{ Cell["Morin 3.69", "Subsection", CellChangeTimes->{{3.4360043137390003`*^9, 3.436004315868*^9}, 3.436179723259*^9, {3.436180164182*^9, 3.436180164796*^9}, { 3.4361802044379997`*^9, 3.436180208902*^9}}], Cell["\<\ Consider a particle moving in two dimensions under the influence of force \ \>", "Text", CellChangeTimes->{{3.4361802660699997`*^9, 3.436180344672*^9}, { 3.436219390958729*^9, 3.436219429741729*^9}}], Cell[BoxData[ RowBox[{"F\[Theta]", " ", "=", " ", RowBox[{"2", " ", "m", " ", RowBox[{ RowBox[{"r", "'"}], "[", "t", "]"}], " ", RowBox[{ RowBox[{"\[Theta]", "'"}], "[", "t", "]"}]}]}]], "Input", CellChangeTimes->{{3.4362194311647286`*^9, 3.4362194442617283`*^9}}], Cell["\<\ Equating with m a\[Theta] = m (r \[Theta]'' + 2 r' \[Theta]') we learn that\ \>", "Text", CellChangeTimes->{{3.4362194585617285`*^9, 3.4362195115617285`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"r", "[", "t", "]"}], "*", RowBox[{ RowBox[{"\[Theta]", "''"}], "[", "t", "]"}]}], " ", "\[Equal]", " ", "0"}]], "Input", CellChangeTimes->{{3.4362195155137286`*^9, 3.436219523550729*^9}}], Cell["\<\ So unless we are at r=0 (when the angle \[Theta] is ill-defined anyway) we \ conclude that \[Theta]''[t] = 0, which is a differential equation we can \ easily solve:\ \>", "Text", CellChangeTimes->{{3.4362195328987284`*^9, 3.4362196011207285`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"DSolve", "[", RowBox[{ RowBox[{ RowBox[{ RowBox[{"\[Theta]", "''"}], "[", "t", "]"}], "\[Equal]", "0"}], ",", RowBox[{"\[Theta]", "[", "t", "]"}], ",", "t"}], "]"}]], "Input", CellChangeTimes->{{3.436219602831729*^9, 3.4362196213957286`*^9}}], Cell[BoxData[ RowBox[{"{", RowBox[{"{", RowBox[{ RowBox[{"\[Theta]", "[", "t", "]"}], "\[Rule]", RowBox[{ RowBox[{"C", "[", "1", "]"}], "+", RowBox[{"t", " ", RowBox[{"C", "[", "2", "]"}]}]}]}], "}"}], "}"}]], "Output", CellChangeTimes->{3.4362196226117287`*^9}] }, Open ]], Cell["\<\ In short, \[Theta](t) = \[Omega] t (choosing to start at \[Theta]=0 at t=0). \ This also tells us where this kind of force might arise---consider attaching a bead to a frictionless rod which is then spun at a constant rate \ \[Omega]. The normal force must then be exactly this funny force Morin introduces out of thin air. How ever we get there, \ the radial equation then reads:\ \>", "Text", CellChangeTimes->{{3.4362196316817284`*^9, 3.4362196595207286`*^9}, { 3.436219812871729*^9, 3.4362198912227287`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"eqnr", " ", "=", " ", RowBox[{ RowBox[{"(", RowBox[{"Fr", " ", "/.", " ", RowBox[{"{", RowBox[{"Fr", "\[Rule]", "0"}], "}"}]}], ")"}], " ", "\[Equal]", " ", RowBox[{"m", " ", RowBox[{"(", " ", RowBox[{ RowBox[{ RowBox[{"r", "''"}], "[", "t", "]"}], " ", "-", " ", RowBox[{ RowBox[{"\[Omega]", "^", "2"}], " ", RowBox[{"r", "[", "t", "]"}]}]}], ")"}]}]}]}]], "Input", CellChangeTimes->{{3.436180346907*^9, 3.436180412758*^9}}], Cell[BoxData[ RowBox[{"0", "\[Equal]", RowBox[{"m", " ", RowBox[{"(", RowBox[{ RowBox[{ RowBox[{"-", SuperscriptBox["\[Omega]", "2"]}], " ", RowBox[{"r", "[", "t", "]"}]}], "+", RowBox[{ SuperscriptBox["r", "\[Prime]\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], ")"}]}]}]], "Output", CellChangeTimes->{3.436219662180729*^9}] }, Open ]], Cell["\<\ which is a familar looking differential equation---what sort of functions are \ their own 2nd derivatives?\ \>", "Text", CellChangeTimes->{{3.4362196708167286`*^9, 3.4362197072217283`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"DSolve", "[", RowBox[{ RowBox[{"{", "eqnr", "}"}], ",", RowBox[{"r", "[", "t", "]"}], ",", "t"}], "]"}]], "Input", CellChangeTimes->{{3.4362197092517285`*^9, 3.4362197219727287`*^9}}], Cell[BoxData[ RowBox[{"{", RowBox[{"{", RowBox[{ RowBox[{"r", "[", "t", "]"}], "\[Rule]", RowBox[{ RowBox[{ SuperscriptBox["\[ExponentialE]", RowBox[{"t", " ", "\[Omega]"}]], " ", RowBox[{"C", "[", "1", "]"}]}], "+", RowBox[{ SuperscriptBox["\[ExponentialE]", RowBox[{ RowBox[{"-", "t"}], " ", "\[Omega]"}]], " ", RowBox[{"C", "[", "2", "]"}]}]}]}], "}"}], "}"}]], "Output", CellChangeTimes->{3.4362197224717283`*^9}] }, Open ]], Cell["\<\ Oh yeah, any linear combination of growing and decay exponential, as long as \ it is \[Omega]t or \[Dash]\[Omega]t in the exponent. (See the previously assigned Morin 3.38 for the long derivation of this.) In \ terms of \[Theta]=\[Omega]t, the curve traced out by our particle is:\ \>", "Text", CellChangeTimes->{{3.4362197278577285`*^9, 3.4362197983107285`*^9}, { 3.436219900432729*^9, 3.4362199079277287`*^9}, {3.4362199447917285`*^9, 3.4362199566927285`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"r", "[", "\[Theta]_", "]"}], " ", "=", " ", RowBox[{ RowBox[{"A", " ", RowBox[{"Exp", "[", "\[Theta]", "]"}]}], " ", "+", " ", RowBox[{"B", " ", RowBox[{"Exp", "[", RowBox[{"-", "\[Theta]"}], "]"}]}]}]}]], "Input", CellChangeTimes->{{3.4362199162427287`*^9, 3.4362199419267282`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"B", " ", SuperscriptBox["\[ExponentialE]", RowBox[{"-", "\[Theta]"}]]}], "+", RowBox[{"A", " ", SuperscriptBox["\[ExponentialE]", "\[Theta]"]}]}]], "Output", CellChangeTimes->{3.4362199772697287`*^9}] }, Open ]], Cell["For some choice of parameters this is the curve:", "Text", CellChangeTimes->{{3.4362199607707286`*^9, 3.4362199703697286`*^9}}], Cell[CellGroupData[{ 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Since r[t]=R is constant, the F=ma component equation reads:\ \>", "Text", CellChangeTimes->{{3.4360101984849997`*^9, 3.436010222223*^9}, { 3.436010306623*^9, 3.436010327203*^9}, {3.4361798467019997`*^9, 3.436179866332*^9}, {3.4362202591627283`*^9, 3.4362202854457283`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"eqnv", " ", "=", " ", RowBox[{ RowBox[{"m", " ", RowBox[{ RowBox[{"v", "'"}], "[", "t", "]"}]}], " ", "\[Equal]", " ", RowBox[{"(", RowBox[{ RowBox[{ RowBox[{ RowBox[{"-", "\[Mu]"}], " ", "N0"}], " ", "/.", " ", "s"}], " ", "/.", " ", RowBox[{"v", "\[Rule]", RowBox[{"v", "[", "t", "]"}]}]}], ")"}]}]}]], "Input", CellChangeTimes->{{3.436220288173729*^9, 3.4362203402687283`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"m", " ", RowBox[{ SuperscriptBox["v", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "\[Equal]", RowBox[{ RowBox[{"-", "\[Mu]"}], " ", RowBox[{"(", RowBox[{ RowBox[{"g", " ", "m", " ", RowBox[{"Sin", "[", "\[Theta]", "]"}]}], "-", FractionBox[ RowBox[{"m", " ", RowBox[{"Cos", "[", "\[Theta]", "]"}], " ", SuperscriptBox[ RowBox[{"v", "[", "t", "]"}], "2"]}], "R"]}], ")"}]}]}]], "Output", CellChangeTimes->{3.4362203416437283`*^9}] }, Open ]], Cell["\<\ This is a differential equation which can be solved by separating and \ integrating, or just DSolve:\ \>", "Text", CellChangeTimes->{{3.4361799774890003`*^9, 3.4361800066549997`*^9}, { 3.4362203496587286`*^9, 3.4362203642007284`*^9}, {3.436220430437729*^9, 3.4362204600807285`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"sv", " ", "=", " ", RowBox[{ RowBox[{"DSolve", "[", RowBox[{ RowBox[{"{", RowBox[{"eqnv", ",", RowBox[{ RowBox[{"v", "[", "0", "]"}], "\[Equal]", "v0"}]}], "}"}], ",", RowBox[{"v", "[", "t", "]"}], ",", "t"}], "]"}], "[", RowBox[{"[", "1", "]"}], "]"}]}]], "Input", CellChangeTimes->{{3.4362204623257284`*^9, 3.436220489227729*^9}, { 3.436220520963729*^9, 3.4362205213547287`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"Solve", "::", "\<\"ifun\"\>"}], RowBox[{ ":", " "}], "\<\"Inverse functions are being used by \\!\\(Solve\\), so \ some solutions may not be found; use Reduce for complete solution \ information. \\!\\(\\*ButtonBox[\\\"\[RightSkeleton]\\\", \ ButtonStyle->\\\"Link\\\", ButtonFrame->None, \ ButtonData:>\\\"paclet:ref/message/Solve/ifun\\\", ButtonNote -> \ \\\"Solve::ifun\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{{3.4362204757097287`*^9, 3.4362204897307286`*^9}, 3.4362205252837286`*^9}], Cell[BoxData[ RowBox[{"{", RowBox[{ RowBox[{"v", "[", "t", "]"}], "\[Rule]", FractionBox[ RowBox[{ SqrtBox["g"], " ", SqrtBox["R"], " ", SqrtBox[ RowBox[{"Sin", "[", "\[Theta]", "]"}]], " ", RowBox[{"Tanh", "[", FractionBox[ RowBox[{ RowBox[{ SqrtBox["R"], " ", RowBox[{"ArcTanh", "[", FractionBox[ RowBox[{"v0", " ", SqrtBox[ RowBox[{"Cos", "[", "\[Theta]", "]"}]]}], RowBox[{ SqrtBox["g"], " ", SqrtBox["R"], " ", SqrtBox[ RowBox[{"Sin", "[", "\[Theta]", "]"}]]}]], "]"}]}], "-", RowBox[{ SqrtBox["g"], " ", "t", " ", "\[Mu]", " ", SqrtBox[ RowBox[{"Cos", "[", "\[Theta]", "]"}]], " ", SqrtBox[ RowBox[{"Sin", "[", "\[Theta]", "]"}]]}]}], SqrtBox["R"]], "]"}]}], SqrtBox[ RowBox[{"Cos", "[", "\[Theta]", "]"}]]]}], "}"}]], "Output", CellChangeTimes->{{3.4362204757547283`*^9, 3.4362204897377286`*^9}, 3.4362205252947283`*^9}] }, Open ]], Cell["\<\ This seems messy, but most of the mess is just the constants. Solving for \ the time at which v(t) vanishes:\ \>", "Text", CellChangeTimes->{{3.4362205327647285`*^9, 3.4362206288137283`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"st", " ", "=", " ", RowBox[{ RowBox[{"Solve", "[", RowBox[{ RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"v", "[", "t", "]"}], "/.", "sv"}], ")"}], "\[Equal]", "0"}], ",", "t"}], "]"}], "[", RowBox[{"[", "1", "]"}], "]"}]}]], "Input", CellChangeTimes->{{3.4362206331707287`*^9, 3.4362206542577286`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"Solve", "::", "\<\"ifun\"\>"}], RowBox[{ ":", " "}], "\<\"Inverse functions are being used by \\!\\(Solve\\), so \ some solutions may not be found; use Reduce for complete solution \ information. \\!\\(\\*ButtonBox[\\\"\[RightSkeleton]\\\", \ ButtonStyle->\\\"Link\\\", ButtonFrame->None, \ ButtonData:>\\\"paclet:ref/message/Solve/ifun\\\", ButtonNote -> \ \\\"Solve::ifun\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{{3.436220649708729*^9, 3.4362206545127287`*^9}}], Cell[BoxData[ RowBox[{"{", RowBox[{"t", "\[Rule]", FractionBox[ RowBox[{ SqrtBox["R"], " ", RowBox[{"ArcTanh", "[", FractionBox[ RowBox[{"v0", " ", SqrtBox[ RowBox[{"Cos", "[", "\[Theta]", "]"}]]}], RowBox[{ SqrtBox["g"], " ", SqrtBox["R"], " ", SqrtBox[ RowBox[{"Sin", "[", "\[Theta]", "]"}]]}]], "]"}]}], RowBox[{ SqrtBox["g"], " ", "\[Mu]", " ", SqrtBox[ RowBox[{"Cos", "[", "\[Theta]", "]"}]], " ", SqrtBox[ RowBox[{"Sin", "[", "\[Theta]", "]"}]]}]]}], "}"}]], "Output", CellChangeTimes->{{3.4362206497817287`*^9, 3.4362206545197287`*^9}}] }, Open ]], Cell["\<\ As Morin says, kind of messy, but it checks out. If there is no friction, for \ example, it takes forever to stop:\ \>", "Text", CellChangeTimes->{{3.43618001198*^9, 3.4361801039820004`*^9}, { 3.4362206611347284`*^9, 3.4362207248307285`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Limit", "[", RowBox[{ RowBox[{"(", RowBox[{"t", "/.", "st"}], ")"}], ",", RowBox[{"\[Mu]", "\[Rule]", "0"}]}], "]"}]], "Input", CellChangeTimes->{{3.4362206778807287`*^9, 3.4362206888607283`*^9}}], Cell[BoxData[ FractionBox[ RowBox[{ SqrtBox["R"], " ", RowBox[{"ArcTanh", "[", FractionBox[ RowBox[{"v0", " ", SqrtBox[ RowBox[{"Cos", "[", "\[Theta]", "]"}]]}], RowBox[{ SqrtBox["g"], " ", SqrtBox["R"], " ", SqrtBox[ RowBox[{"Sin", "[", "\[Theta]", "]"}]]}]], "]"}], " ", "\[Infinity]"}], RowBox[{ SqrtBox["g"], " ", SqrtBox[ RowBox[{"Cos", "[", "\[Theta]", "]"}]], " ", SqrtBox[ RowBox[{"Sin", "[", "\[Theta]", "]"}]]}]]], "Output", CellChangeTimes->{3.4362206894437284`*^9}] }, Open ]], Cell["\<\ Or if the \"cone\" is in fact a flat table (\[Theta]=\[Pi]/2) when the normal \ force is just the good old mg independent of velocity, so we constantly decelerate at \[Mu]g, so the time to go from v0 to 0 is:\ \>", "Text", CellChangeTimes->{{3.4362207269197283`*^9, 3.4362207855207286`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Limit", "[", RowBox[{ RowBox[{"(", RowBox[{"t", "/.", "st"}], ")"}], ",", RowBox[{"\[Theta]", "\[Rule]", RowBox[{"\[Pi]", "/", "2"}]}]}], "]"}]], "Input", CellChangeTimes->{{3.4362207869937286`*^9, 3.4362208010237284`*^9}}], Cell[BoxData[ FractionBox["v0", RowBox[{"g", " ", "\[Mu]"}]]], "Output", CellChangeTimes->{3.436220803173729*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Morin 5.56", "Subsection", CellChangeTimes->{{3.4360043137390003`*^9, 3.436004315868*^9}, 3.436179723259*^9, {3.436180164182*^9, 3.436180164796*^9}, { 3.4361802044379997`*^9, 3.436180208902*^9}, {3.436180886426*^9, 3.43618092334*^9}, {3.4362208472527285`*^9, 3.436220848253729*^9}}], Cell["\<\ This one sort of belongs with the \"Speedy Travel\" from last week. A planet \ of mass density \[Rho] spins at \[Omega]. How fast can it spin before material flies off, e.g. at the equator? That is, for what \[Omega] is the \ gravity force just large enough to keep material in orbit at R? If m is a small piece of material at the equator, then we want that:\ \>", "Text", CellChangeTimes->{{3.4362208501707287`*^9, 3.4362208810477285`*^9}, { 3.436220918687729*^9, 3.4362210410417285`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"s", "=", RowBox[{"Solve", "[", " ", RowBox[{ RowBox[{ RowBox[{ RowBox[{"-", " ", "m"}], " ", RowBox[{"\[Omega]", "^", "2"}], " ", "R"}], " ", "\[Equal]", " ", RowBox[{ RowBox[{"-", "G"}], " ", "M", " ", RowBox[{"m", " ", "/", " ", RowBox[{"R", "^", "2"}]}]}]}], ",", "\[Omega]"}], "]"}]}]], "Input", CellChangeTimes->{{3.4362210179477286`*^9, 3.4362210759997287`*^9}}], Cell[BoxData[ RowBox[{"{", RowBox[{ RowBox[{"{", RowBox[{"\[Omega]", "\[Rule]", RowBox[{"-", FractionBox[ RowBox[{ SqrtBox["G"], " ", SqrtBox["M"]}], SuperscriptBox["R", RowBox[{"3", "/", "2"}]]]}]}], "}"}], ",", RowBox[{"{", RowBox[{"\[Omega]", "\[Rule]", FractionBox[ RowBox[{ SqrtBox["G"], " ", SqrtBox["M"]}], SuperscriptBox["R", RowBox[{"3", "/", "2"}]]]}], "}"}]}], "}"}]], "Output", CellChangeTimes->{{3.4362210616707287`*^9, 3.4362210765337286`*^9}}] }, Open ]], Cell["\<\ Phrasing the answer in terms of \[Rho] the period is then:\ \>", "Text", CellChangeTimes->{{3.4362210791717286`*^9, 3.4362210901777287`*^9}, { 3.4362211480317287`*^9, 3.4362211512257285`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"T", " ", "=", " ", RowBox[{ RowBox[{ RowBox[{"2", " ", RowBox[{"\[Pi]", "/", "\[Omega]"}]}], " ", "/.", " ", RowBox[{"s", "[", RowBox[{"[", "2", "]"}], "]"}]}], " ", "/.", " ", RowBox[{"M", "\[Rule]", " ", RowBox[{ RowBox[{"(", RowBox[{"4", "/", "3"}], ")"}], " ", "\[Pi]", " ", RowBox[{"R", "^", "3"}], " ", "\[Rho]", " "}]}]}]}]], "Input", CellChangeTimes->{{3.436221095015729*^9, 3.4362211572307286`*^9}}], Cell[BoxData[ FractionBox[ RowBox[{ SqrtBox[ RowBox[{"3", " ", "\[Pi]"}]], " ", SuperscriptBox["R", RowBox[{"3", "/", "2"}]]}], RowBox[{ SqrtBox["G"], " ", SqrtBox[ RowBox[{ SuperscriptBox["R", "3"], " ", "\[Rho]"}]]}]]], "Output", CellChangeTimes->{{3.436221115366729*^9, 3.4362211258897285`*^9}, 3.436221162102729*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"T0", "=", RowBox[{"Simplify", "[", RowBox[{"T", ",", RowBox[{"Assumptions", "\[Rule]", RowBox[{"R", ">", "0"}]}]}], "]"}]}]], "Input", CellChangeTimes->{{3.4362211704547286`*^9, 3.4362211931277285`*^9}, { 3.436221480628729*^9, 3.436221481432729*^9}}], Cell[BoxData[ FractionBox[ SqrtBox[ RowBox[{"3", " ", "\[Pi]"}]], RowBox[{ SqrtBox["G"], " ", SqrtBox["\[Rho]"]}]]], "Output", CellChangeTimes->{3.4362211939697285`*^9, 3.4362214820047283`*^9}] }, Open ]], Cell["Numerically, let's load up the constants we need:", "Text", CellChangeTimes->{{3.4362212164207287`*^9, 3.436221224142729*^9}, { 3.4362213493557286`*^9, 3.4362213652007284`*^9}}], Cell[BoxData[ RowBox[{"Needs", "[", "\"\\"", "]"}]], "Input", CellChangeTimes->{{3.435622893804*^9, 3.435622913704*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"T0", "/.", " ", RowBox[{"{", RowBox[{ RowBox[{"G", "\[Rule]", "GravitationalConstant"}], ",", RowBox[{"\[Rho]", "\[Rule]", RowBox[{"EarthMass", "/", RowBox[{"(", RowBox[{ RowBox[{"(", RowBox[{"4", "/", "3"}], ")"}], "\[Pi]", " ", RowBox[{"EarthRadius", "^", "3"}]}], ")"}]}]}]}], "}"}]}], "/.", " ", RowBox[{"{", RowBox[{"Newton", "\[Rule]", RowBox[{"Kilogram", " ", RowBox[{"Meter", " ", "/", RowBox[{"Second", "^", "2"}]}]}]}], "}"}]}]], "Input", CellChangeTimes->{{3.436221366930729*^9, 3.4362214460347285`*^9}, { 3.4362214854177284`*^9, 3.436221489899729*^9}}], Cell[BoxData[ FractionBox["5068.492989278671`", RowBox[{ SqrtBox[ FractionBox["Kilogram", SuperscriptBox["Meter", "3"]]], " ", SqrtBox[ FractionBox[ SuperscriptBox["Meter", "3"], RowBox[{"Kilogram", " ", SuperscriptBox["Second", "2"]}]]]}]]], "Output", CellChangeTimes->{{3.4362214131707287`*^9, 3.4362214419397287`*^9}, 3.4362214914577284`*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Convert", "[", RowBox[{ RowBox[{"5068.49", " ", "Second"}], ",", " ", "Hour"}], "]"}]], "Input", CellChangeTimes->{{3.436221501956729*^9, 3.4362215169367285`*^9}}], Cell[BoxData[ RowBox[{"1.4079138888888887`", " ", "Hour"}]], "Output", CellChangeTimes->{3.4362215172437286`*^9}] }, Open ]], Cell["\<\ And oh yeah, this is exactly the time for speedy travel through the rabbit \ hole.\ \>", "Text", CellChangeTimes->{{3.4362215197827287`*^9, 3.4362215433377285`*^9}}] }, Open ]], Cell[CellGroupData[{ Cell["Morin 5.83", "Subsection", CellChangeTimes->{{3.4360043137390003`*^9, 3.436004315868*^9}, 3.436179723259*^9, {3.436180164182*^9, 3.436180164796*^9}, { 3.4361802044379997`*^9, 3.436180208902*^9}, {3.436180886426*^9, 3.43618092334*^9}}], Cell["\<\ Consider a standard 2D elastic collision: initially particle one (M) is \ moving at v0 toward particle 2 (m at rest). After a collision, particle 2 heads off at an angle \[Theta] while m1 goes at \ \[Phi]. We are asked to find \[Theta] when the vertical component of particle 2's \ final state velocity is as large as it can be.\ \>", "Text", CellChangeTimes->{{3.436180956528*^9, 3.436181015651*^9}, { 3.4361810520550003`*^9, 3.436181183084*^9}, {3.4362216073547287`*^9, 3.436221718845729*^9}, {3.4362218091297283`*^9, 3.4362218413397284`*^9}, { 3.436295377951*^9, 3.43629542444*^9}}], Cell["\<\ Thinking about it in the center of mass frame makes the problem easy. First, \ observe that the CM is moving purely in the x-direction, so the lab and CM frame agree on the y-direction \ velocities of a given object, and in particular maximizing v2y in one frame will maximize in the other. So \ how do we maximize v2y in the CM frame? By having the particles fly off purely in the y-direction \ obviously. So in the CM description, particle 2 comes in with velocity (-Vcm) in the x-direction, then scatters at \ the same speed (Vcm) in the y-direction. In the lab frame then, the resulting velocity vector is (Vcm,Vcm), i.e. at \ angle \[Pi]/4 relative to the x-axis.\ \>", "Text", CellChangeTimes->{{3.436295465727*^9, 3.436295776028*^9}}], Cell["\<\ Now let's solve this problem purely in the lab frame. The conservation \ equations are:\ \>", "Text", CellChangeTimes->{{3.436295796453*^9, 3.43629579796*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"eqnpx", " ", "=", " ", RowBox[{ RowBox[{"M", " ", "v0"}], " ", "\[Equal]", " ", RowBox[{ RowBox[{"M", " ", "v1", " ", RowBox[{"Cos", "[", "\[Phi]", "]"}]}], " ", "+", " ", RowBox[{"m", " ", "v2", " ", RowBox[{"Cos", "[", "\[Theta]", "]"}]}]}]}]}]], "Input", CellChangeTimes->{{3.436181184146*^9, 3.436181235847*^9}, { 3.436221844404729*^9, 3.4362218486157284`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"M", " ", "v0"}], "\[Equal]", RowBox[{ RowBox[{"m", " ", "v2", " ", RowBox[{"Cos", "[", "\[Theta]", "]"}]}], "+", RowBox[{"M", " ", "v1", " ", RowBox[{"Cos", "[", "\[Phi]", "]"}]}]}]}]], "Output", CellChangeTimes->{3.4362218663687286`*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"eqnpy", " ", "=", " ", RowBox[{"0", " ", "\[Equal]", " ", RowBox[{ RowBox[{"M", " ", "v1", " ", RowBox[{"Sin", "[", "\[Phi]", "]"}]}], " ", "-", " ", 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RowBox[{"Cos", "[", RowBox[{"2", " ", "\[Theta]"}], "]"}]}]}]]], "]"}]}]}], "}"}]}], "}"}]], "Output", CellChangeTimes->{{3.436221895515729*^9, 3.4362219168087287`*^9}}] }, Open ]], Cell["\<\ which is messy enough, but taking out the solution we seek we define\ \>", "Text", CellChangeTimes->{{3.436221926555729*^9, 3.4362219548627286`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"v2y", "[", "\[Theta]_", "]"}], " ", "=", " ", RowBox[{ RowBox[{"(", RowBox[{"v2", "/.", RowBox[{"s", "[", RowBox[{"[", "7", "]"}], "]"}]}], ")"}], " ", RowBox[{"Sin", "[", "\[Theta]", "]"}]}]}]], "Input", CellChangeTimes->{{3.4362219561317286`*^9, 3.4362219770647287`*^9}}], Cell[BoxData[ FractionBox[ RowBox[{"2", " ", "M", " ", "v0", " ", RowBox[{"Cos", "[", "\[Theta]", "]"}], " ", RowBox[{"Sin", "[", "\[Theta]", "]"}]}], RowBox[{"m", "+", "M"}]]], "Output", CellChangeTimes->{{3.4362219696177287`*^9, 3.4362219786607285`*^9}}] }, Open ]], Cell["\<\ For what value of \[Theta] is this maximized?\ \>", "Text", CellChangeTimes->{{3.4362219824867287`*^9, 3.4362219924657288`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Solve", "[", RowBox[{ RowBox[{ RowBox[{ RowBox[{"v2y", "'"}], "[", "\[Theta]", "]"}], "\[Equal]", "0"}], ",", "\[Theta]"}], "]"}]], "Input", CellChangeTimes->{{3.4362219978157287`*^9, 3.436222007114729*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"Solve", "::", "\<\"ifun\"\>"}], RowBox[{ ":", " "}], "\<\"Inverse functions are being used by \\!\\(Solve\\), so \ some solutions may not be found; use Reduce for complete solution \ information. \\!\\(\\*ButtonBox[\\\"\[RightSkeleton]\\\", \ ButtonStyle->\\\"Link\\\", ButtonFrame->None, \ ButtonData:>\\\"paclet:ref/message/Solve/ifun\\\", ButtonNote -> \ \\\"Solve::ifun\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.4362220079477286`*^9}], Cell[BoxData[ RowBox[{"{", RowBox[{ RowBox[{"{", RowBox[{"\[Theta]", "\[Rule]", RowBox[{"-", FractionBox[ RowBox[{"3", " ", "\[Pi]"}], "4"]}]}], "}"}], ",", RowBox[{"{", RowBox[{"\[Theta]", "\[Rule]", RowBox[{"-", FractionBox["\[Pi]", "4"]}]}], "}"}], ",", RowBox[{"{", RowBox[{"\[Theta]", "\[Rule]", FractionBox["\[Pi]", "4"]}], "}"}], ",", RowBox[{"{", RowBox[{"\[Theta]", "\[Rule]", FractionBox[ RowBox[{"3", " ", "\[Pi]"}], "4"]}], "}"}]}], "}"}]], "Output", CellChangeTimes->{3.4362220079907284`*^9}] }, Open ]], Cell["\<\ Oh yeah, that becomes obvious e.g. when one remembers trig ID's such as\ \>", "Text", CellChangeTimes->{{3.436222010220729*^9, 3.4362220296907287`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"TrigExpand", "[", RowBox[{"Sin", "[", RowBox[{"2", " ", "\[Theta]"}], "]"}], "]"}]], "Input", CellChangeTimes->{{3.4362220310217285`*^9, 3.436222036812729*^9}}], Cell[BoxData[ RowBox[{"2", " ", RowBox[{"Cos", "[", "\[Theta]", "]"}], " ", RowBox[{"Sin", "[", "\[Theta]", "]"}]}]], "Output", CellChangeTimes->{3.4362220371537285`*^9}] }, Open ]], Cell["\<\ and then notes that Sin[2 \[Theta]] is maximized when 2\[Theta] = \[Pi]/2. \ Note that maximal v2y is then:\ \>", "Text", CellChangeTimes->{{3.4362221205637283`*^9, 3.4362221575717287`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"v2y", "[", RowBox[{"\[Pi]", "/", "4"}], "]"}]], "Input", CellChangeTimes->{{3.4362221586857285`*^9, 3.4362221626977286`*^9}}], Cell[BoxData[ FractionBox[ RowBox[{"M", " ", "v0"}], RowBox[{"m", "+", "M"}]]], "Output", CellChangeTimes->{3.436222163036729*^9}] }, Open ]], Cell["which just happens to be the center of mass velocity.", "Text", CellChangeTimes->{{3.4362221691987286`*^9, 3.436222179246729*^9}}] }, Open ]], Cell[CellGroupData[{ Cell["Morin 5.89", "Subsection", CellChangeTimes->{{3.4360043137390003`*^9, 3.436004315868*^9}, 3.436179723259*^9, {3.436180164182*^9, 3.436180164796*^9}, { 3.4361802044379997`*^9, 3.436180208902*^9}, {3.436180886426*^9, 3.43618092334*^9}, {3.4362222518037286`*^9, 3.4362222527197285`*^9}}], Cell["\<\ A plate of mass M moves at v. A mass m is dropped on the plate and skids to \ rest relative to the plate. How much energy must be pump in to the system to make the plate (together \ with mass m) move again at v?\ \>", "Text", CellChangeTimes->{{3.4362222599977283`*^9, 3.4362223607577286`*^9}}], Cell["\<\ Well, there are no external forces (frictionless table) so momentum is \ conserved in the plate/pea collision, and the new speed is given by:\ \>", "Text", CellChangeTimes->{{3.4362223643847284`*^9, 3.4362224332117286`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"s", "=", RowBox[{"Solve", "[", RowBox[{ RowBox[{ RowBox[{"M", " ", "v"}], " ", "\[Equal]", " ", RowBox[{ RowBox[{"(", RowBox[{"M", "+", "m"}], ")"}], " ", "vnew"}]}], ",", "vnew"}], "]"}]}]], "Input", CellChangeTimes->{{3.4362224353307285`*^9, 3.4362224562667284`*^9}}], Cell[BoxData[ RowBox[{"{", RowBox[{"{", RowBox[{"vnew", "\[Rule]", FractionBox[ RowBox[{"M", " ", "v"}], RowBox[{"m", "+", "M"}]]}], "}"}], "}"}]], "Output", CellChangeTimes->{{3.4362224513767285`*^9, 3.4362224582997284`*^9}}] }, Open ]], Cell["That means that we have kinetic energy", "Text", CellChangeTimes->{{3.4362224617157288`*^9, 3.436222470536729*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Knew", " ", "=", " ", RowBox[{ RowBox[{ RowBox[{"(", RowBox[{"1", "/", "2"}], ")"}], " ", RowBox[{"(", RowBox[{"M", "+", "m"}], ")"}], " ", RowBox[{ RowBox[{"(", "vnew", ")"}], "^", "2"}]}], " ", "/.", " ", RowBox[{"s", "[", RowBox[{"[", "1", "]"}], "]"}]}]}]], "Input", CellChangeTimes->{{3.4362224737957287`*^9, 3.4362224874747286`*^9}, { 3.436222570536729*^9, 3.4362225712627287`*^9}, {3.4362226239577284`*^9, 3.4362226310457287`*^9}}], Cell[BoxData[ FractionBox[ RowBox[{ SuperscriptBox["M", "2"], " ", SuperscriptBox["v", "2"]}], RowBox[{"2", " ", RowBox[{"(", RowBox[{"m", "+", "M"}], ")"}]}]]], "Output", CellChangeTimes->{3.4362226314687285`*^9}] }, Open ]], Cell["\<\ And if we want to end up in a state with both M and m moving at the original \ v we must supply an energy\ \>", "Text", CellChangeTimes->{{3.4362226428157287`*^9, 3.436222672173729*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"\[CapitalDelta]K", " ", "=", " ", RowBox[{ RowBox[{ RowBox[{ RowBox[{"(", RowBox[{"1", "/", "2"}], ")"}], " ", RowBox[{"(", RowBox[{"M", "+", "m"}], ")"}], " ", RowBox[{"v", "^", "2"}]}], " ", "-", " ", "Knew"}], " ", "//", "Simplify"}]}]], "Input", CellChangeTimes->{{3.4362226754827285`*^9, 3.4362227041007285`*^9}}], Cell[BoxData[ FractionBox[ RowBox[{"m", " ", RowBox[{"(", RowBox[{"m", "+", RowBox[{"2", " ", "M"}]}], ")"}], " ", SuperscriptBox["v", "2"]}], RowBox[{"2", " ", RowBox[{"(", RowBox[{"m", "+", "M"}], ")"}]}]]], "Output", CellChangeTimes->{{3.4362226848917284`*^9, 3.4362227048137283`*^9}}] }, Open ]], Cell["\<\ Notice that when M is large this becomes:\ \>", "Text", CellChangeTimes->{{3.4362227157907286`*^9, 3.4362227277287283`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Limit", "[", RowBox[{"\[CapitalDelta]K", ",", RowBox[{"M", "\[Rule]", "Infinity"}]}], "]"}]], "Input", CellChangeTimes->{{3.4362227288797283`*^9, 3.4362227371037283`*^9}}], Cell[BoxData[ RowBox[{"m", " ", SuperscriptBox["v", "2"]}]], "Output", CellChangeTimes->{3.4362227379017286`*^9}] }, Open ]], Cell["\<\ The \"intuitive\" explanation is that from beginning to end the plate is in \ the same state, but we have added the energy necessary to bring mass m from speed 0 up to v (i.e (1/2) m v^2), AND we've \ replaced the energy lost to heat in the scraping, sliding, stopping process as the little mass interacts with the \ plate. In the limit that M is large, let's view that process in the CM frame (i.e. the plate frame). In that frame, the \ little m lands with relative velocity v, then skids to a stop, losing *all* of its kinetic energy. Thus (1/2) m v^2 is \ lost to heat, and altogether our energy bill is mv^2. \ \>", "Text", CellChangeTimes->{{3.4362227444177284`*^9, 3.4362227592247286`*^9}, { 3.4362229238117285`*^9, 3.4362231231847286`*^9}}] }, Open ]], Cell[CellGroupData[{ Cell["Morin 5.94", "Subsection", CellChangeTimes->{{3.4360043137390003`*^9, 3.436004315868*^9}, 3.436179723259*^9, {3.436180164182*^9, 3.436180164796*^9}, { 3.4361802044379997`*^9, 3.436180208902*^9}, {3.436180886426*^9, 3.43618092334*^9}, {3.4362222518037286`*^9, 3.4362222527197285`*^9}, { 3.4362231444127283`*^9, 3.4362231469287286`*^9}}], Cell["\<\ An essentially massless dustpan starts from rest at the top of a incline \ (parameterized by the angle \[Theta]). Along the ramp there is a layer of dust with density \[Sigma] kg/meter. The \ dustpan slides (frictionlessly) down the hill, sweeping up the dust along the way. We see the motion x(t) where x \ is the distance from the start.\ \>", "Text", CellChangeTimes->{{3.4362231483867283`*^9, 3.4362231695387287`*^9}, { 3.436295827025*^9, 3.4362959424309998`*^9}}], Cell["\<\ We derive the differential equation in the usual \[CapitalDelta]t way: at \ time t we have a dustpan of mass m(t) = \[Sigma] x(t) (that being the amount of dust which has been swept up so far) colliding with \ an amount \[CapitalDelta]m of dust at rest. If we are moving at v(t) = x'(t), the amount of dust we are about to pick up \ in \[CapitalDelta]t is \[Sigma] \[CapitalDelta]x = \[Sigma] v \ \[CapitalDelta]t. After the collision, both dustpan and dust continue on at v+\[CapitalDelta]v. \ Thus the change in momentum is:\ \>", "Text", CellChangeTimes->{{3.436295937676*^9, 3.4362961189119997`*^9}}], Cell[BoxData[ RowBox[{"\[CapitalDelta]p", " ", "=", " ", RowBox[{ RowBox[{ RowBox[{"(", RowBox[{"m", " ", "+", " ", "\[CapitalDelta]m"}], ")"}], RowBox[{"(", RowBox[{"v", "+", "\[CapitalDelta]v"}], ")"}]}], " ", "-", " ", RowBox[{"m", " ", "v"}]}]}]], "Input", CellChangeTimes->{{3.436296122541*^9, 3.4362961707019997`*^9}}], Cell["Dividing by \[CapitalDelta]t we get ", "Text", CellChangeTimes->{{3.436296309774*^9, 3.436296320358*^9}, {3.436296350942*^9, 3.436296351682*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"(", RowBox[{"\[CapitalDelta]p", "/", "\[CapitalDelta]t"}], ")"}], " ", "/.", " ", RowBox[{"{", RowBox[{ RowBox[{"m", "\[Rule]", RowBox[{"\[Sigma]", " ", RowBox[{"x", "[", "t", "]"}]}]}], ",", " ", RowBox[{"\[CapitalDelta]m", " ", "\[Rule]", " ", RowBox[{"\[Sigma]", " 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How do we solve this one \"by hand\"?! The argument runs backward from the answer: we ask \"what could x(t) look \ like\"? We have the following building blocks from which to build an expression for \ x(t): t, g, \[Theta], \[Sigma]. Given the physical dimensions, we observe that the answer for x(t) cannot \ involve \[Sigma], since it is the only parameter which has kilograms in its units, and there is \ nothing to cancel those in an expression which must end up with dimensions of meters. This \ \"explains\" why the \[Sigma]'s cancelled out of our differential equation. Then from the \ rest the only expression with units of distance is: g t^2. Thus x(t) = f(\[Theta]) g t^2 where all that \ remains is to find the dimensionless constant. So we try the ansatz x(t) = (1/2) a t^2:\ \>", "Text", CellChangeTimes->{{3.436296503623*^9, 3.436296538546*^9}, {3.436296608157*^9, 3.436296914429*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"eqn", " ", "/.", " ", RowBox[{"{", RowBox[{"x", "\[Rule]", RowBox[{"Function", "[", RowBox[{ RowBox[{"{", "t", "}"}], ",", RowBox[{ RowBox[{"(", RowBox[{"1", "/", "2"}], ")"}], " ", "a", " ", RowBox[{"t", "^", "2"}]}]}], "]"}]}], "}"}]}]], "Input", CellChangeTimes->{{3.436223299746729*^9, 3.4362233099627285`*^9}, { 3.4362233449087286`*^9, 3.4362234042597284`*^9}}], Cell[BoxData[ RowBox[{"a", "\[Equal]", RowBox[{ RowBox[{ RowBox[{"-", "2"}], " ", "a"}], "+", "g\[Theta]"}]}]], "Output", CellChangeTimes->{{3.4362233700747285`*^9, 3.4362234052317286`*^9}, 3.436296919638*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Solve", "[", RowBox[{"%", ",", "a"}], "]"}]], "Input", CellChangeTimes->{{3.436296923863*^9, 3.436296927924*^9}}], Cell[BoxData[ RowBox[{"{", RowBox[{"{", RowBox[{"a", "\[Rule]", FractionBox["g\[Theta]", "3"]}], "}"}], "}"}]], "Output", CellChangeTimes->{3.4362969285769997`*^9}] }, Open ]], Cell["\<\ In other words, the acceleration is constant and equal to (1/3) g \ Sin[\[Theta]]. As a last check:\ \>", "Text", CellChangeTimes->{{3.436296970942*^9, 3.436296995767*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"eqn", " ", "/.", " ", RowBox[{"{", RowBox[{"x", "\[Rule]", RowBox[{"Function", "[", RowBox[{ RowBox[{"{", "t", "}"}], ",", RowBox[{ RowBox[{"(", RowBox[{"1", "/", "6"}], ")"}], " ", "g\[Theta]", " ", RowBox[{"t", "^", "2"}]}]}], "]"}]}], "}"}]}]], "Input", CellChangeTimes->{{3.436297004604*^9, 3.436297028487*^9}}], Cell[BoxData["True"], "Output", CellChangeTimes->{{3.436297011304*^9, 3.436297028976*^9}}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["The Raindrop", "Subsection", CellChangeTimes->{{3.4360043137390003`*^9, 3.436004315868*^9}, 3.436179723259*^9, {3.436180164182*^9, 3.436180164796*^9}, { 3.4361802044379997`*^9, 3.436180208902*^9}, {3.4365399879997597`*^9, 3.436539991749736*^9}}], Cell["\<\ Suppose a raindrop forms with initial mass M0 and then falls through a cloud, \ picking up speed and accumulating mass. Obviously, the larger the drop, the more water vapor it encounters, and the \ faster it accumulates mass. In problem 5.31 considers spherical raindrops, where the mass grows like r^3 \ of course. Here let's consider the simpler problem of \"pancakes\", where the area and \ the mass are proportional. Specifically we suppose that between time t and t+\[CapitalDelta]t we \ accumulate a mass:\ \>", "Text", CellChangeTimes->{{3.436539995249713*^9, 3.4365402485762167`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"\[CapitalDelta]m", " ", "=", " ", RowBox[{"k", " ", RowBox[{"m", "[", "t", "]"}], " ", RowBox[{"v", "[", "t", "]"}], " ", "\[CapitalDelta]t"}]}]], "Input", CellChangeTimes->{{3.4365402514668236`*^9, 3.436540285263482*^9}}], Cell[BoxData[ RowBox[{"k", " ", "\[CapitalDelta]t", " ", RowBox[{"m", "[", "t", "]"}], " ", RowBox[{"v", "[", "t", "]"}]}]], "Output", CellChangeTimes->{3.4365404347469006`*^9}] }, Open ]], Cell["\<\ where v[t] is the downward velocity. The change in momentum in the collision \ process is:\ \>", "Text", CellChangeTimes->{{3.436540288294713*^9, 3.436540314372671*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"\[CapitalDelta]p", " ", "=", " ", RowBox[{ RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"m", "[", "t", "]"}], "+", "\[CapitalDelta]m"}], ")"}], RowBox[{"(", RowBox[{ RowBox[{"v", "[", "t", "]"}], "+", RowBox[{ RowBox[{ RowBox[{"v", "'"}], "[", "t", "]"}], "\[CapitalDelta]t"}]}], ")"}]}], " ", "-", " ", RowBox[{ RowBox[{"m", "[", "t", "]"}], " ", RowBox[{"v", "[", "t", "]"}]}]}]}]], "Input", CellChangeTimes->{{3.4365403158414116`*^9, 3.4365403629973593`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"-", RowBox[{"m", "[", "t", "]"}]}], " ", RowBox[{"v", "[", "t", "]"}]}], "+", RowBox[{ RowBox[{"(", RowBox[{"\[CapitalDelta]m", "+", RowBox[{"m", "[", "t", "]"}]}], ")"}], " ", RowBox[{"(", RowBox[{ RowBox[{"v", "[", "t", "]"}], "+", RowBox[{"\[CapitalDelta]t", " ", RowBox[{ SuperscriptBox["v", "\[Prime]", 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may not be found; use Reduce for complete solution \ information. \\!\\(\\*ButtonBox[\\\"\[RightSkeleton]\\\", \ ButtonStyle->\\\"Link\\\", ButtonFrame->None, \ ButtonData:>\\\"paclet:ref/message/Solve/ifun\\\", ButtonNote -> \ \\\"Solve::ifun\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.436540563839824*^9, 3.43654062926128*^9, 3.436540719573202*^9}], Cell[BoxData[ RowBox[{"{", RowBox[{ RowBox[{"v", "[", "t", "]"}], "\[Rule]", FractionBox[ RowBox[{ SqrtBox["g"], " ", RowBox[{"Tanh", "[", RowBox[{ SqrtBox["g"], " ", SqrtBox["k"], " ", "t"}], "]"}]}], SqrtBox["k"]]}], "}"}]], "Output", CellChangeTimes->{{3.4365405536211395`*^9, 3.436540563855449*^9}, 3.436540629276905*^9, 3.436540719588827*^9}] }, Open ]], Cell["Note that there is a terminal velocity:", "Text", CellChangeTimes->{{3.4365405951052494`*^9, 3.4365406231519446`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Limit", "[", RowBox[{ RowBox[{ RowBox[{"v", "[", "t", "]"}], "/.", "sv"}], ",", RowBox[{"t", "\[Rule]", 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