(Newton s 2 nd Law for linear motion)

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1 PHYSICS 3 Final Exaination ( Deeber Tie liit 3 hours Answer all 6 questions You and an assistant are holding the (opposite ends of a long plank when oops! the butterfingered assistant drops his end If the plank s weight is W and if the plank was level when it was dropped, at that (initial instant the weight you feel is? (Hint: use Newton s nd Law for rotations & linear otion W f N = I θ (Newton s nd Law for rotations l W = Ml θ 3 l W f = Ma = M θ Eliinating θ we have l W = Ml W f 3 M l W = ( W f 3 W f = (Newton s nd Law for linear otion The surfae gravity of the Moon is alost exatly 6 of the Earth s, and its urvature is suh that in 87 eters of forward progress the surfae drops eter How fast would a projetile launhed horizontally have to ove in order to ake a losed (irular orbit around the Moon? (Hint: WWGD [what would Galileo do?] The projetile ust fall eter for every 87 eters of forward otion Thus, y= g M t = x 87 t = = v v and so

2 g 98 v = 87 M = 87 = 69k/se 3 A pendulu onsists of a assless rigid rod of length R with a weight of ass attahed to its end The bob is released fro a height h above a table, upon whih is resting a blok of ass, whose end is exatly below the point of suspension There is no frition between blok and table The bob ollides elastially with the blok To what height does it rise, and does it rebound or ontinue in the sae diretion? Desribe the result if the bob had ass and the blok ass Use onservation of kineti energy and linear oentu to desribe the ollision Then v = v + ( u v= v + u and eliinating u we find v v ( v v ( v+ v = ( v v = ( v v This gives either the uninteresting solution v= v (no ollision or the interesting one v v = 3 The latter represents a rebound ( v < Sine the veloity of the bob before the ollision is given by energy onservation as v gh =, v h we see h = = The bob rebounds to /9 of its original height g 9 If the bob weighed and the blok weighed, the obination of energy and oentu onservation would give ( v v ( v+ v = ( v v = ( v v v v =, u = v 3 3 That is, the bob ontinues in the sae diretion at /3 of its original speed (The blok oves fast enough to get out of the way! The bob again rises to height h = h 9

3 A bead of ass slides fritionlessly on a hoop of radius R that is oriented vertially and rotates about a diaeter with onstant angular veloity ω a Write the Lagrangian of the syste in appropriate generalized oordinates, and find the equation(s of otion b Show that there is a ritial value ω of the rotational veloity suh that for ω<ω the equilibriu position is at the botto enter, whereas for position is at a non-zero angle θ ω>ω the equilibriu Exaine sall osillations about equilibriu in both ases and deterine whether the equilibria are stable or unstable, and if stable, deterine their osillation frequenies The Lagrangian is L = ( x + y + z gz = R ( θ + sin θϕ + grosθ = R ( θ +ω sin θ + gr osθ and the equation of otion is d L = R θ= V ( eff θ = R Ω +ω os θ sin θ dt θ θ where Ω = g R and ( θ = R Ω osθ+ ω sin θ At equilibriu, θ= Thus if Ω >ω the only equilibria are θ=, π, whereas if Ω <ω, there is also a third equilibriu point at that g ω =Ω= R If we expand V eff θ about the equilibriu points we find Ω + Ω ω θ, θ θ R Ω Ω +ω θ π, θ π Ω θ=θ = os We onlude ω

4 and for ω>ω, ( θ R Ω ( ω Ω + ( ω Ω ( θ θ ω For any ω the equilibriu at θ=π is unstable the potential has a axiu there For ω<ω the equilibriu at θ= is stable, and the angular frequeny of sall osillations about that point is Ω ω For ω>ω the equilibriu at θ= is unstable (the oeffiient of sign, turning a iniu into a axiu, but the equilibriu at The angular frequeny of sall osillations about that point is ω Ω θ has hanged θ=θ is stable 5 A blok of ass is attahed to two fixed posts by springs of onstant k and negligible ass, as shown, and exeutes siple haroni otion in the horizontal diretion (It slides fritionlessly on a table k k a What is the frequeny of osillation? b A lup of putty, also of ass, falls on the blok and stiks to it What is the effet on the aplitude and frequeny of the subsequent osillation if the lup hits just when the blok is at an extree end of its osillation; the lup hits when the blok is exatly in the iddle of its osillation? The equation of otion of the osillator before the putty falls is x + kx = k so the angular frequeny of osillation is ω= If the putty falls at the liit of the osillation, ie x= xax, the veloity of the blok is, so all that happens is the ass, and hene the frequeny hanges The new frequeny is k ω ω= =, and the aplitude reains the sae If the putty falls when x = the veloity is axiu Then, by onservation of linear oentu the new veloity is v = v, so that the kineti energy falls by a fator of ( the ass, ¼ of the veloity squared But at the extrea of the otion, all the energy is potential (KE= so that the square of the new aplitude ust be half the

5 square of the old That is, we have x ax = xax The frequeny will be the sae as in ase above 6 A partile of ass oving in one diension is subjet to the fore F= F ax sinh a What is the potential energy orresponding to this fore? b Does this potential have a stable equilibriu position? If the answer to b above is Yes, what is the period of sall osillations about equilibriu? d If the answer to b above is Yes, find an expression for the period of osillations that are not sall (Do not attept to evaluate this expression! dv Sine F =, we ay iediately integrate and, disregarding a onstant of integration, find dx F V( x = osh ( ax a Sine this potential has a single iniu and beoes infinite at x =±, we see that there is a stable equilibriu position at x = If we expand about this point we find V x = F osh ax F + F ax = F + Fax a a a a Fa The frequeny of sall osillations is therefore ω= and the period is τ= π Fa To find the period of non-sall osillations, we use onservation of energy: F F x + osh( ax = E= osh ( al a a where l is the aplitude of the osillation This equation is separable, giving or a dx osh F dt =± ( a l osh ( ax a dx du τ= = = τ l la dt F l osh osh osh osh The latter integral annot be evaluated in losed for af la al ax ( al ( u

ME357 Problem Set The wheel is a thin homogeneous disk that rolls without slip. sin. The wall moves with a specified motion x t. sin..

ME357 Problem Set The wheel is a thin homogeneous disk that rolls without slip. sin. The wall moves with a specified motion x t. sin.. ME357 Proble Set 3 Derive the equation(s) of otion for the systes shown using Newton s Method. For ultiple degree of freedo systes put you answer in atri for. Unless otherwise speified the degrees of freedo