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# CLASSICAL MECHANICS PHYS 105B

UCSB

GPA 3.8

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This 3 page Class Notes was uploaded by Hailey Halvorson on Thursday October 22, 2015. The Class Notes belongs to PHYS 105B at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/227146/phys-105b-university-of-california-santa-barbara in Physics 2 at University of California Santa Barbara.

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Date Created: 10/22/15

91 In the noninertial frame of a car we have two forces gravity and the inertial force which combine to give an effective gravityli If A is the accelaration o the car with respect to the ground gaff g i A The important point is that the buoyant force directly opposes this effective gravityi In the case of a helium balloon the buoyant force is greater than the effective gravitational force so the displacement of the balloon will be in the direction of the buoyant force upward and forward The equilibrium angle is the same as the angle of the effective gravity 9811 arctan 9 An additional and unneccessary complication the above treatment of the buoyant force depends on the car being enclosedi If the car is not enclosed the buoyant force will align with the true gravitional eld In that case the helium balloon will tilt backwards even ignoring air resistance 98 a The centrifugal force always points radially outward from the axis of rotation Near the North pole that direction is south The coriolis force on someone increasing their radius will be against the direction of rotation Since the earth rotates to the east the coriolis force in this case is westwar b East is the direction of rotation so the person on the equator is moving tangen tially in the direction of rotational motion The centrifugal force is as always radially outward which from the perspective of someone standing on the equa tor is directly up The coriolis force is also radially outward or up since at the equator here An object moving in the direction of rotational motion requires a stronger centripetal acceleration to maintain circular motion That increase in net centripetal force is produced by a reduction in the normal force from the ground This is felt as an effective force in the up directioni C The centrifugal force is independent of motion so as above the centrifugal force is upwardi In this case there is no coriolis force however This can be seen easily in the formula for the coriolis force Fcor Zmr X Q If one is walking south at the equator r is antiparallel to Q and the cross product gives zeroi 910 The derivation of the ctitious forces7 came from the expansion of the acceler ation in 932 lt gtSlltgts l 394 In the simpli cation that follows 9 is assumed constant therefore its time derivative is zero Relaxing that assumption gives us the additional term d2r 7 previous Q X r d t2 S 0 Which is then carried through to equation 934 where it appears as a third ctitious force the change in order in the cross product contains the negative sign picked up in this manipulation Fezmrgtlt l Problem 5 First consider the problem without including the rotational motion In this case we have projectile motion as described below 1 2t vot 7 g t2 From this we can also get that the time of impact is ti 21109 and the maximum height is h 70 We want to consider the effect of rotational motion on the eastwest motion of the projectile This motion will be impacted only by the coriolis force the centrifugal force is radially outward and does not act in the tangential or eastwest direction Our coordinate system for our location in the northern hermisphere is arranged tangentially to the ground with 2 up I the eastwest axis with east positive and y the northsouth axis with north positive This is a righthanded coordinate system To a rst approximation our projectile motion holds for the 2 direction However accounting for the coriolis force gives Fcor 2m X 2 2mvoigt2gtlt Q 2m v0 7 g t 7 cos A The cos A comes from the cross product remember that the earth7s latitude is measured from the equator not from the pole as we normally do in physics Now we just need to solve the equation of motion for the z or eastwest direction For the above force and initial conditions of 10 0 and 0 1t its 7 votQ 9 cos A So to nd how far in the eastwest direction the projectile moves before hitting the ground just plug in ti from above 3 2 g 2110 2110 A ti 9 A 7 7 7 7 z cos 3ltggt v0ltg 4 v3 7E9 cos A 3 7 9 503A amp 3 V 9 Remember that the minus sign above is indicating that this displacement is to the west as per our de nition of the positive z direction Problem 6 Let7s begin by considering the motion tangential to the surface rather than the strictly radial motion so we can avoid explicit consideration of the normal force a The angle of the curve at a given point 10 is given by tan 1f 10 So for our parabolic surface in which yT k T tant9 216 T Where 9 is the angle between horizontal and the surface at that point The tangential forces are then F lil 7mgsint9 Fall mQQTCOS6 Since there is no initial velocity we can leave out the coriolis force for the time being b Now balancing those tangential forces F lil Fall m 7gsint9 Q2Tcust9 0 gtant9 Q2T 29 k T Q2T N29 9

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