Class Note for PH 101 at UA-Generl Physics II (2)
Class Note for PH 101 at UA-Generl Physics II (2)
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This 10 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Alabama - Tuscaloosa taught by a professor in Fall. Since its upload, it has received 25 views.
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Date Created: 02/06/15
Chapter 7 Rotational Motion and the Law of Gravity This chapter deals with rotational kinematics 7 the relationships between rotational position velocity acceleration and time Rotational dynamics will be discussed in the next chapter Dynamics deals with energy momentum and forces Universal gravitation is also discussed in this chapter in part since satellite and planetary motion involves rotation Angular displacement velocity and acceleration s Angle 9 can be de ned in terms of radius r and arc length A s on a circle as S 6 Unless otherwise stated 639 is usually measured in the counterclockwise direction from the positive XaXis If s and r are measured in the same units eg m then the above equation gives His in radians Angle can also be measured in degrees or revolutions lrev 24 rad 360 Average angular velocity is given by Al At a The instantaneous angular velocity is just the average angular velocity in the limit of very short time interval Depending on units for 9 and t units for n can be rads degs revs revmin rpm etc Average angular acceleration is given by A1 At a Units can be rads2 degs2 revs2 revminz etc The relationships above are mathematically similar to those for motion in 1D Thus we can get the equations for constant angular acceleration from those for constant linear acceleration by appropriately changing the variables lD motion with constant a Rot motion with constant at L 2 J l 322quot Ax vot2at xgt6 AQ azot2at vv0at v gta wi 700 gtt v2 v02 2an 0 gt 0 gmz39 a02 52 05rAQ It is important that the units in the above equations be compatible For example if at is in rads2 then 03 should be in rads t in s and you will get 9 in rad Also don t mix the left and right equations above For example the equation A6 v0 t a t2 would not make sense Also don t confuse 0c and a although they look similar Example A wheel increases its rotational velocity from 200 rpm to 300 rpm in 10 sec What is its angular acceleration in rads2 mo 200 revminlmin 6OS27r radrev 209 rads a 300 revminlmin 6OS27139 radrev 314 rads Aw w wo 314rads 209rads At 2 At 2 10s a l05raals2 How many turns did the wheel make 6 m0 tat2 209rads10sl05rad321032 2615 rad naturns i 16 27139 Oruse 9a7t Relationship between angular and linear motion Since 3 7639 then it follows that As A6 iri 0r wmz At Av Am 7 r7 or far 2w At At In the above v1 is the tangential speed of a point going around a circle and at is the component of acceleration tangent to the circle Example A merrygoround rotates at a constant angular speed It takes 20 sec to make a complete revolution What is the speed of a rider who is 4 m from the center a A76 27rrad At 203 VI M 4m03l4rads 126ms 03l4rads The speed of a rider increases as he moves further from the center Centripetal acceleration An object moving in a circle can have a component of acceleration tangent to its path if it has an angular acceleration at rot This is related to the change in the speed In addition the object can have a radial component towards or away from the center because it is moving in a curved path even is its speed is constant This is also called the centripetal acceleration These two components of acceleration are A 2 V1 V1 v v arf gabi V2 A Av 39 At r A V2 The diagram to the right shows how the centripetal acceleration is obtained for uniform circular motion constant speed Because of the change in direction the velocity changes by an amount AV which is directed toward the center of the circle From the similarity of the triangles formed by sides As and r and by Av and v we V lVll lV2 have V MAA V r v AVLAS r 2 v Av vAs v v aci 77 Vi At rAt r r So for uniform circular motion the acceleration is always directed toward the center of the circle even though the speed is constant Example A car goes around a circular track of circumference 2000 m with constant speed of 1 minute What is the magnitude of the acceleration of the car 2000 quot 333ms At 603 r 2000 318m 27 27 2 2 ac L 33393MS 35ms2 r 318m 7 Combined tangential and centripetal acceleration As mentioned previously an object can have both tangential and radial centripetal components of acceleration if its speed is changing and it is changing direction Since these are mutually perpendicular components of the acceleration then the magnitude of the total acceleration is Av 39aC2 where at 7 and ac 7 At Centripetal force By Newton s 2quotd law an object moving in a circle with constant speed must have a net force directed toward the center of the circle given by V239 Fc39 3mab Example Suppose that the car going around the track in the above example has a mass of 1500 kg The net force acting on the car is then Fc mac 1500kg35ms2 5250N What is the origin of the centripetal force on the car going around the track I 11 If the track is at with no banking then the centripetal force is entirely due to friction between the tires and F the track This would require a coefficient of friction c given by Fc 5250N 7 2 036 n mg 1500kg98ms us Note This might be considered static friction even though the car is moving if the tires are not sliding skidding because of not enough friction the track can be banked so that the normal force has a radial component Then at just nsil the right bank angle no friction is required and 11g To avoid the possibility ofthe car 11 110099 2 Fc m nsin 639 r mg ncos6 which gives for the optimum angle of bank 2 myquot L g r Example What is the effective weight of a person when at the top and when at the bottom of a Ferris wheel Bottom m n Weff T 1 u mg The effective weight is the normal force exerted by the Ferris wheel seat on the person At the bottom V2 n mg mac m7 r v2 v2 Weff n mg 7 mgli r gr At the top V2 mg n mac m7 r 2 2 We nmg V7mg1 7 So the person weighs more at the bottom and less at the top The fractional change in 2 v h1s effect1ve we1ght 1s i 7 gr Example A Ferris wheel has radius r 8 m and it takes 30 sec to make one revolution What is the effective change in the weight of a 160 lb person at the bottom and top of the Ferris wheel V 27H 27r8m At 30s 168ms 168ms2 gr 98m328m 2 AW imgV i1601b0036 57 gr 7 0036 Universal law of gravity Newton s universal law of gravity which was rst introduced in chapter 4 states that two masses are attracted by a force given by Gm2 2 F where G 667 x 103911 Nmzkg2 not g 98 msz For spherical masses r is the distance between the centers of the masses The weight of a mass m on the surface of earth is then GM Em R E2 where RE is the radius of the earth Since we can also write W mg then WF GME R Example Calculate g from the above formula g 667x10 1117mzkg2598x2024 kg 6 2 98ms2 638x10 m At a distance of 638 x 106 m above the surface ofthe earth g 9822 245 ms2 and a persons weight would be 1A that on the surface Gravitational potential energy From the gravitational force formula one can obtain using calculus a general expression for the potential energy of two attracting masses as Iquot This expression is more general than the expression PE mgy which is valid for values of y that are small compared to earth s radius The above formula assumes that PE 0 when r 00 For finite separations the PE is negative In applications we only worry about changes in PE which can be positive or negative Example What is the escape speed of an object from a planet By escape speed we mean the minimum speed to launch an object such that it never returns to the surface of the planet This would require that it go an in nite distance from the planet where it eventually comes to rest We use conservation of energy Vesc 1 00 v 0 1 RE For earth the escape speed is ill 24 vesc w 112x104 ms 25000mph 638x106 The actual escape speed is larger because of atmospheric resistance Kepler s laws of planetary motion Kepler s laws are l Planets move in elliptical orbits with the sun at one of the focal points 2 A line from the sun to a planet sweeps out equal areas in equal times 3 The square of the orbital period of the planets is proportional to the cube of the average distance from the planet to the sun Johannes Kepler deduced these laws by carefully studying data on planetary motion obtained by Tycho Brahe Isaac Newton was able to explain Kepler s laws from the laws of motion and the law of universal gravitation Kepler s lst law An ellipse is an oblong closed curve with two focal points as shown to the right The ellipse can be traced out by following the path such that the sum of the distances from the focal points to any point on the curve is constant That is r1 r constant A circle is a special case of an ellipse where the two focal points are the same Kepler s 2nd law lanet The gure to the right is meant to illustrate B area p C Kepler s 2quotd law A planet goes in an elliptical quot path around the sun The time to move from A to B is the same as the time to move from C to A D From A to B the planet moves faster than from C to D so that the area swept out by the line from the planet to the sun is the same for both time intervals sun Kepler s 2quotd law is a direct consequence of conservation of angular momentum and the fact that the force of attraction is directed alone the line connecting the two bodies Kepler s 3rd law The 3rd law can most easily be obtained for a circular orbit as follows Fmac 2 2 WW zm zmwzmm r2 r r T2 where M 5 mass of sun m mass of planet T period of orbit of planet and r sun planet distance Solving the above for T 2 we have 2 p T2 439 r3 GM This equation would apply for any object orbiting a fixed body For example for a satellite orbiting the earth M 5 would be replaced by M E the mass of the earth Example What would be the period of a satellite in orbit just above the surface of earth Of course such an orbit could not be sustained because of atmospheric resistance 2 2 T2 4 RE3 f1 24 638x1063 GME 667x10 598x10 T 5070s 85min What is the speed of a satellite in a circular orbit or For such a low earth orbit ill 24 V V 7900ms 17700mph 638x10 Questions How does the period of orbit change as the radius increases How does the satellite speed change with increasing radius How does the period or orbit depend on the mass of the satellite 10
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