Acceleration due to Gravity1 ObjectTo determine the acceleration due to gravity by different methods.2 ApparatusBalance, ball bearing, clamps, electric timers, meter stick, paper strips, precision pulley, r amps,weights, metal track.3 TheoryAccording to Newton any two objects will attract one anothe r if they both have mass. Thisattraction, F , is given by the formulaF =GM1M2r2(1)where M1and M2are the masses of the two objects and r is the distance betwee n the centers ofmass of the two. G is a constant of propor ti onal i ty, 6.67 × 10−11Nm2/kg2.If one of these objects is the Earth (ME) and the other is a relatively small object near the surfaceof the Earth (Mo), the distance between t he two is essentially the radiu s of the Earth, RE. Thenequation 1 becomesF =GMEMoR2E(2)where MEis the mass of the Earth (5.983 × 1024kg), REis its radius (6.371 × 106m) and Moisthe mass of the object.If the object is in free fall, this is the only force acting on it andF = Moa = Mog (3)where g is the acceleration due to the gravity of the Eart h.Combining the last two equations givesMog =GMEMoR2E(4)g =GMER2E(5)org = 9.807 m/s2(6)In si t uati on s where the object is not in free fall it will still interact with the Eart h and this willcause the object to have a weight, W , which is given by Mog.1The weight of an object as well as the acceleration due to gravity are variabl es , but if one staysnear the surface of the Earth both may be consi de r e d to be constant. In this experiment g will bedetermined by a number of methods.3.1 Atwood’s MachineA common device used to calculate accelerations is Atwood’s apparatus. In this case, it consistsof two masses, m1and m2, attached by a string (assumed massless) which in turn is on a pulley.Schematically it looks like Fig. 1. Applying Newton’s Secon d Law to each object (assumingRm12m1m2(+)(+)mpT T1 2m1m2g gmFigure 1: Atwood’s machine (left) and fr e e body diagrams for each mass (right) with forces labeled.m1> m2) one hasm1a = m1g − Tm2a = T − m2g (7)The accelerations are eq ual since the objects ar e tied together, and it is customary to assume thetension equal and evenly distributed throughout a string. Solving the above equations for theacceleration due to gravity gives:g =m1+ m2m1− m2a (8)If the accelerati on a has an error associated with it, δa, then the corresponding error on g, δg, willbe given by:δg =m1+ m2m1− m2δa (9)The above discussion ignores the inertia of the pulley. If it is to be taken into acc ount one mustconsider the rotational acceleration, α, caused by the string:τ = Iα (10)2Where I =12MR2is the rotational inertia of a disk, α = a/R for the edge of the pull e y, mpis themass of the pulley and R is its radius.Now, however, the two tensions are differe nt and equat i ons 7 becomem1a = m1g − T1m2a = T2− m2g (11)and these two tensions cause the pulley to rotate by producing torques.The total torque, τ, produced by the strings is:τ = T1R − T2R (12)Equations 10 and 12 giveτ = Iα =12mpR2aR= T1R − T2R (13)or12mpa = T1− T2(14)Combining equations 11 and 14 and solving for g yieldsg =m1+m2+0.5mpm1−m2a(15)with an associated error on g ofδg =m1+m2+0.5mpm1−m2δa(16)3.2 Galileo’s RampGalileo measured the acceleration of objects as they rolled down a r amp . He did this inste adof working with falling objects at least partially because the balls on ramps roll e d much slowerand were more easily observed. The acceleration of the ball on the ramp can be related to theacceleration due to gravi ty. Consider a ball rolling down the ramp shown in Figure 2. UsingNewton’s Second Law of motion, F = ma, in two dimensi ons gives: parallel to inclineW sin θ − fs= ma (17)perpendicular to inclineN − W cos θ = 0 (18)(no motion and thus no acceleration ⊥ to incline)Applying the rotational equivalent of Newton’s Second Law about the center of mass yieldsτ = Iα (19)fsR =25mR2αIspher e=25mR2(20)because only the friction force produces a torque. Sofs=25mRα (21)3wsfθθNFigure 2: Galileo’s ramp with a ball rolling down without slipping. The three forces on the ballare shown. The fricti onal f orc e pr oduces t he torque about the center of the ball.but for rolling motion α = a/R. This lead s tofs=25ma (22)Using this and W = mg, eq uat i on 17 becomesmg sin θ −25ma = mamg sin θ =75ma (23)Solving for g yieldsg =75 sin θa(24)δg =75 sin θδa(25)where a is the acceleration of the ball down the pl ane and θ is the angl e the plane makes with thehorizontal.4 ProcedureNote on measuring acceler ati on: motion in one directi on is given by d = vot +12at2, where d isthe distance the object moves, vois the initial velo c i ty, a is the constant acceleration, and t is theelapsed time. If the object starts from rest, then vo= 0, and:d =12at2(26)Solving for a gives:a =2dt2δa = 2aδtt(27)If d and t are known, then a can be calculated.44.1 Atwood’s MachinePut a string over the pulley and attach the pulley firmly to a support apparatus. Attach knownmasses to both ends of the string and record the time nece ssar y for the heavier mass to fall a k nowndistance. No masses should come in contact with the pulley, or the floor. Information recordedshould be the two masses (m1heavier), the mass of the pull e y (mp= 114 g for the solid grey pulley;mp= 68 g for the orange spoked pulley), the dis tan ce traveled by the masses, and the time interval.Repeat for a total of three (3) trials, an d redo the trials for a total of three different mass combi-nations.4.2 Galileo’s RampNote the ball that is to roll down the incline must be a sol i d ball. Set up the track so that it i s atan incline. By measuring and usi n g trigonometry, determine the angle θ of your ramp with respectto the horizontal. Also record the distance d that the ball will traverse while b e i ng timed. Thenrecord the time for the ball to travel this distance, starting from rest, for a total of three (3) times.Use total of three (3) different balls and repeat the trials.Change the inclination of the track and the length d and repeat the procedure just outli ned above.5 Calculations5.1 Atwood’s MachineCalculate the mean time tavgand its associated error δt for the time of fall of the heavier mass.From these results, calculate the mean acceleration aavgand its associated error δa usi ng equations27. Then calculate the mean val u e of the acceleration of gravity g and its associated error δg