07_freefall

# 07_freefall - UCSD Physics 10 Falling Stuff UCSD Physics 10...

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Unformatted text preview: UCSD Physics 10 Falling Stuff UCSD Physics 10 Do Falling Objects Accelerate? It sure seems like it! Starts from rest, goes faster and faster.... What about a feather, though? Air resistance, drag Terminal velocity What if we could get rid of the air? What's responsible for the downwards force? If it's accelerating, then a force is acting: F = ma Spring 2008 2 UCSD Physics 10 Mass vs. Weight Mass is reluctance to accelerate (mass inertia) Weight is the force exerted by gravity Go to the moon: Does your mass change? Does your weight change? Spring 2008 3 UCSD Physics 10 Acceleration Due to Gravity At the earth's surface, all objects experience the same acceleration from gravity! "g"= 9.8m/s2 = 32 ft/s2 If the acceleration due to gravity is indeed universal, then... Since F = ma, the gravitational force must be proportional to mass. Spring 2008 4 UCSD Physics 10 Golf Ball vs. Bowling Ball Which one is more massive? Which one experiences more gravitational force? Which one is most reluctant to accelerate? How do they respond to a gravitational force? 5 Spring 2008 UCSD Physics 10 A Logical Argument Consider two identical falling objects Now imagine two more, connected together with a tiny thread Mass of each = M Total Mass = 2 M But they all have the same acceleration! Force on the joined balls must be twice the force on one of them, since the mass doubled but the acceleration stayed the same. Conclusion: Gravitational force must be proportional to mass Spring 2008 6 UCSD Physics 10 Spring 2008 7 UCSD Physics 10 How Do We Know the Accelerations are the Same? Experimental tests show the Universality of Free Fall is the same for different materials to within 0.00000000001% Spring 2008 8 UCSD Physics 10 Force Exerted by Gravity If the gravitationally induced acceleration is the same for all objects at the surface of the Earth, then Force exerted by gravity = (mass) (acceleration due to gravity) Fgravity = m g = WEIGHT, where g = 9.8 m/s2 For a mass of 100 kg, force from gravity at Earth's surface is F = 100 kg 9.8 m/s2 = 980 Newtons. (An apple weighs about 2 N, a golf ball weighs 1.4 N) Spring 2008 9 UCSD Physics 10 Said Another Way.... Gravitational force is proportional to mass F = ma gives an object's responding acceleration Divide both sides of the equation by "m" a = F/m Both numerator and denominator are proportional to "m", if force is gravity SO....acceleration is the same, regardless of the mass We'll return to this point when we consider General Relativity! Spring 2008 10 UCSD Physics 10 Falling Objects Accelerate Ignoring air resistance, falling objects near the surface of the Earth experience a constant acceleration of 9.8 m/s2. That means if you drop something it goes faster and faster, increasing its speed downwards by 9.8 m/s in each passing second. Spring 2008 11 UCSD Physics 10 Gravitational Force is Acting All the Time! Consider a tossed ball.... Does gravity ever switch off? As a ball travels in an arc, does the gravitational force change? Spring 2008 12 UCSD Physics 10 An Example of the Reductionist Approach By breaking the motion into independent parts, analysis is simplified! The horizontal and vertical motions are independent Spring 2008 13 UCSD Physics 10 Components of Motion Break the motion into 2 aspects, "components" Horizontal Vertical Is there a force acting in the horizontal direction? Is there a force acting in the vertical direction? Does the ball accelerate in the horizontal direction? Does its horizontal velocity change? Does the ball accelerate in the vertical direction? Does its vertical velocity change? Spring 2008 14 UCSD Physics 10 Projectile Motion All objects released at the same time (with no vertical initial velocity) will hit the ground at the same time, regardless of their horizontal velocity The horizontal velocity remains constant throughout the motion (since there is no horizontal force) Spring 2008 15 UCSD Physics 10 Some Exercises A ball falls from rest for 4 seconds. Neglecting air resistance, during which of the 4 seconds does the ball's speed increase the most? If you drop a ball from a height of 4.9 m, it will hit the ground 1 s later. If you fire a bullet exactly horizontally from a height of 4.9 m, it will also hit the ground exactly 1 s later. Explain. If a golf ball and a bowling ball (when dropped from the same height) will hit your foot at the same speed, why does one hurt more than the other? Spring 2008 16 UCSD Physics 10 Doing the Numbers Imagine dropping an object, and measuring how fast it's moving over consecutive 1 second intervals The vertical component of velocity is changing by 9.8 m/s in each second, downwards Let's approximate this acceleration as 10 m/s2 Spring 2008 17 UCSD Physics 10 Starting from rest, letting go: Time Interval Acceleration (m/s2 down) Vel. at end of interval (m/s down) After an interval t, the velocity changes by an amount at, so that vfinal = vinitial + at How fast was it going at the end of 3 sec? vinitial was 20 m/s after 2 sec a was 10 m/s (as always) t was 1 sec (interval) vfinal = 20 m/s + 10 m/s2 1 s = 30 m/s 01s 12s 23s 34s 45s 10 10 10 10 10 10 20 30 40 50 Spring 2008 18 UCSD Physics 10 Starting from rest, letting go: Time Interval Acceleration (m/s2 down) InitFinal Velocity (m/s down) Average Velocity (m/s down) The average velocity in the interval is just Vavg = (vinitial + vfinal) 01s 12s 23s 34s 45s 10 10 10 10 10 0 10 10 20 20 30 30 40 40 50 5 For the 1 2 s interval, 15 25 35 45 vi = 10 m/s vf = 20 m/s So vavg = (10+20) m/s = 15 m/s Spring 2008 19 UCSD Physics 10 Starting from rest, letting go: Time Interva l 01s 12s 23s 34s 45s Acceleration Final (m/s2 down) Velocity (m/s down) 10 10 10 10 10 10 20 30 40 50 Average Velocity (m/s down) 5 15 25 35 45 Dist. moved (m down) 5 15 25 35 45 + Final Position (m down) 5 20 45 80 125 Spring 2008 20 UCSD Physics 10 This can all be done in shorthand The velocity at the end of an interval is just (the starting velocity) plus (the time interval times the acceleration): vfinal = v(t) = vinit + at The position at the end of an interval is just (the starting position) plus (the time interval times the average velocity over the interval): x(t) = xinit + vavgt Since vavg = (vinit + vfinal), and vfinal = vinit + at, = xinit + (vinit + vinit + at) t, or x(t) = xinit + vinitt + at2 Spring 2008 21 x(t) UCSD Physics 10 An aside on units and cancellation What happens if you multiply an acceleration by a time? Units of acceleration are m/s2, units of time are s m Result is m s = s s s = m s s2 And this has units of velocity Spring 2008 22 UCSD Physics 10 Summary Velocity refers to both speed and direction Acceleration means a change in velocity Mass is a property of objects that represents their reluctance to accelerate If an object is accelerating, it's being acted on by an unbalanced force, and F = ma Gravity causes all objects to suffer the same acceleration, regardless of their mass or composition Gravitational acceleration only affects the vertical component of motion think in terms of components Spring 2008 23 UCSD Physics 10 Assignments HW 2: due Friday (4/18): Hewitt 11.E.16, 11.E.20, 11.E.32, 11.P.5, 2.E.6, 2.E.11, 2.E.14, 2.E.36, 2.E.38, 3.E.4, 3.E.5, 3.E.6, 3.E.19 turn in at lecture, or in box outside SERF 336 by 3PM Read Hewitt Chapters 2, 3, 4 suggested order/skipping detailed on website Spring 2008 24 ...
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## This note was uploaded on 02/12/2012 for the course PHYSICS 104 taught by Professor Staff during the Fall '10 term at Rutgers.

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