chapter 6: force & motion ii - department of physics€¦ · 23/09/2009 · force & motion...
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Chapter 6: Force & Motion II
Lecture 139/23/09
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The Drag Force &The Centripetal ForceGoals for this Lecture:
Describe the drag force between two objects.Relate the rag force to the terminal speed of falling objectsBriefly review uniform circular motionRelate the centripetal force to uniform circular motion
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Drag force is the non-solid equivalent of friction
When do we deal with the drag force: Whenever an object moves in a fluid.
A fluid can be a liquid (water, molasses, ...) or a gas (air, ...)
“Viscous” fluid = more drag force
Application point:At the surface of the moving objects
Direction:Opposite to the direction of motion
The Drag Force: D
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Magnitude:
C: Drag coefficient (Depends on the shape on an object)A: The effective cross section of the object (Depends on the shape on an object)ρ: The density of the fluidv: The velocity of the object through the fluid
Note: If v = 0, D = 0 To keep an object moving at a speed v requires a force F that is ∝ to v2 (since Fnet = F - D = 0)
The Drag Force: DD = ½CρAv2
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ExampleAuto fuel consumption vs velocity
The frictional forces on a moving automobile consist of two major components a) a constant rolling resistance term and b) a velocity dependent drag force
A Hummer H2 has the following propertiesFrolling = 300 NDrag coefficient: C = 0.57 (CHonda-Insight = 0.25)Area (looking from the front): A = 2.5 m2 Density of air: ρ = 1.3 kg/m3
How much more force must be supplied by the engine to keep the H2 moving at 100 mph, compared to 60 mph
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ExampleAuto fuel consumption vs velocity
60 mph = 60 miles/hour*1600m/mile*1 hour/3600 s = 27 m/sFdrag-60 = ½CρAv2 = ½(0.57)(1.3 kg/m3)(2.5 m2)(27 m/s)2 = 675 NFdrag-100 = ½CρAv2 = ½(0.57)(1.3 kg/m3)(2.5 m2)(45 m/s)2 = 1876 NFtot-60 = 300 N + 675 N = 975 NFtot-100 = 300 N + 1876 N = 2176 NFtot-100 / Ftot-60 = 1976/775 = 2.23 <-- 2.23x worse fuel consumption
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Terminal SpeedConsider an object in free fall through air
The forces acting on it are gravity and the drag force1. The object has just been dropped
v = 0, D = 0, Fnet,1 = Fg - D = Fg : object accelerates2. The object has fell for a little time
v > 0, D > 0, Fnet,2 = Fg - D < Fg : object still accelerates, but at a slower rate
3. The object has fell for more timev >> 0, D = Fg , Fnet,2 = Fg - D = 0 : object no longer accelerates, but falls at a constant speed
D
Fg
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When the object has reached terminal speedD = Fg --> 1/2 CρAv2 = mgmt.
Terminal speed depends on:Shape of the object (C, A)Size of the object (A)Mass of the object (m)Density of the air (ρ)
Terminal Speed
D
Fg
vterm =!
2mg
C!A
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Object Mass Area Terminal SpeedTerminal SpeedObject Mass Aream/s mph
Skydiver 75 kg 0.7 m2 60 134Baseball 145 g 42 cm2 33 74Golf ball 46 g 14 cm2 32 72Hail stone 0.48 g 0.79 cm2 14 31Raindrop 0.034 g 0.13 cm2 9 20
Terminal Speeds
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Uniform Circular MotionAn object moving with uniform circular motion has:
Constant angular velocity: ωConstant speed (i.e. magnitude of velocity): v = ωRChanging velocity vectorAcceleration towards the center (centripetal acceleration): ac = v2/r = ω2RConstant magnitude of accelerationChanging acceleration vector
C
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Centripetal Force: FcWhen do we deal with the centripetal force:
Whenever an object moves in a curved trajectory
Direction:Points towards the center of curvature, and always perpendicular to the velocity vector
Magnitude: Newton 2nd law: Fc = mac --> |F| = mv2/R = mω2R
Note: The centripetal force in the NET force perpendicular to the direction of motion
C
Fc
FcFc
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The Physical Cause of Centripetal Forces
Any force can be a centripetal forceAs long as it is perpendicular to the velocity
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The Physical Cause of Centripetal Forces
Any force can be a centripetal forceGravity: Fg
Fg
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The Physical Cause of Centripetal Forces
Any force can be a centripetal forceFriction: f
CC
x
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Any force can be a centripetal forceTension: T
The Physical Cause of Centripetal Forces
T
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The Physical Cause of Centripetal Forces
Any force can be a centripetal forceNormal force: N
C
y
x
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ExampleArched bridge vs. Suspended bridge
A car with mass m is driving at constant speed v across a bridge with radius of curvature R. What is the force which the car exerts on the bridge?
Hint: The force which the car exerts on the bridge is equal to (in magnitude) to the force which the bridge exerts on the car (i.e. The normal force)
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ExampleArched bridge vs. Suspended bridge:
Fg
N
Fg
N
FN + Fg = maFN - Fg = -mac
|FN| = Fg - mac = m(g - v2/R)x
y
FN + Fg = maFN - Fg = +mac
|FN| = Fg + mac = m(g + v2/R)
A suspension bridge feels more stress from passing traffic
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ExampleSan Francisco car chase:
How fast does a car have to moving in order to fly off the top of a curved road / hill with a radius of curvature of R meters?
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ExampleSan Francisco car chase:
Fg
NFN + Fg = maFN - Fg = -mac
0 - mg = -mac
mac = mgac = g = v2/Rv = Rac
x
y