Chapter 6: Force & Motion II

Lecture 139/23/09

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

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

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

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

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

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

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

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

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

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

The Physical Cause of Centripetal Forces

Any force can be a centripetal forceAs long as it is perpendicular to the velocity

The Physical Cause of Centripetal Forces

Any force can be a centripetal forceGravity: Fg

Fg

The Physical Cause of Centripetal Forces

Any force can be a centripetal forceFriction: f

CC

x

Any force can be a centripetal forceTension: T

The Physical Cause of Centripetal Forces

T

The Physical Cause of Centripetal Forces

Any force can be a centripetal forceNormal force: N

C

y

x

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)

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

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?

ExampleSan Francisco car chase:

Fg

NFN + Fg = maFN - Fg = -mac

0 - mg = -mac

mac = mgac = g = v2/Rv = Rac

x

y