chapter 6 – force and motion ii drag forces and terminal speed.drag forces and terminal speed....

Chapter 6 – Force and Motion IIChapter 6 – Force and Motion II

• Drag forces and terminal speed.Drag forces and terminal speed.

• Uniform circular motion.Uniform circular motion.

• Other Applications of Newton’s LawsOther Applications of Newton’s Laws

Uniform Circular MotionUniform Circular Motion

• A force, A force, FFrr , is directed , is directed

toward the center of toward the center of the circlethe circle

• This force is associated This force is associated with an acceleration, with an acceleration, aacc

• Applying Newton’s Applying Newton’s Second Law along the Second Law along the radial direction givesradial direction gives

2

c

vF ma m

r

Uniform Circular MotionUniform Circular Motion

• A force causing a A force causing a centripetal acceleration acts centripetal acceleration acts toward the center of the toward the center of the circlecircle

• It causes a change in the It causes a change in the direction of the velocity direction of the velocity vectorvector

• If the force vanishes, the If the force vanishes, the object would move in a object would move in a straight-line path tangent to straight-line path tangent to the circlethe circle

Centripetal ForceCentripetal Force

• The force causing the centripetal acceleration The force causing the centripetal acceleration is sometimes called the is sometimes called the centripetal forcecentripetal force

• This is not a new force, it is a new This is not a new force, it is a new rolerole for a for a forceforce

• It is a force It is a force acting in the role of a force that acting in the role of a force that causes a circular motioncauses a circular motion

r

vmmaF c

2

Conical PendulumConical Pendulum

• The object is in equilibrium in The object is in equilibrium in the vertical direction and the vertical direction and undergoes uniform circular undergoes uniform circular motion in the horizontal motion in the horizontal directiondirection(1) (1) TcosTcosθθ = mg = mg

(2) (2) TsinTsinθθ = ma = macc= mv= mv22/r/r

Conical PendulumConical Pendulum

• Dividing (2) by (1) and using Dividing (2) by (1) and using sinsinθθ/cos/cosθθ = tan = tanθθ we we eliminate eliminate TT and find: and find:

• vv is independent of is independent of mm

sin tanv Lg

tantan2

rgvrg

v

Using Using r = Lsinr = Lsinθθ

Consider a conical pendulum with an Consider a conical pendulum with an 80.0-kg80.0-kg bob on a bob on a 10.0-m 10.0-m wire making an angle of wire making an angle of 5.005.00 with the vertical. with the vertical. Determine (a) the horizontal and vertical components of the Determine (a) the horizontal and vertical components of the force exerted by the wire on the pendulum and (b) the radial force exerted by the wire on the pendulum and (b) the radial acceleration of the bob.acceleration of the bob.

Consider a conical pendulum with an Consider a conical pendulum with an 80.0-kg80.0-kg bob on a bob on a 10.0-m 10.0-m wire making an angle of wire making an angle of 5.005.00 with the vertical. with the vertical. Determine (a) the horizontal and vertical components of the Determine (a) the horizontal and vertical components of the force exerted by the wire on the pendulum and (b) the radial force exerted by the wire on the pendulum and (b) the radial acceleration of the bob.acceleration of the bob.

toward the center of the circle.

2cos5.00 80.0 kg 9.80 m sT mg

(a)(a) : ˆ ˆ68.6 N 784 N T i j787 NT

sin5.00 cT ma (b)(b)

The length of the wire is unnecessary information. We could, The length of the wire is unnecessary information. We could, on the other hand, use it to find the radius of the circle, the on the other hand, use it to find the radius of the circle, the speed of the bob, and the period of the motion speed of the bob, and the period of the motion

ac= (T sin5.00/)m = .857 m/s2

A A 4.00-kg 4.00-kg object is attached to a vertical rod by two strings, object is attached to a vertical rod by two strings, as in Figure. The object rotates in a horizontal circle at as in Figure. The object rotates in a horizontal circle at constant speed constant speed 6.00 m/s6.00 m/s. Find the tension in (a) the upper . Find the tension in (a) the upper string and (b) the lower string.string and (b) the lower string.


24 kg 9.8 m s 39.2 NgF mg

1.5 msin

2 m48.6

2 m cos48.6 1.32 mr

2

24 kg 6 m s

cos48.6 cos48.61.32 m

109 N165 N

cos48.6

x x

a b

a b

mvF ma

r

T T

T T

39.2 N

T a

T b forces

v a c

motion


2

24 kg 6 m s

cos48.6 cos48.61.32 m

109 N165 N

cos48.6

x x

a b

a b

mvF ma

r

T T

T T

sin48.6 sin48.6 39.2 N 0

39.2 N52.3 N

sin48.6

y y

a b

a b

F ma

T T

T T

39.2 N

T a

T b forces

v a c

motion


39.2 N

T a

T b forces

v a c

motion

(a)To solve simultaneously, we add the equations in Ta and Tb :

(b)(b)

165 N 52.3 Na b a bT T T T

217 N108 N

2aT

165 N 165 N 108 N 56.2 Nb aT T

How fast can it spin?How fast can it spin?• The speed at which the The speed at which the

object moves depends object moves depends on the mass of the object on the mass of the object and the tension in the and the tension in the cord. The centripetal cord. The centripetal force is supplied by the force is supplied by the tension:tension:

Trv

m

r

vmT

2

This shows tat v increases with T This shows tat v increases with T and decreases with larger m.and decreases with larger m.

Horizontal (Flat) CurveHorizontal (Flat) Curve• The force of static friction The force of static friction

supplies the centripetal forcesupplies the centripetal force

• The maximum speed at which the The maximum speed at which the car can negotiate the curve iscar can negotiate the curve is

• Note, this does not depend on Note, this does not depend on the mass of the carthe mass of the car

v grFr

r

vmfS

2

Banked CurveBanked Curve

• These are designed with These are designed with friction equaling zerofriction equaling zero

• There is a component of the There is a component of the normal force that supplies the normal force that supplies the centripetal force (1), and centripetal force (1), and component of the normal component of the normal force that supplies the force that supplies the gravitational force (2).gravitational force (2).

• Dividing (1) by (2) gives:Dividing (1) by (2) gives:

)2(cos

)1(sin2

mgn

r

mvn

2

tanv

rg

Loop-the-LoopLoop-the-Loop• A pilot in a jet aircraft A pilot in a jet aircraft

executes a loop-the loop. executes a loop-the loop. This is an example of a This is an example of a vertical circlevertical circle

• At the bottom of the loop, At the bottom of the loop, (b), the upward constant (b), the upward constant force, that keeps the pilot force, that keeps the pilot moving in a circular path moving in a circular path at a constant speed, is at a constant speed, is greater than its weight, greater than its weight, because the normal and because the normal and gravitational forces act in gravitational forces act in opposite direction: opposite direction:

rg

vmgnbottom

2

1r

vmmgnF bot

2

Loop-the-LoopLoop-the-Loop

• At the top of the circle, At the top of the circle, (c), the force exerted on (c), the force exerted on the object is less than the object is less than its weight, because its weight, because both the gravitational both the gravitational force and the normal force and the normal force, nforce, ntoptop, exerted on , exerted on

the pilot by the seat act the pilot by the seat act in same direction:in same direction:

2

1top

vn mg

rg

r

vmmgnF top

2

A A 0.400-kg0.400-kg object is swung in a vertical circular path on a object is swung in a vertical circular path on a string string 0.500 m0.500 m long. If its speed is long. If its speed is 4.00 m/s4.00 m/s at the top of the at the top of the circle, what is the tension in the string there?circle, what is the tension in the string there?

A A 0.400-kg0.400-kg object is swung in a vertical circular path on a object is swung in a vertical circular path on a string string 0.500 m0.500 m long. If its speed is long. If its speed is 4.00 m/s4.00 m/s at the top of the at the top of the circle, what is the tension in the string there?circle, what is the tension in the string there?

At the top of the vertical circle,At the top of the vertical circle,2v

T m mgR

24.00

0.400 0.400 9.80 8.88 N0.500

T

Non-Uniform Circular MotionNon-Uniform Circular Motion• When the force acting on a

particle moving in a circular path has a tangential component , the particles speed changes.

• The acceleration has a tangential components

• FFrr produces the centripetal acceleration

• Ft produces the tangential acceleration.

• The total force is the vector sum or the radial force and tangential force: tr FFF

tF

Vertical Circle with Non-Uniform SpeedVertical Circle with Non-Uniform Speed

• The gravitational force The gravitational force exerts a tangential force on exerts a tangential force on the objectthe object– Look at the components of Look at the components of FFgg

• The gravitational force The gravitational force resolve into a tangential resolve into a tangential component component mg sinmg sinθθ and a and a radial component radial component mg cosmg cosθθ..

Vertical Circle with Non-Uniform SpeedVertical Circle with Non-Uniform Speed

The tension at any point The tension at any point can be found:can be found:

2

cosv

T m gR

r

mvmgTF

gamamgF

r

ttt

2

cos

sinsin

•Applying II NL to the tangential and Applying II NL to the tangential and radial directions:radial directions:

Tarzan (Tarzan (mm = 85.0 kg = 85.0 kg) tries to cross a river by swinging from ) tries to cross a river by swinging from a vine. The vine is a vine. The vine is 10.0 m10.0 m long, and his speed at the bottom long, and his speed at the bottom of the swing (as he just clears the water) will be of the swing (as he just clears the water) will be 8.00 m/s8.00 m/s. . Tarzan doesn't know that the vine has a breaking strength Tarzan doesn't know that the vine has a breaking strength of of 1 000 N1 000 N. Does he make it safely across the river?. Does he make it safely across the river?

Tarzan (Tarzan (mm = 85.0 kg = 85.0 kg) tries to cross a river by swinging from ) tries to cross a river by swinging from a vine. The vine is a vine. The vine is 10.0 m10.0 m long, and his speed at the bottom long, and his speed at the bottom of the swing (as he just clears the water) will be of the swing (as he just clears the water) will be 8.00 m/s8.00 m/s. . Tarzan doesn't know that the vine has a breaking strength Tarzan doesn't know that the vine has a breaking strength of of 1 000 N1 000 N. Does he make it safely across the river?. Does he make it safely across the river?

Let the tension at the lowest point be Let the tension at the lowest point be TT..

2

2

22

:

8.00 m s85.0 kg 9.80 m s 1.38 kN 1000 N

10.0 m

cmv

F ma T mg mar

vT m g

r

T

He doesn’t make it across the river because the vine breaks.

Motion in accelerated FramesMotion in accelerated Frames

When Newton’s laws of motion was When Newton’s laws of motion was introduced in Chapter 5, we emphasized that they introduced in Chapter 5, we emphasized that they are valid only for inertial frames of reference.are valid only for inertial frames of reference.

In this section, we will analyze the In this section, we will analyze the noninertial frames, that is, on that is accelerating.noninertial frames, that is, on that is accelerating.

Example: Let’s consider a hockey puck on at table Example: Let’s consider a hockey puck on at table in a moving train. The train moving with a constant in a moving train. The train moving with a constant velocity represents an inertial frame. The puck at velocity represents an inertial frame. The puck at rest remains at rest, and Newton’s I low is obeyed. rest remains at rest, and Newton’s I low is obeyed.

Motion in accelerated FramesMotion in accelerated FramesExample: Let’s consider a hockey puck on at table in a Example: Let’s consider a hockey puck on at table in a moving train. The train moving with a constant velocity moving train. The train moving with a constant velocity represents an inertial frame. The puck at rest remains at represents an inertial frame. The puck at rest remains at rest, and Newton’s I low is obeyed. rest, and Newton’s I low is obeyed.

The accelerating train is not an inertial frame. For the The accelerating train is not an inertial frame. For the observer on the train, there appears to be no visible force observer on the train, there appears to be no visible force on the puck, but it will accelerate from rest toward the back on the puck, but it will accelerate from rest toward the back of the train, as the train start to accelerate. The Newton’s I of the train, as the train start to accelerate. The Newton’s I law is violated.law is violated.

The observer on the accelerating train, if he applied The observer on the accelerating train, if he applied the N II law to the puck, might conclude that a force has the N II law to the puck, might conclude that a force has acted on the puck to cause it to accelerate. acted on the puck to cause it to accelerate.

We call an apparent force such as this a We call an apparent force such as this a fictitious fictitious force.force.

Motion in Accelerated FramesMotion in Accelerated Frames• A A fictitious forcefictitious force results from an results from an

accelerated frame of referenceaccelerated frame of reference– A fictitious force appears to act on an object in A fictitious force appears to act on an object in

the same way as a real force, but you cannot the same way as a real force, but you cannot identify a second object for the fictitious forceidentify a second object for the fictitious force

““Centrifugal” ForceCentrifugal” Force

• From the frame of the passenger (b), a From the frame of the passenger (b), a force appears to push her toward the doorforce appears to push her toward the door

• From the frame of the Earth, the car From the frame of the Earth, the car applies a leftward force on the passengerapplies a leftward force on the passenger

• The outward force is often called a The outward force is often called a centrifugalcentrifugal force force– It is a fictitious force due to the acceleration It is a fictitious force due to the acceleration

associated with the car’s change in directionassociated with the car’s change in direction

If the coefficient of static friction between your coffee cup If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is and the horizontal dashboard of your car is μμss = 0.800 = 0.800, how , how

fast can you drive on a horizontal roadway around a right fast can you drive on a horizontal roadway around a right turn of radius turn of radius 30.0 m30.0 m before the cup starts to slide? If you before the cup starts to slide? If you go too fast, in what direction will the cup slide relative to the go too fast, in what direction will the cup slide relative to the dashboard?dashboard?

If the coefficient of static friction between your coffee cup If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is and the horizontal dashboard of your car is μμss = 0.800 = 0.800, how , how

fast can you drive on a horizontal roadway around a right fast can you drive on a horizontal roadway around a right turn of radius turn of radius 30.0 m30.0 m before the cup starts to slide? If you before the cup starts to slide? If you go too fast, in what direction will the cup slide relative to the go too fast, in what direction will the cup slide relative to the dashboard?dashboard?

We adopt the view of an inertial observer. If it is on the verge of sliding, the cup is moving on a circle with its centripetal acceleration caused by friction.

2

: 0

:

y y

x x s s

F ma n mg

mvF ma f n mg

r

20.8 9.8 m s 30 m 15.3 m ssv gr

If you go too fast the cup will begin sliding straight across the dashboard to the left.

• A person stands on a scale in an elevator. As the A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of elevator starts, the scale has a constant reading of 591 N591 N. . As the elevator later stops, the scale reading is As the elevator later stops, the scale reading is 391 N391 N. . Assume the magnitude of the acceleration is the same Assume the magnitude of the acceleration is the same during starting and stopping, and determine (a) the weight of during starting and stopping, and determine (a) the weight of the person, (b) the person's mass, and (c) the acceleration the person, (b) the person's mass, and (c) the acceleration of the elevator.of the elevator.


max 591 NgF F ma

min 391 NgF F ma

2 982 NgF

+

491 NgF (a)

(b)2

491 N50.1 kg

9.80 m sm


max 591 NgF F ma

min 391 NgF F ma -

(c)

2 200 Nma 22.00 m sa

“Coriolis Force”Coriolis Force”

• This is an apparent This is an apparent force caused by force caused by changing the radial changing the radial position of an object position of an object in a rotating in a rotating coordinate systemcoordinate system

• The result of the The result of the rotation is the curved rotation is the curved path of the ballpath of the ball

The Earth rotates about its axis with a period of The Earth rotates about its axis with a period of 24.0 h24.0 h. . Imagine that the rotational speed can be increased. If an Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight, (a) object at the equator is to have zero apparent weight, (a) what must the new period be? (b) By what factor would what must the new period be? (b) By what factor would the speed of the object be increased when the planet is the speed of the object be increased when the planet is rotating at the higher speed? rotating at the higher speed?

Note that the apparent weight of the object becomes Note that the apparent weight of the object becomes zero when the normal force exerted on it is zero. zero when the normal force exerted on it is zero.

The Earth rotates about its axis with a period of The Earth rotates about its axis with a period of 24.0 h24.0 h. . Imagine that the rotational speed can be increased. If an object Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight, (a) what must at the equator is to have zero apparent weight, (a) what must the new period be? (b) By what factor would the speed of the the new period be? (b) By what factor would the speed of the object be increased when the planet is rotating at the higher object be increased when the planet is rotating at the higher speed? Note that the apparent weight of the object becomes speed? Note that the apparent weight of the object becomes zero when the normal force exerted on it is zero. zero when the normal force exerted on it is zero.

(a) r rF ma22

2

2

2 63

2

2

4

4 6.37 10 m2 5.07 10 s 1.41 h

9.80 m s

mv m Rmg

R R T

Rg

T

RT

g

(b) speed increase factor

new current current

current new new

2 24.0 h17.1

2 1.41 hv T TRv T R T

Fictitious Forces, examplesFictitious Forces, examples

• Although fictitious forces are not real Although fictitious forces are not real forces, they can have real effectsforces, they can have real effects

• Examples:Examples:– Objects in the car do slideObjects in the car do slide– You feel pushed to the outside of a rotating You feel pushed to the outside of a rotating

platformplatform– The Coriolis force is responsible for the The Coriolis force is responsible for the

rotation of weather systems and ocean rotation of weather systems and ocean currentscurrents

Fictitious Forces in Linear SystemsFictitious Forces in Linear Systems

• The inertial observer (a) The inertial observer (a) seessees

• The noninertial observer The noninertial observer (b) sees(b) sees

sin

cos 0

x

y

F T ma

F T mg

' sin

' cos 0

x fictitious

y

F T F ma

F T mg

Fictitious Forces in a Rotating SystemFictitious Forces in a Rotating System

• According to the inertial observer (a), the tension is the According to the inertial observer (a), the tension is the centripetal forcecentripetal force

• The noninertial observer (b) seesThe noninertial observer (b) sees

2mvT

r

2

0fictitious

mvT F T

r

Motion with Resistive ForcesMotion with Resistive Forces

• Motion can be through a mediumMotion can be through a medium– Either a liquid or a gasEither a liquid or a gas

• The medium exerts a The medium exerts a resistive force, resistive force, RR,, on an on an object moving through the mediumobject moving through the medium

• The magnitude of The magnitude of RR depends on the medium depends on the medium• The direction of The direction of RR is opposite the direction of is opposite the direction of

motion of the object relative to the mediummotion of the object relative to the medium• RR nearly always increases with increasing speed nearly always increases with increasing speed

Motion with Resistive ForcesMotion with Resistive Forces

• The magnitude of The magnitude of RR can depend on the can depend on the speed in complex waysspeed in complex ways

• We will discuss only twoWe will discuss only two– R R is proportional to is proportional to vv

• Good approximation for slow motions or small Good approximation for slow motions or small objectsobjects

– R R is proportional to is proportional to vv22

• Good approximation for large objectsGood approximation for large objects

RR Proportional To Proportional To vv

• The resistive force can be expressed as The resistive force can be expressed as R R == - - b b vv

• bb depends on the property of the medium, depends on the property of the medium, and on the shape and dimensions of the and on the shape and dimensions of the objectobject

• The negative sign indicates The negative sign indicates RR is in the is in the opposite direction to opposite direction to vv

RR Proportional to Proportional to vv, Example, Example

• Analyzing the motion Analyzing the motion results inresults in

dvmg bv ma m

dtdv b

a g vdt m

RR Proportional to Proportional to vv

• Initially, Initially, vv = 0 = 0 and and dvdv//dtdt = = gg• As As tt increases, increases, RR increases increases

and and aa decreasesdecreases• The acceleration approaches The acceleration approaches

0 0 when when RR →→ mgmg• At this point, At this point, vv approaches approaches

the the terminal speedterminal speed of the of the objectobject

Terminal SpeedTerminal Speed

To find the terminal speed, let To find the terminal speed, let aa = 0 = 0

Solving the differential Solving the differential equation givesequation gives

is the is the time constanttime constant and and = = mm//bb

T

mgv

b

1 1bt m tT

mgv e v e

b

• For objects moving at high speeds through air, For objects moving at high speeds through air, the resistive force is approximately equal to the the resistive force is approximately equal to the square of the speedsquare of the speed

R = ½ DR = ½ DAvAv22

– DD is a dimensionless empirical quantity that called the is a dimensionless empirical quantity that called the drag coefficientdrag coefficient

is the density of airis the density of air– AA is the cross-sectional area of the object is the cross-sectional area of the object– vv is the speed of the object is the speed of the object

RR Proportional to Proportional to vv22

R Proportional to vR Proportional to v22

• Analysis of an object falling Analysis of an object falling through air accounting for through air accounting for air resistanceair resistance

2

2

1

2

2

F mg D Av ma

D Aa g v

m

RR Proportional to Proportional to vv22, Terminal Speed, Terminal Speed

• The terminal speed will The terminal speed will occur when the occur when the acceleration goes to zeroacceleration goes to zero

• Solving the equation Solving the equation givesgives

2T

mgv

D A

Some Terminal SpeedsSome Terminal Speeds

Process for Problem-SolvingProcess for Problem-Solving

• Analytical MethodAnalytical Method– The process used so far involves the The process used so far involves the

identification of well-behaved functional identification of well-behaved functional expressions generated from algebraic expressions generated from algebraic manipulation or techniques of calculusmanipulation or techniques of calculus

Analytical MethodAnalytical Method

• Apply the method using this procedure:Apply the method using this procedure:– Sum all the forces acting on the particle to Sum all the forces acting on the particle to

find the net force, find the net force, F.F.– Use this net force to determine the Use this net force to determine the

acceleration from the relationship acceleration from the relationship aa = =FF//mm– Use this acceleration to determine the velocity Use this acceleration to determine the velocity

from the relationship from the relationship dvdv//dtdt = = aa– Use this velocity to determine the position Use this velocity to determine the position

from the relationship from the relationship dxdx//dtdt = = vv

Analytic Method, ExampleAnalytic Method, Example

• Applying the procedure:Applying the procedure:

FFgg = = mamayy = - = - mgmg

aayy = - = -g g and and dvdvyy//dtdt = - = -gg

vvyy(t)(t) = = vvyiyi – – gtgt

y(t)y(t) = = yyii + + vvyiyi t t – ½ – ½ gtgt22

Numerical ModelingNumerical Modeling

• In many cases, the analytic method is not In many cases, the analytic method is not sufficient for solving “real” problemssufficient for solving “real” problems

• Numerical modeling can be used in place Numerical modeling can be used in place of the analytic method for these more of the analytic method for these more complicated situationscomplicated situations

• The The Euler methodEuler method is one of the simplest is one of the simplest numerical modeling techniquesnumerical modeling techniques

Euler MethodEuler Method

In the Euler Method, derivatives are In the Euler Method, derivatives are approximated as ratios of finite differencesapproximated as ratios of finite differences

tt is assumed to be very small, such that is assumed to be very small, such that the change in acceleration during the time the change in acceleration during the time interval is also very smallinterval is also very small

Equations for the Euler MethodEquations for the Euler Method

( ) ( )

( ) ( ) ( )

and

( ) ( )( )

( ) ( )

v v t t v ta t

t tv t t v t a t t

x x t t x tv t

t tx t t x t v t t

Euler MethodEuler Method

• It is convenient to set up the numerical It is convenient to set up the numerical solution to this kind of problem by solution to this kind of problem by numbering the steps and entering the numbering the steps and entering the calculations into a tablecalculations into a table

• Many small increments can be taken, and Many small increments can be taken, and accurate results can be obtained by a accurate results can be obtained by a computercomputer

Euler Method Set UpEuler Method Set Up

Euler Method FinalEuler Method Final• One advantage of the method is that the dynamics

are not obscured– The relationships among acceleration, force, velocity

and position are clearly shown

• The time interval must be small– The method is completely reliable for infinitesimally small

time increments– For practical reasons a finite increment must be chosen– A time increment can be chosen based on the initial

conditions and used throughout the problem

• In certain cases, the time increment may need to be changed within the problem

Accuracy of the Euler MethodAccuracy of the Euler Method

• The size of the time increment influences the The size of the time increment influences the accuracy of the resultsaccuracy of the results

• It is difficult to determine the accuracy of the It is difficult to determine the accuracy of the result without knowing the analytical solutionresult without knowing the analytical solution

• One method of determining the accuracy of the One method of determining the accuracy of the numerical solution is to repeat the solution with a numerical solution is to repeat the solution with a smaller time increment and compare the resultssmaller time increment and compare the results– If the results agree, the results are correct to the If the results agree, the results are correct to the

precision of the number of significant figures of precision of the number of significant figures of agreementagreement

Euler Method, Numerical ExampleEuler Method, Numerical Example

II. Drag force and terminal speedII. Drag force and terminal speed

-Fluid:Fluid: anything that can flow. anything that can flow. ExampleExample: gas, liquid.: gas, liquid.

-Drag force: DDrag force: D - Appears when there is a relative velocity between a - Appears when there is a relative velocity between a fluid and a bodyfluid and a body..

OpposesOpposes the relative motion of a body in a fluid. the relative motion of a body in a fluid.

-- Points in the direction in which the fluid flows.Points in the direction in which the fluid flows.

2

2

1AvCD

AC

Fv gt

2

-Terminal speed: vTerminal speed: vtt

02

10 2 gg FAvCaifmaFD

AssumptionsAssumptions::* Fluid = air.* Fluid = air.* Body is blunt (baseball).* Body is blunt (baseball).* Fast relative motion * Fast relative motion turbulent air. turbulent air.

CC = drag coefficient (0.4-1). = drag coefficient (0.4-1).ρρ = air density (mass/volume). = air density (mass/volume). A A = effective body’s cross sectional area = effective body’s cross sectional area area area perpendicular to perpendicular to vv

- - Reached when the acceleration of an object that experiences a Reached when the acceleration of an object that experiences a vertical movement through the air becomes zero vertical movement through the air becomes zero FFgg=D=D

III. Uniform circular motionIII. Uniform circular motion

-Centripetal accelerationCentripetal acceleration::

R

vmF

2

r

va

2

A centripetal force accelerates a body by changing the A centripetal force accelerates a body by changing the direction of the body’s velocity without changing its speed. direction of the body’s velocity without changing its speed.

v, av, a = const, but direction changes during motion = const, but direction changes during motion.

-Centripetal force:Centripetal force:

aa,, FF are directed toward the center are directed toward the center of curvature of the particle’s path.of curvature of the particle’s path.

49. A puck of mass m slides on a frictionless table while A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord attached to a hanging cylinder of mass M by a cord

through a hole in the table. What speed through a hole in the table. What speed keeps the cylinder at rest?keeps the cylinder at rest?

Mg

mg T

TN

49. A puck of mass m slides on a frictionless table while A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord attached to a hanging cylinder of mass M by a cord

through a hole in the table. What speed through a hole in the table. What speed keeps the cylinder at rest?keeps the cylinder at rest?

Mg

mg T

TN

MgTMFor :

m

Mgrv

r

vmMg

r

vmTmFor

22

:

33E. Calculate the drag force on a missile 53cm in diameter cruising with a speed of 250m/s at low altitude, where the density of air is 1.2kg/m3. Assume C=0.75

kNsmmmkgAvCD 2.6/250)2/53.0()/2.1(75.05.02

1 2232

32. The terminal speed of a ski diver is 160 km/h in the spread eagle position and 310 km/h in the nose-dive position. Assuming that the diver’s drag coefficient C does not change from one point to another, find the ratio of the effective cross sectional area A in the slower position to that of the faster position.

7.32

2

/310

/1602

D

E

E

D

D

g

E

g

gt A

A

A

A

AC

F

AC

F

hkm

hkm

AC

Fv

32. The terminal speed of a ski diver is The terminal speed of a ski diver is 160 km/h160 km/h in the spread in the spread eagle position and eagle position and 310 km/h310 km/h in the nose-dive position. in the nose-dive position. Assuming that the diver’s drag coefficient Assuming that the diver’s drag coefficient CC does not change does not change from one point to another, find the ratio of the effective cross from one point to another, find the ratio of the effective cross sectional area sectional area A A in the slower position to that of the faster in the slower position to that of the faster position.position.

7.32

2

/310

/1602

D

E

E

D

D

g

E

g

gt A

A

A

A

AC

F

AC

F

hkm

hkm

AC

Fv

Block B weighs 711N. The coefficient of static friction between the block and the table is 0.25; assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary.

FgA

N

FgB

T1

f

T3

T1

T2

T3

NfstationarySystem ss max,

NNTfT

gmNBBlock

s

B

75.17771125.00 1max,1

322

2221

30sin

25.20530cos

75.17730cos

TTT

NN

TTTTKnot

y

x

NNTgmTABlock A 62.10225.2055.030sin23

Two blocks of weights 3.6N and 7.2N, are connected by a massless string and slide down a 30º inclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.10; that between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the string. (c) Describe the motion if, instead, the heavier block leads.

A

B

FgA

FgB

NA

NB

T

T

fk,B

fk,A

MovementBlock Block AA Block Block BBNA NB

fkAfkB

TFgxA

FgyA

FgxB

FgyB

T

Light block Light block AA leads leads

Light block A leadsLight block A leads

Heavy block Heavy block BB leads leads

aTaNTNamfTF

NNNf

NgmFNBBlock

aTaTNNamTfF

NNNf

NgmFNABlock

BkBgxB

BkBkB

BgyBB

AkAgxA

AkAkA

AgyAA

73.035.273.025.130sin)2.7(

25.1)23.6)(2.0(

23.630cos

37.049.137.0312.030sin)6.3(

312.0)12.3)(1.0(

12.330cos

NT

sma

2.0

/49.3 2

NWW

WWT kAkB

BA

BA 2.0cos

The above set of equations is not valid in this circumstance The above set of equations is not valid in this circumstance aaAA≠ a≠ aBB The blocks move independently from each other.The blocks move independently from each other.

Reversing the blocks is equivalent to switching the labels. This would Reversing the blocks is equivalent to switching the labels. This would give give TT~(~(μμkAkA--μμkBkB)<0)<0 impossible!!! impossible!!!

NNNf

NNNf

kk

ss

8.22)60(38.0

33)60(55.0max,

74. A block weighing A block weighing 22N 22N is held against a vertical wall by a horizontal force is held against a vertical wall by a horizontal force FF of magnitude of magnitude 60N60N. The coefficient of static friction between the wall . The coefficient of static friction between the wall and the block isand the block is 0.55 0.55 and the coefficient of kinetic friction between them and the coefficient of kinetic friction between them is is 0.380.38. A second force . A second force P P acting parallel to the wall is applied to the acting parallel to the wall is applied to the block. For the following magnitudes and directions of block. For the following magnitudes and directions of PP, , determine whether the block moves, the direction of motion, determine whether the block moves, the direction of motion,

and the magnitude and direction of the and the magnitude and direction of the frictional force acting on the block: frictional force acting on the block: (a) (a) 34N 34N up, (b) up, (b) 12N12N up, (c) up, (c) 48N48N up, up, (d) (d) 62N 62N up, (e) up, (e) 10N10N down, (f) 1 down, (f) 18N 8N down.down.

F=60N

(a) P=34N, up(a) P=34N, up

mg=22N

N

N=F=60N

P

22Nf

Without Without PP, the block is at rest , the block is at rest

movenotdoesBlockNff

downNffNN

affassumeweIf

mafmgP

s

s

33

122234

0

max,

P

74. A block weighing 22N is held against a vertical wall by a horizontal force F of magnitude 60N. The coefficient of static friction between the wall and the block is 0.55 and the coefficient of kinetic friction between them is 0.38. A second P acting parallel to the wall is applied to the block. For the following magnitudes and directions of P, determine whether the block moves, the direction of motion, and the magnitude and direction of the frictional force acting on the block: (a) 34N up (b) 12N up, (c) 48N up, (d) 62N up, (e) 10N down, (f) 18N down.

F=60N

mg=22N

N

N=F=60N

P

(b) P=12N, up

P

22N

f

movingNotNff

upNNNf

mamgfP

s

33

101222

0

max,

(c) P=48N, up

P

22Nf

movingNotNff

downNNNf

mamgfP

s

33

262248

0

max,

(d) P=62N, up P

22Nf

downNffwithmamgfP

wrongAssumptionupmovesBlockNff

upNNNfmgfP

k

s

8.22

(*)33

402262(*)0

max,

(e) P=10N, down

P22N

f

movingNotNff

upNNNf

mamgPf

s

33

321022

0

max,

(f) P=18N, down

P22N

f

upNff

movesNff

upNNNf

mamgPf

k

s

8.22

33

402218

0

max,

WA=44N

P

28. Blocks A and B have weights of 44N and 22N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μs between A and the table is 0.2. (b) Block C suddenly is lifted of A. What is the acceleration of block A if μk between A and the table is 0.15?

T

T

WB=22N

WA=44N

f

Wc

N

)1(00 max,

max,

NTfTaABlock

Nff

ss

ss

)2(220 NTgmTBBlock B

NNT

Ns

1102.0

22)2()1(

NNNWWWNBABlocks CCA 6644110,

(a)

28. Blocks A and B have weights of 44N and 22N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μs between A and the table is 0.2. (b) Block C suddenly is lifted of A. What is the acceleration of block A if μk between A and the table is 0.15?

T

T

WB=22N

WA=44N

f

Wc

N

amTgm

amNT

NgmNdisappearsC

BB

Ak

A

44(b)

NTaT

smaaT

172.222

/3.25.46.6 2

29. The two blocks (with m=16kg and m=88kg) shown in the figure below are not attached. The coefficient of static friction between the blocks is: μs=0.38 but the surface beneath the larger block is frictionless.

What is the minimum value of the horizontal force F required to keep the smaller block from slipping

down the larger block?

Mm

FaamF total

)2(0'0

)1('

mgFmgf

Mm

FmmaFFblockSmall

ss

NM

MmmgFmg

Mm

FM

ss 488)2()1(

Mg

N

F’

mg

fF’

Movement

FFminmin required to keep m from sliding down? required to keep m from sliding down?

Treat both blocks as a single system sliding across a frictionless floor

mg

44. An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weigh of the car and riders is 5kN, and the radius of the circle is 10m.

What are the magnitude and the direction of the force of the boom on the car at the top of the circle if the car’s speed is (a) 5m/s (b) 12m/s?

Rg

vWF

R

vmWF BB

22

1

The force of the boom on the car is capable of pointing any direction

FB

W

y

downFsmvbupNFsmva BB 3.2/12)(7.3/5)(

chapter 6 – force and motion ii drag forces and terminal speed.drag forces and terminal speed....

Documents

new force

circlethis force

horizontal circle

vertical rod

upper string

lower string

constant speed

vertical direction