Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular...

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Chapter 6.2. Uniform Chapter 6.2. Uniform Circular Motion Circular Motion

Transcript of Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular...

Page 1: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Chapter 6.2. Uniform Circular Chapter 6.2. Uniform Circular MotionMotion

Page 2: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Centripetal Centripetal forces keep forces keep

these children these children moving in a moving in a

circular path.circular path.

Page 3: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Uniform Circular MotionUniform Circular MotionUniform circular motionUniform circular motion is motion along a circular path in which there is no change in speed, only a change in direction.

Constant force toward center.

Constant velocity tangent to path.

vFc

Question: Is there an outward force on the ball?

Page 4: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Uniform Circular Motion Uniform Circular Motion (Cont.)(Cont.)

The question of an The question of an outwardoutward force can be force can be resolved by asking what happens when resolved by asking what happens when the string breaks! the string breaks!

When central force is removed, ball continues in straight line.

v

Ball moves tangent Ball moves tangent to path, to path, NOTNOT outward as might outward as might be expected.be expected.

Centripetal forceCentripetal force is needed to change is needed to change direction.direction.

Page 5: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Examples of Centripetal Examples of Centripetal ForceForce

• Car going around Car going around a curve. a curve.

You are sitting on the seat next You are sitting on the seat next to the outside door. What is to the outside door. What is the direction of the resultant the direction of the resultant force on you as you turn? Is it force on you as you turn? Is it awayaway from center or from center or toward toward centercenter of the turn? of the turn?

You are sitting on the seat next You are sitting on the seat next to the outside door. What is to the outside door. What is the direction of the resultant the direction of the resultant force on you as you turn? Is it force on you as you turn? Is it awayaway from center or from center or toward toward centercenter of the turn? of the turn?

Force Force ON ON you is you is towardtoward the the center.center.

FFcc

Page 6: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Car Example ContinuedCar Example Continued

There There isis an outward force, but it does not an outward force, but it does not act ON you. It is the reaction force act ON you. It is the reaction force exerted exerted BYBY you you ONON the door. It affects the door. It affects only the door.only the door.

There There isis an outward force, but it does not an outward force, but it does not act ON you. It is the reaction force act ON you. It is the reaction force exerted exerted BYBY you you ONON the door. It affects the door. It affects only the door.only the door.

The centripetal The centripetal force is force is

exerted exerted BYBY the the door door ONON you. you.

(Centrally)(Centrally)

Fc

F’ReactionReaction

Page 7: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Spin Cycle on a WasherSpin Cycle on a Washer

How is the water How is the water removed from clothes removed from clothes during the spin cycle of during the spin cycle of a washer?a washer?Think carefully before answering . . . Think carefully before answering . . .

Does the centripetal force throw water Does the centripetal force throw water off the clothes?off the clothes?

NO.NO. Actually, it is the Actually, it is the LACKLACK of a force of a force that allows the water to leave the that allows the water to leave the clothes through holes in the circular clothes through holes in the circular wall of the rotating washer.wall of the rotating washer.

NO.NO. Actually, it is the Actually, it is the LACKLACK of a force of a force that allows the water to leave the that allows the water to leave the clothes through holes in the circular clothes through holes in the circular wall of the rotating washer.wall of the rotating washer.

Page 8: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Deriving Central Deriving Central AccelerationAcceleration

Consider initial velocity at A and final velocity at B:

Rvo

vfvf

-vo

A

B

Rvo

v s

Page 9: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Deriving Acceleration Deriving Acceleration (Cont.)(Cont.)

vf

-vo

Rvo

v s

ac =v

tDefinitio

n:

=v

v

s

R

Similar Triangle

s

=v

t

vs

Rtac = =

vv

R

mass m

Centripetal acceleration

:

Centripetal acceleration

:

2 2

; c c c

v mva F ma

R R

Page 10: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Circular Motion is often described in Circular Motion is often described in terms of frequency and period.terms of frequency and period.

• The frequency f – the number of The frequency f – the number of revolutions per second.revolutions per second.

• The period T – time required for one The period T – time required for one complete revolution.complete revolution.

T=1/fT=1/f

Page 11: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 1:Example 1: A A 3-kg3-kg rock swings in a rock swings in a circle of radius circle of radius 5 m5 m. If its constant . If its constant speed is speed is 8 m/s8 m/s, what is the , what is the centripetal acceleration?centripetal acceleration?

R

vm

R = 5 m; v = 8 m/s

m = 3 kg

F = (3 kg)(12.8 m/s2)

Fc = 38.4 NFc = 38.4 N2

c c

mvF ma

R

2

c c

mvF ma

R

22(8 m/s)

5 12.8 /s

mmca

2

c

va

R

Page 12: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Car Negotiating a Flat TurnCar Negotiating a Flat Turn

R

v

What is the direction of the What is the direction of the force force ON the car?the car?

Ans.Ans. Toward CenterToward Center

FFcc

This central force is This central force is exerted exerted BY BY the road the road ONON

the car.the car.

Page 13: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Car Negotiating a Flat TurnCar Negotiating a Flat Turn

R

v

Is there also an Is there also an outward outward force acting force acting ON ON the car?the car?

Ans.Ans. No,No, but the car does exert a but the car does exert a outward outward reactionreaction force force ON ON the the road. road.

FFcc

Page 14: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Car Negotiating a Flat TurnCar Negotiating a Flat Turn

The centripetal force The centripetal force FFcc is that of static is that of static friction ffriction fss::

The central force The central force FFCC and the friction force and the friction force ffss are not two different forces that are are not two different forces that are equal. There is just equal. There is just oneone force on the car. force on the car. The The naturenature of this central force is static of this central force is static friction.friction.

The central force The central force FFCC and the friction force and the friction force ffss are not two different forces that are are not two different forces that are equal. There is just equal. There is just oneone force on the car. force on the car. The The naturenature of this central force is static of this central force is static friction.friction.

Fc = fsR

v

mFc

n

mg

fs

R

Page 15: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Finding the maximum speed for negotiating a turn without slipping.Finding the maximum speed for

negotiating a turn without slipping.

Fc = fsfs = smgFc =

mv2

R

The car is on the verge of slipping when The car is on the verge of slipping when FFC C is equal to the maximum force of is equal to the maximum force of

static friction static friction ffss. .

R

vm

Fc

Fc = fsn

mg

fs

R

Page 16: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Maximum speed without slipping (Cont.)Maximum speed without slipping (Cont.)

Fc = fs

mv2

R= smg

v = sgR

Velocity v is maximum speed for no slipping.

Velocity v is maximum speed for no slipping.

n

mg

fs R

R

vm

Fc

Page 17: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 4:Example 4: A car negotiates a turn A car negotiates a turn of radius of radius 70 m70 m when the coefficient when the coefficient of static friction is of static friction is 0.70.7. What is the . What is the maximum speed to avoid slipping?maximum speed to avoid slipping?

v = 21.9 m/s

v = 21.9 m/s(0.7)(9.8)(70m)sv gR

R

v

m Fc

s = 0.7

fs = smgFc = mv2

R

From which:v = sgR

g = 9.8 m/s2; R = 70 m

Page 18: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

slow speedslow speed

Optimum Banking AngleOptimum Banking AngleBy banking a curve at By banking a curve at thethe optimum angle, the optimum angle, the

normal force normal force nn can can provide the necessary provide the necessary centripetal force without centripetal force without the need for a friction the need for a friction force.force.

fast speedfast speed optimumoptimum

nfs = 0

w w

nfs

w

nfs

R

v

mFc

Page 19: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Free-body DiagramFree-body Diagram

nn

mgmg

Acceleration Acceleration aa is toward is toward the center. Set the center. Set xx axis axis along the direction of along the direction of aacc , i. e., horizontal (left , i. e., horizontal (left to right).to right).

nn

mgmg

nn sin sin

nn cos cos

++ aacc

n

mg

xx

Page 20: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Optimum Banking Angle Optimum Banking Angle (Cont.)(Cont.)

n

mgmg

nn sin sin

nn cos cos

FFxx = m = maacc

FFyy = 0 = 0 n n cos cos = mg= mg

mvmv22

RRnn sin sin Apply Apply

Newton’s Newton’s 2nd Law to x 2nd Law to x and y axes.and y axes.

n

mg

Page 21: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Optimum Banking Angle Optimum Banking Angle (Cont.)(Cont.)

n cos = mg

mv2

Rn sin

n

mg

2

2

tan

1

mvvR

mg gR

n

mg

n sin

n cos

sintan

cos

nn

Page 22: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Optimum Banking Angle Optimum Banking Angle (Cont.)(Cont.)

n

mg

n sin

n cos

n

mg

2

tanv

gR

Optimum Banking Angle

Page 23: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 5:Example 5: A car negotiates a turn A car negotiates a turn of radius 80 m. What is the of radius 80 m. What is the optimum banking angle for this optimum banking angle for this curve if the speed is to be equal to curve if the speed is to be equal to 12 m/s?12 m/s?

n

mg

n sin

n cos

tan = =v2

gR

(12 m/s)2

(9.8 m/s2)(80 m)

tan = 0.184

n

mg

2

C

mvF

R

2

C

mvF

R

How might you find the centripetal force on the car, knowing its mass?

= 10.40

Page 24: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

The Conical PendulumThe Conical PendulumA conical pendulum consists of a mass m revolving in a horizontal circle of radius R at the end of a cord of length L.

h

T

L

R mg

T

T sin

T cos

Note: The inward component of Note: The inward component of tension tension T sin T sin gives the needed gives the needed

central force.central force.

Page 25: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Angle Angle and velocity and velocity vv::

h

T

L

R mg

T

T sin

T cos

T cos = mg

mv2

RT sin Solve two

equations to find angle

tan = v2

gR

Page 26: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 6:Example 6: A A 2-kg2-kg mass swings in mass swings in a horizontal circle at the end of a a horizontal circle at the end of a cord of length cord of length 10 m10 m. What is the . What is the constant speed of the mass if the constant speed of the mass if the rope makes an angle of rope makes an angle of 303000 with with the vertical?the vertical?

R = L sin 300 = (10 m)(0.5)

R = 5 m

1. Draw & label 1. Draw & label sketch.sketch.2. Recall formula for 2. Recall formula for pendulum.pendulum.

2

tanv

gR

2

tanv

gR Find: Find: v = ?v = ?

3. To use this formula, we need to find R 3. To use this formula, we need to find R = ?= ?

hT

L

R

Page 27: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 6(Cont.): Example 6(Cont.): Find Find vv for for = = 303000

R = R = 5 5 mm

v = 5.32 m/s

v = 5.32 m/s

g = g = 9.8 9.8 m/m/ss22

Solve for Solve for v = ?v = ?

2

tanv

gR

2

tanv

gR

4. Use given info to find 4. Use given info to find the velocity at the velocity at 303000..

2 tanv gR tanv gR

2 0(9.8 m/s )(5 m) tan 30v

hT

L

R

R = R = 5 m5 m

Page 28: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 7:Example 7: Now find the tension Now find the tension TT in the cord if in the cord if mm = 2 kg= 2 kg, , = 30= 3000, and , and LL = 10 m= 10 m..

h

T

L

R mg

T

T sin

T cos

Fy = 0: T cos - mg = 0; T cos = mg

T = =mg

cos

(2 kg)(9.8 m/s2)

cos 300

T = 22.6 NT = 22.6 N

2 kg

Page 29: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 8:Example 8: Find the centripetal Find the centripetal force force FFcc for the previous example. for the previous example.

h

T

L

R mg

T

T sin

T cos

m = 2 kg; v = 5.32 m/s; R = 5 m; T = 22.6 N

Fc = 11.3 NFc = 11.3 N

2 kg

Fc = mv2

R

or Fc = T sin 300

Fc

= 30= 3000

Page 30: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Swinging Seats at the FairSwinging Seats at the Fair

h

T

L

R

d

This problem is This problem is identical to the other identical to the other examples except for examples except for finding R.finding R.

R = d + b

R = L sin + b

tan = v2

gRand v = gR tan

b

Page 31: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 9.Example 9. If If bb = 5 m = 5 m and and LL = 10 m = 10 m, , what will be the speed if the angle what will be the speed if the angle = = 262600??

R = d + b

R = 4.38 m + 5 m = 9.38 m

tan = v2

gR

TL

Rd

bd = d = (10 m) sin 26(10 m) sin 2600 = 4.38 = 4.38

mm

2 tanv gR tanv gR

2 0(9.8 m/s )(9.38 m) tan 26v v = 6.70 m/s

v = 6.70 m/s

Page 32: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Motion in a Vertical CircleMotion in a Vertical CircleConsider the forces on a Consider the forces on a ball attached to a string ball attached to a string as it moves in a vertical as it moves in a vertical loop.loop.Note also that the Note also that the positive positive direction is direction is always along always along acceleration, i.e., acceleration, i.e., toward the centertoward the center of of the circle.the circle.Note changes as you Note changes as you click the mouse to click the mouse to show new positions.show new positions.

+

T

mg

v

BottomBottom

Maximum Maximum tension T, W tension T, W opposes Fopposes Fcc

+v

T

mg

Top Top RightRight

Weight has Weight has no effect on no effect on

TT

+

Tmg

v

Top RightTop Right

Weight Weight causes small causes small decrease in decrease in tension Ttension T

vT

mg

+

Left SideLeft Side

Weight has Weight has no effect on no effect on

TT

+

T

mg

v

BottomBottom

v

T

mg

Top of PathTop of PathTension is Tension is

minimum as minimum as weight helps weight helps

FFcc force force

+

Page 33: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

R

v

v

As an exercise, As an exercise, assume that a central assume that a central force of force of FFcc = = 40 N40 N is is required to maintain required to maintain circular motion of a circular motion of a ball and ball and W = 10 NW = 10 N..

The tension T must adjust so that

central resultant is 40 N.

The tension T must adjust so that

central resultant is 40 N.

At top: At top: 10 N + 10 N + TT = 40 = 40 NN

Bottom: Bottom: TT – 10 N = 40 – 10 N = 40 NN

T = T = __?_____?___T = T = 50 N50 N

T = T = 30 N30 NT = T = __??__

T

10 N

++

++T

10 N

Page 34: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Motion in a Vertical CircleMotion in a Vertical Circle

R

v

v

Resultant force toward

centerFc =

mv2

R

Consider TOP of circle:

AT TOP:

T

mg

T

mg

+

mg + T = mv2

R

T = - mg mv2

R

Page 35: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Vertical Circle; Mass at Vertical Circle; Mass at bottombottom

R

v

v

Resultant force toward

centerFc =

mv2

R

Consider bottom of circle:

AT Bottom:

Tmg

+

T - mg = mv2

R

T = + mg

mv2

R

T

mg

Page 36: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Visual Aid:Visual Aid: Assume that the centripetal Assume that the centripetal force required to maintain circular force required to maintain circular motion is motion is 20 N20 N. Further assume that the . Further assume that the weight is weight is 5 N5 N..

R

v

v

2

20 NC

mvF

R

Resultant central force FFCC at every point in

path!FFCC = = 2020 NNWeight vector W W is

downward at every point.

W = 5 N, down

FFC C = = 2020 NN at top AND at

bottom.

Page 37: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Visual Aid:Visual Aid: The resultant force (The resultant force (20 N20 N) is ) is the vector sum of the vector sum of TT and and WW at ANY point at ANY point in path. in path.

R

v

v

FFC C = = 2020 NN at top AND at

bottom.

Top: T + W = FC

T + 5 N = 20 N

T = 20 N - 5 N = 15 N

T - W = FC

T - 5 N = 20 N

T = 20 N + 5 N = 25 N

Bottom:

WT

++

T W

Page 38: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

For Motion in CircleFor Motion in Circle

R

v

v

AT TOP:

T

mg

+

AT BOTTOM:

Tmg

+

T = - mg mv2

R

T = + mg

mv2

R

Page 39: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 10:Example 10: A A 2-kg2-kg rock swings in a rock swings in a vertical circle of radius vertical circle of radius 8 m8 m. The speed of . The speed of the rock as it passes its highest point is the rock as it passes its highest point is 10 10 m/sm/s. What is tension . What is tension TT in rope?in rope?

R

v

v

T

mg

mg + T = mv2

RAt Top:

T = - mg mv2

R

T = 25 N - 19.6 N T = 5.40 NT = 5.40 N

22(2kg)(10m/s)

2 kg(9.8 m/s )8 m

T

Page 40: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 11:Example 11: A A 2-kg2-kg rock swings in a rock swings in a vertical circle of radius vertical circle of radius 8 m8 m. The speed of . The speed of the rock as it passes itsthe rock as it passes its lowest lowest point is point is 10 10 m/sm/s. What is tension . What is tension TT in rope? in rope?

R

v

vT

mg

T - mg = mv2

R

At Bottom:

T = + mg

mv2

R

T = 25 N + 19.6 N T = 44.6 NT = 44.6 N

22(2kg)(10m/s)

2 kg(9.8 m/s )8 m

T

Page 41: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 12:Example 12: What is the critical speed What is the critical speed vvcc at the top, if the at the top, if the 2-kg2-kg mass is to mass is to continue in a circle of radius continue in a circle of radius 8 m8 m??

R

v

v

T

mgmg + T =

mv2

RAt Top:

vc = 8.85 m/svc = 8.85 m/s

vc occurs when T = 0

0

mg = mv2

R

v = gR = (9.8 m/s2)(8 m)

vc = gR

Page 42: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

The Loop-the-LoopThe Loop-the-LoopSame as cord, Same as cord, nn replaces replaces TT

AT TOP:

n

mg

+

AT BOTTOM:

nmg

+

n = - mg mv2

R

n= + mg

mv2

R

R

v

v

Page 43: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

The Ferris WheelThe Ferris Wheel

AT TOP:

n

mg

+

AT BOTTOM:

nmg

+

mg - n = mv2

R

n = + mg

mv2

R

R

v

vn = mg -

mv2

R

Page 44: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Example 13:Example 13: What is the What is the apparent weight of a apparent weight of a 60-60-kgkg person as she moves person as she moves through the highest point through the highest point when when RR = 45 m = 45 m and the and the speed at that point is 6 speed at that point is 6 m/s?m/s?

n

mg

+

R

v

v

mg - n = mv2

Rn = mg -

mv2

R

Apparent weight will be Apparent weight will be the normal force at the the normal force at the

top:top:

22 (60kg)(6m/s)

60 kg(9.8 m/s )45 m

n n = 540 Nn = 540 N

Page 45: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

SummarySummaryCentripetal acceleration

:

Centripetal acceleration

:

2 2

; c c c

v mva F ma

R R

v = sgR

tan = v2

gR

v = gR tan Conical

pendulum:

Page 46: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Summary: Motion in CircleSummary: Motion in Circle

v

R

v AT TOP:

T

mg

+ T = - mg mv2

R

AT BOTTOM:

Tmg

+T = +

mg mv2

R

Page 47: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

Summary: Ferris WheelSummary: Ferris Wheel

AT TOP:

n

mg

+

AT BOTTOM:

nmg

+

mg - n = mv2

R

n = + mg

mv2

R

R

v

vn = mg -

mv2

R

Page 48: Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.

CONCLUSION: Chapter 10CONCLUSION: Chapter 10Uniform Circular MotionUniform Circular Motion