Chapter 6 Circular Motion, Orbits, and Gravity. Slide 6-23 Uniform Circular Motion.

Chapter 6

• Circular Motion, Orbits, and Gravity

23

Uniform Circular Motion

For uniform circular motion, the acceleration

A. is parallel to the velocityB. is directed toward the center of

the circleC. is large for a larger orbital the

same speedD. is always due to gravity

is paral

lel to

the ve

locity

is dire

cted to

ward th

e cente

r...

is lar

ge for a

large

r orb

ital th

...

is alw

ays due to

gravit

y

0% 0%0%

100%

Forces in Circular Motion

Slide 6-24

22

2

net , toward center of circle

v = r

va = r

rmv

F = ma = r

© 2015 Pearson Education, Inc.

Time of Circular Motion

𝑓 =1𝑇

The frequency is just the inverse of the period

Example of frequency units:revmin

𝑣=2𝜋𝑟𝑇

𝑣=2𝜋 𝑟𝑓


What happens when speed decreases and radius decreases?

𝑎=𝑣2

𝑟


Comparing centripetal accelerations

Rank the acceleration of the following uniform circular motions from smallest to largest:

𝑎=𝑣2

𝑟

1 2 3 4 5

𝑣=2𝜋 𝑟𝑓

𝑎=(2𝜋𝑟𝑓 )2

𝑟=4𝜋 2𝑟 𝑓 2

They’re all going to have so we just need to find for each one


Comparing centripetal accelerations

Rank the acceleration of the following uniform circular motions from smallest to largest

1 2 3 4 5

𝑟1 𝑓 12

𝑎=𝑣2

𝑟𝑣=2𝜋 𝑟𝑓

(10cm )(50revmin )

2

𝑟2 𝑓 22

(20cm )(50revmin )

2

𝑟3 𝑓 32

(20cm )(100revmin )

2

𝑟 4 𝑓 42

( 40cm )(100revmin )

2

𝑟5 𝑓 52

( 40 cm )(200revmin )

2


Driving on a curvy road

𝑎=𝑣2

𝑟

𝑎=𝑣2

𝑟

𝑎=0


Problem solving uniform circular motion

𝑎=𝑣2

𝑟

𝑣=2𝜋 𝑟𝑓

A level curve on a country road has a radius of 150 m. What is the maximum speed at which this curve can be safely negotiated on a rainy day when the coefficient of friction between the tires on a car and the road is 0.40?

Slide 6-28


A level curve on a country road has a radius of 150 m. What is the maximum speed at which this curve can be safely negotiated on a rainy day when the coefficient of friction between the tires on a car and the road is 0.40?

Slide 6-28

𝑓 𝑠=𝜇𝑠𝑁

𝑓 𝑠=𝜇𝑠𝑚𝑐𝑔¿𝑚𝑐𝑣2

𝑟 𝑣=√𝜇𝑠𝑔𝑟


When a car turns a corner on a level road, which force provides the necessary centripetal

acceleration?

A. FrictionB. TensionC. Normal ForceD. Air resistanceE. Gravity

Fricti

on

Tensio

n

Normal F

orce

Air res

istance

Gravity

76%

2%10%

0%

12%

A coin sits on a rotating turntable.

1. At the time shown in the figure, which arrow gives the direction of the coin’s velocity?

Additional Questions

Slide 6-45

AnswerA coin sits on a rotating turntable.

1. At the time shown in the figure, which arrow gives the direction of the coin’s velocity?

A

Slide 6-46


2. At the time shown in the figure, which arrow gives the direction of the frictional force on the coin?


Slide 6-47

AnswerA coin sits on a rotating turntable.

2. At the time shown in the figure, which arrow gives the direction of the frictional force on the coin? D

Slide 6-48


3. At the instant shown, suppose the frictional force disappeared. In what direction would the coin move?


Slide 6-49

Answer

A


3. At the instant shown, suppose the frictional force disappeared. In what direction would the coin move?

Slide 6-50

A ball moves CW in circular motion. What happens when it reaches the opening?

A. AB. BC. CD. D

A B C D

0%0%

95%

5%

A world-class thrower can get the hammer up to a speed of 29 m/s. If an athlete swings the mass in a horizontal circle centered on the handle he uses to hold the chain, what is the tension in the chain?

In the hammer throw, an athlete spins a heavy mass in a circle at the end of a chain, then lets go of the chain. For male athletes, the “hammer” is a mass of 7.3 kg at the end of a 1.2 m chain.

Slide 6-29

𝑚 𝑣2

𝑟

𝑇=𝑚 𝑣2

𝑟=7.3kg

(29ms )

2

1.2m


A car of mass 1500 kg goes over a hill at a speed of 20 m/s. The shape of the hill is approximately circular, with a radius of 60 m, as in the figure at right. When the car is at the highest point of the hill,

a. What is the force of gravity on the car?

b. What is the normal force of the road on the car at this point?

Slide 6-30


𝐹=𝑚𝑔

A car of mass 1500 kg goes over a hill at a speed of 20 m/s. The shape of the hill is approximately circular, with a radius of 60 m, as in the figure at right. When the car is at the highest point of the hill,

a. What is the force of gravity on the car?

b. What is the normal force of the road on the car at this point?

Slide 6-30


∑ 𝐹 𝑦=¿𝑁+𝑚𝑔=−𝑚 𝑣2

𝑟¿

𝑁=−𝑚 (𝑣2

𝑟+𝑔)

vmax gr

Slide 6-31

Maximum walking speed

A roller coaster car goes through a vertical loop at a constant speed. For positions A to E, rank order the:

• centripetal acceleration

• normal force

• apparent weight

Slide 6-32

Example Problem

𝑎=𝑣2

𝑟

A = B = C = D = E

A roller coaster car goes through a vertical loop at a constant speed. For positions A to E, rank order the:

• centripetal acceleration

• normal force

• apparent weight

Slide 6-32

Example Problem

A = E > B = D > C


Net Force causes centripetal acceleration

∑ 𝐹=𝐹 net


Net Force causes centripetal acceleration


Friction force as a Net Force


Problem solving circular motion with Net Force

𝑓 𝑠=𝜇𝑠𝑚𝑐𝑔¿𝑚𝑐𝑣2

𝑟𝑣=√𝜇𝑠𝑔𝑟

A car travels around a corner with the cruise control on. Which vector is the car’s centripetal acceleration?

A. AB. BC. CD. DE. E

A B C D E

0% 2%

94%

2%2%

A car travels around a corner while speeding up. Which vector is the net force on the car?


A B C D E

0% 2%

84%

6%8%

A car travels around a corner while speeding up. Which vector is the force of friction from the road on the car?


A B C D E

2% 0% 2%

96%

0%


Fictitious Forces in circular motion

Feels like something is pushing you out of the

car

A car travels around a corner at a constant speed. Which vector is the fictitious force on the car?


A B C D E

0%

100%

0%0%0%


This one guy you knew from a while back says that in order to avoid loss of bone density on his trip to mars so, the ship must be spinning to simulate gravity. How fast must the craft below spin to simulate gravity?

𝑁

𝑣𝐹 𝑔=𝑁

𝑚𝑔=𝑚 𝑣2

𝑟

𝑣=√𝑔𝑟

Simulating Gravity


Tilted roads in favor of circular motion

Example Problem: Over the TopA handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with diameter 14 ft, what is the minimum speed the skater must have at the very top of the loop?

Slide 6-33

Example Problem: Over the TopA handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with diameter 14 ft, what is the minimum speed the skater must have at the very top of the loop?

Slide 6-33

𝐹 𝑔

𝑁=0∑ 𝐹=𝐹𝑔=𝑚𝑣2

𝑟−𝑚𝑔=−𝑚𝑣2

𝑟

𝑣=√|𝑔|𝑟

help: |𝑔|=32ft

s2


Keeping riders safe in circular motion

When a ball on the end of a string is swung in a vertical circle, the ball’s acceleration is because

A. The speed is changingB. The direction is changingC. The speed and direction are

changingD. The ball is not acceleration

The s

peed is

changin

g

The d

irecti

on is ch

angin

g

The s

peed an

d directi

on are...

The b

all is

not acce

leration

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100%

35

The Force of Gravity

𝑦

𝑥

Same as the familiar y and x

The inverse-square law


versus in Newton’s law of gravitation

is a value unique to the mass that’s pulling

is the same throughout the whole universe!

A typical bowling ball is spherical, weighs 16 pounds, and has a diameter of 8.5 in. Suppose two bowling balls are right next to each other in the rack. What is the gravitational force between the two—magnitude and direction? What is the magnitude and direction of the force of gravity on a 60 kg person?

Example Problem

Slide 6-36

A 60 kg person stands on each of the following planets. Ranks order her weight on the three

bodies, from highest to lowest

A. A>B>CB. B>A>CC. B>C>AD. C>B>AE. C>A>B

A>B>C

B>A>C

B>C>AC>B

>AC>A

>B

77%

0% 0%

19%

4%

Which vector pair matches the car’s motion over the hill?

A. AB. BC. C

A B C

94%

2%4%


The Force of Gravity

𝑣=√𝐺𝑀 e

𝑟 e

17,000 mph

Example Problem: Orbital MotionPhobos is one of two small moons that orbit Mars. Phobos is a very small moon, and has correspondingly small gravity—it varies, but a typical value is about 6 mm/s2. Phobos isn’t quite round, but it has an average radius of about 11 km. What would be the orbital speed around Phobos, assuming it was round with gravity and radius as noted?

Slide 6-34

𝑣=√𝐺𝑀𝑟 e

=√ (6.67 ∙ 10−11 𝑚3

kg ∙ s2 )( .0069.8

∙ 5.97 ∙1024 kg )6.4 ∙106 m

A satellite orbits the earth. A space shuttle crew is sent to boost the satellite into a higher orbit.

Which of these quantities increases?

A. SpeedB. Angular speedC. PeriodD. Centripetal accelerationE. Gravitational force of the

earth Speed

Angular

speed

Period

Centripet

al ac

celerati

on

Gravita

tional fo

rce of t

he earth

0% 2% 0%4%

94%

Summary

Slide 6-41

Summary

Slide 6-42

Chapter 6 Circular Motion, Orbits, and Gravity. Slide 6-23 Uniform Circular Motion.

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Transcript of Chapter 6 Circular Motion, Orbits, and Gravity. Slide 6-23 Uniform Circular Motion.