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0. Physics 1710 Chapter 6—Circular Motion. Answer: a = ∆v/∆ t a = (112 m/s)/(2.5 sec) = 44.8 m/s 2 Thrust = force F = ma = (1800. kg)(44.8 m/s 2 ) ~ 81 kN. 0. Physics 1710 Chapter 6—Circular Motion. 1 ′ Lecture:

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physics 1710 chapter 6 circular motion

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Physics 1710 Chapter 6—Circular Motion

Answer:

a = ∆v/∆t

a = (112 m/s)/(2.5 sec) = 44.8 m/s2

Thrust = force

F = ma = (1800. kg)(44.8 m/s 2 ) ~ 81 kN

physics 1710 chapter 6 circular motion1

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Physics 1710 Chapter 6—Circular Motion

1′ Lecture:

The net force on a body executing circular motion is equal to the mass times the centripetal acceleration of the body. Fcentripetal = m acentripetal

acentripetal = v 2/ R [toward the center]; a = -ω2r

In non-inertial frames of reference one may sense fictitious forces.

At terminal velocity the velocity-dependent resistive forces balance the accelerating forces so that no further acceleration occurs.

physics 1710 chapter 6 circular motion2

vy

vx

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Physics 1710 Chapter 6—Circular Motion

x = r cos θy= r sin θ

vx = (dr/dt) cos θ- r sin θ(d θ/dt)vy = (dr/dt) sin θ+r cos θ(d θ/dt)Let (dr/dt) = 0 vx = - r ω sin θvy = +r ω cos θ

ax = dvx /dt= d(- r ω sin θ)/dt = r ω2cos θ- r ω sin θ(d ω/dt)ay = dvy /dt= d(r ω cos θ)/dt = -r ω2sin θ+ r ω cosθ(d ω/dt)a = - ω2r + r (d vtangential /dt)

θ

Centripetal Acceleration:

physics 1710 chapter 6 circular motion3

vy

vx

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Physics 1710 Chapter 6—Circular Motion

θ

a centripetal= - v 2 / r

atangential= r (d vtangential/dt)

F = m a

F centripetal = m a centripetal= - mv 2 / r

Centripetal Acceleration:

physics 1710 chapter 6 circular motion4

m

Fcentripetal

v

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Physics 1710 Chapter 6—Circular Motion

r

Demonstration: Ball on a String

Fcentripetal = m v 2/r

slide6

Physics 1710 — e-Quiz

Answer Now !

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An athlete swings a 8 kg “hammer” around on a 1.2 m long cable at an angular frequency of 1.0 revolution per second. How much force must he exert on the hammer throw handle?

  • About 12. N
  • About 30. N
  • About 78. N
  • About 380. N
physics 1710 chapter 6 circular motion5

m

Fcentripetal

v

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Physics 1710 Chapter 6—Circular Motion

r

Demonstration: Ball on a String

v = 2π r/T = 2(3.14) (1.2 m)/(1.0 sec) = 7.5 m/s

Fcentripetal = m v 2/r = (8.0 kg)(7.5 m/s)2/(1.2 m)

= 378 N ~ 380 N

physics 1710 chapter 6 circular motion6

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Physics 1710 Chapter 6—Circular Motion

Demonstration:

Marble in a bottle

Why does the marble stay up on the side?

physics 1710 chapter 6 circular motion7

No Talking!

Think!

Confer!

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Physics 1710 Chapter 6—Circular Motion

Why does the marble stay up on the side?

Peer Instruction Time

physics 1710 chapter 6 circular motion8

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Physics 1710 Chapter 6—Circular Motion

Satellites in Orbit

Which orbit is most like that of the Space Shuttle?

physics 1710 unit 1 review

4

3

2

1

No Talking!

Think!

Confer!

0

Physics 1710 Unit 1—Review

Peer Instruction Time

physics 1710 chapter 6 circular motion9

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Physics 1710 Chapter 6—Circular Motion

Satellite Motion:

h ~ 100 km , R⊕ ~ 6300 km

Fg= G Mm/ r 2Gravity

Fg = Fr = mv 2/r

v = √[GM/r] = √[GM/(R⊕ + h)]~ √[GM/R⊕ ] = √[gR⊕ ]= √[(9.8 m/s2)(6.3 x106m)]= 7.9x103 m/s ~17,600 mph

Why do the shuttle astronauts appear “weightless?”

physics 1710 chapter 6 circular motion10

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Physics 1710 Chapter 6—Circular Motion

Satellites in Orbit

Orbiting satellites are in free fall but miss the earth because it curves.

physics 1710 chapter 6 circular motion11

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Physics 1710 Chapter 6—Circular Motion

Summary

The net force on a body executing circular motion is equal to the mass times the centripetal acceleration of the body.

acentripedal = v 2/ R [toward the center]

The “centrifugal” force is a fictitious force due to a non-inertial frame of reference.

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