Find the orbital periods of Mercury (~0.4 AU from the Sun ...humanic/p1200_lecture11.pdfExample:...

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Example: Find the orbital periods of Mercury (~0.4 AU from the Sun) and Neptune (~30 AU from the Sun) and compare with that of the Earth. M = Sun mass, m = planet mass, r = distance to Sun, v = planet speed in orbit 1 AU = distance from Earth to Sun =1.5 x 10 11 m F C = mv 2 r = G mM r 2 v 2 r 2 = G M r 3 τ = orbital period = 2π r v τ = 2π r 3 GM τ Earth = 2π 1.5 × 10 11 ( ) 3 6.67 × 10 11 ( ) 1.99 × 10 30 ( ) = 3.2 × 10 7 s τ Mercury = 2π 0.4 × 1.5 × 10 11 ( ) 3 6.67 × 10 11 ( ) 1.99 × 10 30 ( ) = 8.0 × 10 6 s = 0.25 τ Earth τ Neptune = 2π 30 × 1.5 × 10 11 ( ) 3 6.67 × 10 11 ( ) 1.99 × 10 30 ( ) = 5.2 × 10 9 s = 160τ Earth

Transcript of Find the orbital periods of Mercury (~0.4 AU from the Sun ...humanic/p1200_lecture11.pdfExample:...

Example: Find the orbital periods of Mercury (~0.4 AU from the Sun) and Neptune (~30 AU from the Sun) and compare with that of the Earth. M = Sun mass, m = planet mass, r = distance to Sun, v = planet speed in orbit

1 AU = distance from Earth to Sun =1.5 x 1011 m

FC =mv2

r=G mM

r2 ⇒v2

r2 =GMr3

τ = orbital period = 2πrv

∴τ = 2π r3

GM

τ Earth = 2π1.5×1011( )

3

6.67×10−11( ) 1.99×1030( )= 3.2×107 s

τMercury = 2π0.4×1.5×1011( )

3

6.67×10−11( ) 1.99×1030( )= 8.0×106 s = 0.25τ Earth

τNeptune = 2π30×1.5×1011( )

3

6.67×10−11( ) 1.99×1030( )= 5.2×109 s =160τ Earth

End of material for

Midterm 1

Midterm 1

Physics 1200 – Humanic

xx:xx pm lecture February 21, 2018

Name _________________________________Rec. Instructor_____________________ * Exam is closed book and closed notes. A calculator is allowed. * Write your name and the name of your recitation instructor on every page of the exam. * This exam consists of two sections: Section I – 8 multiple choice questions (40 points) Section II – 2 problems where you are required to show work (50 points) * You have 55 minutes to complete both sections. F = Gm1m2/r2 G = 6.67 x 10-11 Nm2/kg2 g(Earth) = 9.80 m/s2 vx = v0x + axt x = ½ (v0x + vx)t x = v0xt + ½ axt2 vx

2 = v0x2 + 2axx

vy = v0y + ayt y = ½ (v0y + vy)t y = v0yt + ½ ayt2 vy

2 = v0y2 + 2ayy

fk = µkFN h2 = ho

2 + ha2 sin θ = ho/h cos θ = ha/h tan θ = ho/ha

v = 2πr/T F = mg + ma ac = v2/r Fc = mv2/r fs

MAX = µsFN Σ Fx = max Σ Fy = may

Chapter 6

Work and Energy

Work Done by a Constant Force

FsW =( )J joule 1 mN 1 =⋅

Work involves force and displacement.

Work Done by a Constant Force

Work Done by a Constant Force

W = F cosθ( )s

1180cos

090cos

10cos

−=

=

=

More general definition of the work done on an object, W, by a constant force, F, through a displacement, s, with an angle, θ, between F and s:

Note that W is a scalar!

Work Done by a Constant Force

Example: Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0 N, the angle is 50.0 degrees, and the displacement is 75.0 m.

W = F cosθ( )s = 45.0 N( )cos50.0!" #$ 75.0 m( )

= 2170 J

Work Done by a Constant Force

W = F cos0( )s = Fs

W = F cos180( )s = −Fs

A weightlifter doing positive and negative work on the barbell:

Lifting phase --> positive work.

Lowering phase --> negative work.

Example. Work done skateboarding. A skateboarder is coasting down a ramp and there are three forces acting on her: her weight W (675 N magnitude) , a frictional force f (125 N magnitude) that opposes her motion, and a normal force FN (612 N magnitude). Determine the net work done by the three forces when she coasts for a distance of 9.2 m.

W = (F cos θ)s , work done by each force W = (675 cos 65o)(9.2) = 2620 J W = (125 cos 180o)(9.2) = - 1150 J W = (612 cos 90o)(9.2) = 0 J

The net work done by the three forces is: 2620 J - 1150 J +0 J = 1470 J

Work Done by a Constant Force

Example: Accelerating a Crate The truck is accelerating at a rate of +1.50 m/s2. The mass of the crate is 120 kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by all of the forces acting on it?

Work Done by a Constant Force

The angle between the displacement and the normal force is 90 degrees. The angle between the displacement and the weight is also 90 degrees.

W = F cos90( )s = 0Thus, for FN and W,

Work Done by a Constant Force

The angle between the displacement and the friction force is 0 degrees.

W = 180N( )cos0!" #$ 65 m( ) =1.2×104 J

fs =ma = 120 kg( ) 1.5m s2( ) =180N

The Work-Energy Theorem and Kinetic Energy

Consider a constant net external force acting on an object. The object is displaced a distance s, in the same direction as the net force.

The work is simply

s

∑F

W = (Σ F)s = (ma)s

The Work-Energy Theorem and Kinetic Energy

( ) ( ) 2212

2122

21

ofof mvmvvvmasmW −=−==

vf2 = vo

2 + 2 as( )as( ) = 1

2 vf2 − vo

2( )

DEFINITION OF KINETIC ENERGY The kinetic energy KE of and object with mass m and speed v is given by

221KE mv=

constant acceleration kinematics formula

SI unit: joule (J)

The Work-Energy Theorem and Kinetic Energy

THE WORK-ENERGY THEOREM When a net external force does work on an object, the kinetic energy of the object changes according to

2212

f21

of KEKE omvmvW −=−=

(also valid for curved paths and non-constant accelerations)

The Work-Energy Theorem and Kinetic Energy

Example: Deep Space 1 The mass of the space probe is 474 kg and its initial velocity is 275 m/s. If the 56.0 mN force acts on the probe through a displacement of 2.42×109m, what is its final speed?

The Work-Energy Theorem and Kinetic Energy

2212

f21W omvmv −=

W = [(Σ F)cos θ]s

The Work-Energy Theorem and Kinetic Energy

5.60×10-2N( )cos0 2.42×109 m( ) = 12 474 kg( )vf

2 − 12 474 kg( ) 275m s( )2

( )[ ] 2212

f21cosF omvmvs −=∑ θ

sm805=fv

[(Σ F)cos θ]s

The Work-Energy Theorem and Kinetic Energy

Conceptual Example: Work and Kinetic Energy A satellite is moving about the earth in a circular orbit and an elliptical orbit. For these two orbits, determine whether the kinetic energy of the satellite changes during the motion.

Gravitational Potential Energy

W = F cosθ( )s

Wgravity =mg ho − hf( )

Find the work done by gravity in accelerating the basketball downward from ho to hf.

Gravitational Potential Energy

Wgravity =mg ho − hf( )

We get the same result for Wgravity for any path taken between ho and hf.

Gravitational Potential Energy

Example: A Gymnast on a Trampoline The gymnast leaves the trampoline at an initial height of 1.20 m and reaches a maximum height of 4.80 m before falling back down. What was the initial speed of the gymnast?

Gravitational Potential Energy

2212

f21W omvmv −=

( )fo hhmgW −=gravity

( ) 221

ofo mvhhmg −=−

vo = −2g ho − hf( )

vo = −2 9.80m s2( ) 1.20 m − 4.80 m( ) = 8.40m s

Gravitational Potential Energy

fo mghmghW −=gravity

DEFINITION OF GRAVITATIONAL POTENTIAL ENERGY The gravitational potential energy PE is the energy that an object of mass m has by virtue of its position relative to the surface of the earth. That position is measured by the height h of the object relative to an arbitrary zero level:

mgh=PE

( )J joule 1 mN 1 =⋅

The Ideal Spring

HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING The restoring force on an ideal spring is xkFx −=

SI unit for k: N/m

The Ideal Spring

Example: A Tire Pressure Gauge The spring constant of the spring is 320 N/m and the bar indicator extends 2.0 cm. What force does the air in the tire apply to the spring?

The Ideal Spring

( )( ) N 4.6m 020.0mN320 ==

= xkF Appliedx

Potential energy of an ideal spring

A compressed spring can do work.

Potential energy of an ideal spring

Welastic = F cosθ( )s = 12 kxo + kx f( )cos0 xo − x f( )

2212

21

elastic fo kxkxW −=

|F| = kx

x0 xf

F0

Ff

F = 12 F0 +Ff( ) s

F

Potential energy of an ideal spring

DEFINITION OF ELASTIC POTENTIAL ENERGY The elastic potential energy is the energy that a spring has by virtue of being stretched or compressed. For an ideal spring, the elastic potential energy is

221

elasticPE kx=

SI Unit of Elastic Potential Energy: joule (J)