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