Transcript Electromagnetic Waves

Lecture 15 – Drag Force &
Circular Dynamics
Why is the drag force on a piece
of paper so large?
Uniform Circular Motion
v
α
r
α
Circular Motion
• A particle is going around a center at (x,y) = (3m, -4m) in
uniform circular motion with a radius of 3m and a period
of 2 seconds. At t=0 it starts at a point on the negative yaxis on its anti-clockwise motion. Which statement is
correct?
• a. The particle is at (1m, -2m) at t=2s.
• b. The particle’s velocity is constant at 37.7 m/s.
• c. The particle’s velocity points initially in positive ydirection.
• d. The particle’s acceleration points initially in positive xdirection.
Dynamics of circular motion
• Since a = v2/r, we can write down the force
• But: we don’t know what it is , i.e. what
entity provides is
• Centripetal Force has constant strength, but
changing direction
• Especially tricky for non-contact forces
• Examples
– Car turning corner
– Space shuttle orbiting Earth
Pre-lecture Exercise (6.5)
• How is the physics of an object in a vertical circular loop
(Sample problem on page 125) different from the one in a
Ferris wheel (as in problem 3 of Homework #5)?
• a. The normal force on the object at the bottom of the loop (but not
the wheel) is upward.
• b. The normal force on the object at the top of the loop (but not the
wheel) is downward.
• c. The centripetal acceleration at the bottom of the loop (but not the
wheel) is upward.
• d. The centripetal acceleration at the top of the loop (but not the
wheel) is downward.
• e. None of the above.
Example: Simulating gravity in
spaceship with R=10m
Lecture 16 – Energy & Work
Exam
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A > 80%
B > 62%
C > 56%
D > 48%
• Average 60%
Post-lecture Exercise (6.5)
• A normal car is undergoing a banked
circular turn, see the sample problem on
page 128. The bank angle is 10 degrees and
the radius of the curve 200m. What is the
maximal velocity the car can go without
sliding?
• v = 18.59 m/s
– Normal force (tilted 10 degrees towards center) has to
produce the necessary centripetal force, which limits
the velocity, since N is fixed (mass cancels out!)
Work
• Work is “accomplished” by a force applied
over some distance
• Angle matters!
• Work is scalar product between force vector
and displacement vector
• Work is a number or scalar (with units
Joules=J): it can be zero, even negative!
• Maximal work if force is parallel to
displacement
You are holding a book in your
hand, so it doesn’t fall down. Are
you doing work on the book?
• Yes
• No
• Not enough information
Group Work: Page 1 of Workenergy theorem worksheet
Lecture 17 – Work-Energy
Theorem
Reminder: Vectors
• Components vs polar coordinates
• 2D vs 3D vs nD
• Scalar product can be used to calculate
angle in any dimension!
• Need to master vectors (ED is vectors)
Scalar Product
Varied Force
Double index notation: Quo vadis?
• There is a force involved, so there are two
indices!
• Need to keep it straight: who is doing the
work ON what?
• Still 2 indices necessary and same rule:
swap indices, obtain minus sign!
Example: Work done by person
on spring
• F= – kx is Hooke’s law
• It is the force of a spring, i.e. exerted by the
spring on something else: Fp,s = – kx = –Fs,p
• Need to do integral over x from 0 until x???
• Don’t confuse integration variable with
value of variable: x’ = x
• Geometrical interpretation: area of triangle
Energy
• Useful label/name (like force, etc.)
• Rough definition of energy: ability to do
work
• Different forms
– Object in motion: kinetic energy (prop. to
velocity)
– Object in configuration where it could do work:
potential energy
– Chemical energy …
Kinetic Energy
• Object in motion (relative to another) can do
work on contact  train on train in crash, etc.
• Quantitative definition:
W= Fd = (ma)d
; use vf2= vi2+2ad
= m{(vf2 – vi2)/2d}d
= ½ mvf2 – ½ m vi2
= Kf - Ki
Pre-lecture Exercise (7.1-7.4)
Calculate the kinetic energy of a baseball
(110g) going v = – 50 mph, i.e. traveling in
the negative x- direction.
• K= ½ m v2
– Needed to convert to kg and m/s
– Sign of velocity does not matter!
– In higher dimensions v2 = vx2+vy2+vz2
Two marbles, one twice as heavy as the
other, are dropped to the ground from
the top of a building. Just before hitting
the ground, the heavier marble has…
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•
•
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…as much kinetic energy as the lighter one.
…twice as much kinetic energy as the lighter one.
…half as much kinetic energy as the lighter one.
…four times as much kinetic energy as the lighter
one.
Work-energy theorem
• Net work done on an object is equal to
the change of its kinetic energy
• Note that work and hence change of KE can
be zero or even negative!
Pre-lecture Exercise (7.5-7.9)
In the Work and Energy simulation, set the mass to 5 kg, the initial
position to 0, the force to 5 N, and the initial velocity to - 3m/s.
• What is the ratio of the work being done on the block
(from the start to the end of the simulation) and the
difference in kinetic energy (kinetic energy at the end of
the simulation minus the kinetic energy at the start of
the simulation)?
• Answer: 1 (by Work-energy theorem or direct
calculation)