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Exam Review and Friction
Announcements:
•  Exam Thursday at 7:30pm
–  Bring a #2 pencil
–  Will be long answer
questions in addition to
multiple choice
–  Room assignments will be
on web page (Exam info)
–  Calculators and 1 double
sided sheet of notes allowed
The Thinker by August
Rodin – museum in Paris
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Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/
Atwood’s machine
An Atwood's machine is a pulley with two masses
connected by a string as shown. The mass of object
A, mA, is twice the mass of object B, mB. What is the
tension T in the string ?
A
A
B
B
Since the weights are unequal, it should be clear that there
will be some non-zero acceleration. Thus, since
and
are not zero, neither A nor B can be correct.
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Full solution to Atwood’s machine
A
B
A
B
We defined up as positive for both masses which means
&
Solving for acceleration:
give
so the tension is
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Clicker question 1
Set frequency to BA
How does the force exerted on the cart by the string (T) compare with the weight
of body B?
(Assume all surfaces are frictionless)
A: T = mB g.
B: T < mB g.
C: T > mB g.
The system WILL accelerate, Ma will accelerate to the right, Mb will accelerate down
(at the same rate, they're tied together!) So look at Mb: it accelerates down, so the net
force on it is DOWN (by N-II). There are only 2 forces acting on it: mB*g down, and
T up. If the NET force on Mb is down, then mB*g must "win". .
If you think mB*g = T, then that means you think there is ZERO net force on Mb.
Which means it would just sit there.... but there's no friction, so it really will be pulled
down, and start to accelerate.
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Clicker question 2
Set frequency to BA
In the 17th century, Otto von Güricke, a physicist in Magdeburg, fitted two hollow
bronze hemispheres together and removed the air from the resulting sphere with a
pump. Two eight-horse teams could not pull the halves apart even though the
hemispheres fell apart when air was readmitted.
Suppose von Güricke had tied both teams of horses to one side and bolted the other side
to a heavy tree trunk. In this case, the tension on the hemispheres would be
A: twice
B: exactly the same as
C: half
what it was before
Draw a force diagram! Let's call the force of an 8-horse team "F". Before, you had
"F" to the left, and "F" to the right. They balanced, but there was a tension in the
system. Now, you put both teams on one side, which means you have "2F" on that side.
The other side is connected to a tree. Assuming the tree doesn't budge, there must be
"2F" acting the opposite direction. Once again, these forces balance (the sphere doesn't
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accelerate), but the TENSION is twice as much as before.
Clicker question 3
Set frequency to BA
A glider is pulled along an air track with a string at an angle (theta) from the
horizontal, with a constant force.
The direction of the net force on the
glider is
A: Horizontal.
B: At an angle theta above the
horizontal.
C: At some angle above the
horizontal, but not necessarily theta.
Net force = ma, that's Newton's law. "Air track" implies no friction, so this glider
is clearly going to be accelerating. Assuming it stays on the track (implied by
"pulled along an air track"), it must be accelerating to the right. So by N-II, Net
force is horizontal! Apparently, the vertical component of F, plus N (normal
force), minus mg, must all cancel - the net result is OBSERVED, the glider is 6
moving to the right, so the net force must be to the right!
Review of what is covered
Chapter 1
•  Significant figures
•  Units and unit conversion
•  Vectors
–  Getting components from trigonometry
–  Getting magnitude and direction from components
–  Addition and subtraction
The magnitude of
is
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Review of what is covered
Chapter 2 & 3 – Motion
•  Displacement is a vector measuring change in
position
•  Velocity measures displacement per time
Average velocity:
Instantaneous velocity:
•  Acceleration measures change of velocity per time
Average acceleration:
Instantaneous acceleration:
•  Often deal with constant acceleration such as free
fall where
in the downward direction
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Review of what is covered
Chapter 2 & 3 – Motion
One dimensional equations of motion
No displacement in this equation
No final velocity in this equation
No time in this equation
No acceleration in this equation
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Review of what is covered
Chapter 2 & 3 – Motion
Equation for projectile motion
Constant velocity in x direction:
Free fall in y direction:
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Review of what is covered
Chapter 3 – Circular motion & relative velocity
Circular motion, even at constant speed, is an example of
acceleration because the velocity vector direction changes
Can divide acceleration into tangential and radial
components:
and
Relative velocity: A reference frame which moves at
constant velocity (can be 0) is an inertial reference frame.
Use vector addition of velocities to transfer from one
reference frame to another
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Review of what is covered
Chapter 4 & 5 – Forces & Newton’s Laws
Newton’s First Law: A body maintains a constant velocity
(which can be 0) unless acted upon by a net force
Newton’s Second Law:
or
&
Newton’s Third Law: If body A exerts a force on body B then
there is a force which is equal in magnitude but opposite in
direction by body B on body A.
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What chapters are included
Basically, all of chapters 1,2,3, and 4 plus
the first section of chapter 5 are covered
However, some parts were not
covered and will not be on the exam
Chapter 1.10 on vector products is not included
Chapter 2.8 on velocity and position from
integration is not included
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Exam tips
•  Remember that displacement, velocity,
acceleration, and force are vectors and vectors
have a direction and magnitude
•  Understand the CAPA problems, clicker
questions, and tutorials
•  You have nearly two hours to answer multiple
choice plus long answer questions. Take your
time and check your work.
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Clicker question 4
Set frequency to BA
A rider in a "barrel of fun” finds herself stuck with her
back to the wall. Which diagram correctly shows the
forces acting on her?
Pink
A)
Blue
B)
Green
C)
Purple
D)
Yellow
E)
Barrel of Fun
What is going on is that the ride is pushing on the people to
make them go in a circle. For every action there is an equal but
opposite reaction, and so the people push back on the ride. The
centripetal force the ride exerts on people becomes the normal
force that causes the friction that keeps them from sliding down
the walls, and that friction as such must be larger than their
weight, or down the wall they go. Nothing is pushing them to the
outside; acceleration is toward the middle. Remember you feel
"thrown" the opposite direction of the acceleration. (i.e. if your
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car accelerated forward, you are thrown backward)
Clicker question 5
T
N
B
R
Set frequency to BA
A bucket containing a brick is swung in
a circle at constant speed in a vertical
plane as shown. The bucket is swung
fast enough that the brick does not fall
out.
The net force on the brick as it is swung
has maximum magnitude at position.
A) Top.
B) Bottom.
C) Right
D) The net force has the same
magnitude at all positions.
The net force has the same magnitude at all positions. Since the
magnitude of the acceleration is constant (a = v2/R), the net force
must have constant magnitude (Fnet = ma).
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Clicker question 6
Set frequency to BA
A bucket containing a brick is swung in
a circle at constant speed in a vertical
plane as shown.
T
N
B
R
Consider the normal force exerted on the brick by
the bucket when the bucket is at the three positions
shown: R, T, B. The magnitude of the normal force
is a minimum at position...
A) Top.
B) Bottom.
C) Right
D) The net force has the same
magnitude at all positions.
At the top. Draw free-body diagrams and keep in mind the net force
has the same magnitude at the top and the bottom.
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