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Using the

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Using the Clicker If you have a clicker now, and did not do this last time, please enter your ID in your clicker. First, turn on your clicker by sliding the ... – PowerPoint PPT presentation

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Title: Using the


1
Using the Clicker
  • If you have a clicker now, and did not do this
    last time, please enter your ID in your clicker.
  • First, turn on your clicker by sliding the power
    switch, on the left, up. Next, store your student
    number in the clicker. You only have to do this
    once.
  • Press the button to enter the setup menu.
  • Press the up arrow button to get to ID
  • Press the big green arrow key
  • Press the T button, then the up arrow to get a U
  • Enter the rest of your BU ID.
  • Press the big green arrow key.

2
Uniform Circular Motion
  • The path is a circle (radius r, circumference
    2pr).
  • Uniform means constant speed v 2pr / T,
    where the period T is the time to go around the
    circle once.
  • Angle in radians (arc length Ds) / (radius
    r)
  • Angular velocity w Dq/Dt 2p/T rad/sec, is
    also independent of r
  • Note that v (2p/T)r w r m/s, and
    therefore v is proportional to the radius of
    the circle.

Dq Ds1/r1 Ds2/r2 is independent of the
radius r of the circle, and is dimensionless
3
Velocity on circular path
Displacement for large time interval
v Dr/Dt but chord Dr is almost arc s r Dq
So again v (rDq)/Dt r(Dq/Dt) wr
constant
Displacement for small time interval
Direction approaches tangent to circle, which is
perpendicular to r
For uniform circular motion, the velocity vector
has magnitude v wr, and direction is tangent to
the circle at the position of the particle.
4
Magnitude of the acceleration
v2
v1
Dv
For small time intervals, the vector Dv points
toward the center, and has the magnitude Dv v
Dq so a Dv /Dt v (Dq/Dt) v w v2/r
-v1
Dq
Dq
v2
For uniform circular motion, the magnitude of the
acceleration is w2r v2/r, and the direction of
the acceleration is toward the center of the
circle.
5
Coins on a turntable
  • Two identical coins are placed on a flat
    turntable that is initially at rest. One coin is
    closer to the center than the other disk is.
    There is some friction between the coins and the
    turntable. We start spinning the turntable,
    steadily increasing the speed. Which coin starts
    sliding on the turntable first?
  1. The coin closer to the center.
  2. The coin farther from the center.
  3. Neither, both coin start to slide at the same
    time.

6
A general method for solving circular motion
problems
  • Follow the method for force problems!
  • Draw a diagram of the situation.
  • Draw one or more free-body diagrams showing all
    the forces acting on the object(s).
  • Choose a coordinate system. It is often most
    convenient to align one of your coordinate axes
    with the direction of the acceleration.
  • Break the forces up into their x and y
    components.
  • Apply Newton's Second Law in both directions.
  • The key difference use toward the center

7
Coins on a turntable (work together)
  • Sketch a free-body diagram (side view) for one of
    the coins, assuming it is not sliding on the
    turntable.
  • Apply Newtons Second Law, once for each
    direction.

8
Coins on a turntable (work together)
  • Sketch a free-body diagram (side view) for one of
    the coins, assuming it is not sliding on the
    turntable.

FN
Can you tell whether the velocity is into or out
of the screen?
FS
Axis of rotation
mg
9
Coins on a turntable (work together)
  • Apply Newtons Second Law, once for each
    direction.
  • y-direction FN - mg 0 so that FN mg
  • x-direction FS max m(v2/r) both FS and a
    are to left

FN
y
Can you tell whether the velocity is into or out
of the screen?
FS
x
Axis of rotation
mg
As you increase r, what happens to the force of
friction needed to keep the coin on the circular
path?
It is the same diagram and result either way!
10
Trick question!
  • v has a hidden dependence on r, so that the
    obvious dependence on r is not the whole story.
    The two coins have different speeds.
  • Use angular velocity for the comparison, because
    the two coins rotate through the same angle in a
    particular time interval.
  • This gives
  • As you increase r, what happens to the force of
    friction needed to keep the coin staying on the
    circular path?
  • The larger r is, the larger the force of static
    friction has to be. The outer one hits the limit
    first.

11
Conical pendulum
  • A ball is whirled in a horizontal circle by means
    of a string. In addition to the force of gravity
    and the tension, which of the following forces
    should appear on the balls free-body diagram?
  • A normal force, directed vertically up.
  • A centripetal force, toward the center of the
    circle.
  • A centripetal force, away from the center of
    the circle.
  • Both 1 and 2.
  • Both 1 and 3.
  • None of the above.

12
Conical pendulum (work together)
  • Sketch a free-body diagram for the ball.
  • Apply Newtons Second Law, once for each
    direction.

13
Conical pendulum (work together)
  • Sketch a free-body diagram for the ball.
  • Apply Newtons Second Law, once for each
    direction.
  • y-direction T cosq - mg may 0
  • x-direction T sinq max m(v2/r)
  • Solve (mg/cosq)sinq mv2/r
  • (rg tanq )1/2 v

Tsinq
q
q
y
Tcosq
T
Axis of rotation
x
Resolve
mg
Choose
14
Gravitron (or The Rotor)
  • In a particular carnival ride, riders are pressed
    against the vertical wall of a rotating ride, and
    then the floor is removed. Which force acting on
    each rider is directed toward the center of the
    circle?
  • A normal force.
  • A force of gravity.
  • A force of static friction.
  • A force of kinetic friction.
  • None of the above.

15
Gravitron (work together)
  • Sketch a free-body diagram for the rider.
  • Apply Newtons Second Law, once for each
    direction.
  • y direction FS - mg may 0 (he hopes)
  • x direction FN max m (v2/r)

FS
Hes blurry because he is going so fast!
Axis of rotation
FN
y
mg
x
16
Test tonight
  • Go to COM 101. (Lecture section A1)
  • Test is 6-8 pm.
  • Test has more problems than I said, because some
    are shorter or easier.
  • Best wishes! (Good luck implies that you
    might not be fully prepared, and I dont believe
    that for a minute.)
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