12/9 Circular Motion

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12/9 Circular Motion

• Text Chapter 5 Circular Motion
• HW 12/9 Rotating Drum
• HW 12/9 Rotating DiskThese two are for
  practice (will not be collected) and they also
  review friction forces.
• Course Evaluation today

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The Pendulum
Stationary
If the Tension is greater, then acceleration must
point up.
What happens to the Tension?
Is there acceleration?
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Find the acceleration
a ?v/?t as usual
Draw vi, vf, and ?v as usual(Place vs tail to
tail and draw tip to tip.
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In one second
?v
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In one second
?v
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In one second
?v
What is the acceleration?
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In one second
a
What is the acceleration?
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In one second
What if the speed is doubled?
?v
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In one second
a
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In one second
4 times the acceleration for twice the velocity,
same radius.
For v
For 2v
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What if the radius is halved?
In one second
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What if the radius is halved?
In one second
?v
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What if the radius is halved?
In one second
2 times the acceleration for half the radius,
same velocity.
r
a
r/2
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In one second
4 times the acceleration for twice the velocity,
same radius.
2 times the acceleration for half the radius,
same velocity.
For r
For v
a
a
For r/2
For 2v
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Centripetal Acceleration

• perpendicular to the velocity, points towards the
  center of the circle
• ac v2/r v is the instantaneous velocity
  tangent to the path, r is the radius.

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Some Equations and Definitions
The period, T, is the time for one revolution.
The distance for one revolution is 2?r, the
circumference.
The speed, v, is distance time or 2?r/T
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The Pendulum
Stationary
If the Tension is greater, then acceleration must
point up.
a and Fnet both point up
a v2/r and we can get v from energy.
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Example
Fnet,y Ty - WE,B 0 so Ty WE,B
Fnet,x Tx ma mv2/r
v 2?r/T
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What about the centrifugal force?

• There is no such force, regardless of what Mr.
  Wizard says.

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Bucket of water problem

• You swing a bucket of water (m3kg) in a vertical
  circle at constant speed of 5 m/s. The radius of
  the circle is 2 m. What is the normal force by
  the bottom of the bucket on the water ata. the
  top of the circle, andb. the bottom of the
  circle.

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Bucket of water problem at the top
v
Acceleration points
Fnet WE,W NB,W ma
Draw a FBD of the water.
mw 3 kgr 2 mv 5 m/s(constant v)
Want NB,W, find WE,W and ma
Apply Newtons 2nd law.
WE,W mg 3(9.8) 29.4 N
ma mv2/r 3(52)/2 37.5 N
Fnet 37.5 29.4 NB,W
NB,W 8.1 N
If you swing slow enough the water will come out.
How slow do you have to swing?
Slow enough so that NB,W becomes zero.
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Bucket of water problem at the bottom
Acceleration points
Fnet NB,W - WE,W ma
Draw a FBD of the water.
mw 3 kgr 2 mv 5 m/s(constant v)
Want NB,W, find WE,W and ma
Apply Newtons 2nd law.
WE,W mg 3(9.8) 29.4 N
ma mv2/r 3(52)/2 37.5 N
Fnet NB,W - 29.4 37.5
NB,W 66.9 N