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12/9 Circular Motion
Text: Chapter 5 Circular Motion HW 12/9 “Rotating Drum” HW 12/9 “Rotating Disk”
These two are for practice (will not be collected) and they also review friction forces.
Course Evaluation today
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The Pendulum
TS,M
WE,M
Stationary Swinging (moving in a circular arc)
What happens to the Tension?
Is there acceleration?
TS,M
WE,M
?If the Tension is greater, then acceleration must point up.
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Find the acceleration
a = v/t as usual
Draw vi, vf, and v as usual(Place v’s 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?
v = a if t = 1 second
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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:
For 2v
For v
4 times the acceleration for twice the velocity, same radius.
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In one second:What if the radius is halved?
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In one second:What if the radius is halved?
v
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In one second:What if the radius is halved?
r/2
ra
2 times the acceleration for half the radius, same velocity.
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In one second:
For r/2
For ra
2 times the acceleration for half the radius, same velocity.
4 times the acceleration for twice the velocity, same radius.
For 2v
For va
So: ac = r v2
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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 2r, the circumference.The speed, v, is distance ÷ time or 2r/T
Fnet = ma works for centripetal acceleration also and free body diagrams are handled the same way.
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The Pendulum
TS,M
WE,M
Stationary Swinging (moving in a circular arc)
TS,M
WE,M
?If the Tension is greater, then acceleration must point up.
a = v2/r and we can get v from energy.
a and Fnet both point up
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Example
WE,B
TS,B
A 1kg ball is swung in a horizontal circle at constant speed at the end of a string. Draw a FBD. Which way does a point?
y
x
Ty
Tx
Fnet,y = Ty - WE,B = 0 so Ty = WE,B
Fnet,x = Tx = ma = mv2/r
v = 2r/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 (m=3kg) 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 at:
a. the top of the circle, andb. the bottom of the circle.
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Bucket of water problem: at the top
v
mw = 3 kgr = 2 mv = 5 m/s(constant v)
Acceleration points
Draw a FBD of the water.
WE,W
NB,W
Apply Newton’s 2nd law.
Fnet = WE,W + NB,W = ma
Want NB,W, find WE,W and ma
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
v
mw = 3 kgr = 2 mv = 5 m/s(constant v)
Acceleration points
Draw a FBD of the water.
WE,W
NB,W
Apply Newton’s 2nd law.
Fnet = NB,W - WE,W = ma
Want NB,W, find WE,W and ma
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