1 CIRCULAR MOTION 2 r s IN RADIANS length of the arc [ s ] divided by the radius [ r ] subtending...
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![Page 1: 1 CIRCULAR MOTION 2 r s IN RADIANS length of the arc [ s ] divided by the radius [ r ] subtending the arc.](https://reader036.fdocuments.in/reader036/viewer/2022062421/56649e2b5503460f94b196fc/html5/thumbnails/1.jpg)
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CIRCULAR CIRCULAR MOTIONMOTIONCIRCULAR CIRCULAR MOTIONMOTION
CIRCULAR CIRCULAR MOTIONMOTIONCIRCULAR CIRCULAR MOTIONMOTION
CIRCULAR CIRCULAR MOTIONMOTION
CIRCULAR MOTIONCIRCULAR MOTION
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2
r
s
IN RADIANSlength of the arc [ s ] divided by the radius [ r ] subtending the
arc
rs
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© 3
22 r
r
IN A FULL CYCLEIN A FULL CYCLE
RADIANSRADIANS
rs
Hence 3600 = 2 Radians
1800 = radians
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v
v
r
s
A
B
Suppose the distance, s, from A to B takes a time t.r
s rs
tr
t
s
rvv = tangential
velocity
t
= the ANGULAR VELOCITY
[DIVIDING EACH
SIDE BY t]
![Page 5: 1 CIRCULAR MOTION 2 r s IN RADIANS length of the arc [ s ] divided by the radius [ r ] subtending the arc.](https://reader036.fdocuments.in/reader036/viewer/2022062421/56649e2b5503460f94b196fc/html5/thumbnails/5.jpg)
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v
v
r
rv v = tangential velocity t
= the ANGULAR VELOCITY in
RADIANS PER SECOND
Considering a full circle:
T
2 f 2
T = Periodic time,
f = frequency
![Page 6: 1 CIRCULAR MOTION 2 r s IN RADIANS length of the arc [ s ] divided by the radius [ r ] subtending the arc.](https://reader036.fdocuments.in/reader036/viewer/2022062421/56649e2b5503460f94b196fc/html5/thumbnails/6.jpg)
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Newton’s 1st Law “I want to go straight on!”
Come back, I am a F
v
Circular motion requires a force, F, towards the centre of the motion
F
v
A CENTRIPETAL FORCE
Something has to provide it, e.g. the tension in a string
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v
v
Angular Velocity
t
Linear Velocityv = r
Centripetal Acceleration
22
rrv
a
Centripetal Force
22
mrrmv
F
SUMMARY
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F
The necessary centripetal force, F is provided by the friction at the tyres
Question If the track has a radius of 50m and the limiting frictional force is 0.5 of the car’s weight, find the car’s maximum speed before it slides off the track
rmv
F2
50
5.02mv
mg
5081.95.0 v v = 15.7 ms-1
MOTION IN A HORIZONTAL CIRCLE
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MOTION IN
A VERTICAL CIRCLE FORCES ACTING: 1. Weight2. Tension in the string
mg
Tv
rmv
F2
rmv
mgT2
Where F is the resultant force
towards the centre
rmv
mgT2
At the bottom the tension has to provide a centripetal force AND support the weight
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MOTION IN
A VERTICAL CIRCLE FORCES ACTING: 1. Weight2. Tension in the string
mg
T
v
rmv
F2
At the top the tension is less than at the bottom, because the weight provides some of the centripetal force
rmv
mgT2
mgr
mvT
2
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A
B10 m
The radius of the track is 10 m
Using g = 10 ms-2, find the g forces at A and B. [Assume no energy is lost]
20m
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Using EK gain = Ep lost gives velocity at the bottom = 24.5 m s-1
At the bottom
mg
R
v
r
mvmgR
2
This gives R = 70 m
710
70
m
m
mg
RHence
At the top
mg
R r
mvmgR
2
Energy change gives v = 14.1 m-1
R = 10 m and result is 1 g