Download Central Force Post-Lab

Central Force Post-Lab
F N 
mkg 
Since the graph is linear and
contains (0,0)
F m
F  13.3 m 2m

s 
Units :
kgm
N
s2  m

s2
kg
kg
constant radius  30. m
The y - intercept is negligible,
meaning that when the m  0 kg,
constant speed  20. m
the F  0 N.
s
Central Force Post-Lab
F N 
F   0.333 kg v 2
m

 s
constant mass  10. kg
vm
constant radius  30. m
F N 
2  m2
v 

Since the graph is linear and
contains (0,0)
F  v2

s2 
Units :
kgm
2
N
kg
s



m
m2
m2
2
2
s
s
The y - intercept is negligible,
meaning that when v  0 m ,
s
F  0 N.
Central Force Post-Lab
F N 
2

1
kgm

F   4000
 2

 s r
constant mass  10. kg
F N 
r m 
1
1 

r m
Since the graph is linear and
contains (0,0)
constant speed  20. m
Units :
kgm
2
2
N
kgm
s


1
1
s2
m
m
The y - intercept is negligible,
meaning that as the r  ,
the F  0 N .
F1
r
s
Central Force Post-Lab
Since F  m, F  v 2 , and F  1
r
mv 2
F
r
mv 2
Fz
r
where z  constant of proportionality
Central Force Post-Lab
2

1
kg
kgm
2


F   0.333
v F   4000
2 r
m


s 

F  13.3 m 2 m

s 
constant radius  30. m
constant speed  20. m
 v2 
F  z  m
r 

m
v 2 20. s

30. m
r
 13.3 m
z 1

s
constant mass  10. kg
constant mass  10. kg
constant radius  30. m
constant speed  20. m
mv 2
Fz
r
m 2

F  z  v
r
 
1
F  z mv
r
2
2
s2
m 10. kg

r 30. m
 0.333 kg
z 1

mv  10. kg  20. m
2
m
2
kgm
 4000
z 1
s2
s
2
s
Central Force Post-Lab
Since F  m, F  v 2 , and F  1
r
mv 2
F
r
mv 2
Fz
r
z 1
mv 2
F
r
For an object moving in a circle
at constant speed
Uniform Circular Motion (UCM) – Occurs when an object moves in a circle at constant
speed.
Is the object accelerating?
(Acceleration – how quickly the velocity is changing)
Uniform Circular Motion (UCM) – Occurs when an object moves in a circle at constant
speed.
Is the object accelerating?
(Acceleration – how quickly the velocity is changing)
Yes, even though the object is not changing speed it is changing velocity because it
changes direction.

v
Tangential Velocity
r

v

v

v
What then is the magnitude of the acceleration of an object undergoing UCM?
mv 2
F
r
 v2 
F  m  
r 
Fnet  m
v2
a
r
a
Newton’s 2nd Law