The big new thing in 2D: changes in direction
Change in direction;
perpendicular to v
Change in speed; parallel to v
v  vvˆ
dv d (vvˆ ) dv
dvˆ
a


vˆ  v
dt
dt
dt
dt
An object can move at constant speed and still have a ≠0!
This didn’t happen in 1D!!
Graphically: Imagine an object moving along the following
trajectory at constant speed. Take the positions at times t and
Δt and find the average acceleration between them:
v (t )
t
t  t
v (t  t )
v (t )
v (t  t )
v
a
In 2 (or 3) dimensions,
acceleration can occur
both parallel to velocity
or perpendicular to it
(a) Acceleration in the
direction of the velocity
changes the speed.
(b) Acceleration
perpendicular to the
velocity does not change
the speed but shifts the
direction of the motion.
Example: Shown below are the trajectory of a moving object and the snapshots
taken every second. Which of the following is true about the components of the
acceleration?
4s
y
x
3s
2s
1s
A) ax = 0, ay > 0
B) ax > 0, ay > 0
C) ax < 0, ay = 0
Note: Both the speed and the direction of velocity are changing!
4s
3s
2s
1s
v(1)
v(3)
v(2)
Circular motion
T - period
f =1/T - frequency
v(t1 )
r


v(t2 )
-angular frequency,
or angular speed
If   const
l

r
dl
d r 
2r
v

 r 
dt
dt
T
d
dt

a tan

 dv d vvˆ  dv
dvˆ
a


vˆ  v
dt
dt
dt
dt
d
dt
then

2
 2f .
T
- angular acceleration
dv d


r  r
dt
dt
a rad
v2

  2r
r
a  a tan  a rad
2
2
a  a tan
 a rad
Acceleration of uniform circular motion
(centripetal or radial acceleration)
v1
v1
r


v  v 2  v1
v2
2
v
a
r
a   2r
v  2v sin  / 2 
r
t 
v
 2v 2 sin(  / 2) 
v
v2
a  lim
 lim 

t
r
r


lim
2 sin(  / 2)

1
Acceleration of a Point in Circular motion
Tangential
acceleration
Net Acceleration
Radial or
centripetal
acceleration
•
•
•
•
The radial acceleration is given by arad=v²/r=r²
If  is constant there is no tangential component
In general, atan=r
Only the tangential acceleration changes the speed of the point
Example: Period of a satellite motion
g
R
g  9.8m / s 2
R  6400km
v2
a
R
ag
v  gR
v2
g
R
T
2R
R
 2
v
g
v  gR  9.8m / s 2 * 6400 *1000m  8 *103 m / s  8km / s
6400 *1000m
5000s
T  2
 2 * 800s  5000s 
 83 min
2
9.8m / s
60 min
Example: Two balls attached to a string as shown at 0.20 m and 0.40 m
from the center move in circles at a uniform frequency of 20 rpm.
a. What are their linear speeds?
b. What are their periods?
R1  0.20m
f  20min 
R2  0.40m
  2f
v  R  2fR
f  20rpm
v1, 2  ?
T1, 2  ?
T1  T2 
1
1 min 1 1
 s
60 s
3
1
 3s
f


 2  1 3 s  0.40m  0.84m / s
v1  2  1 3 s 1  0.20m  0.42m / s
v2
1
Example: The ferris wheel in the figure rotates counterclockwise at a
uniform rate. What is the direction of the average acceleration of a
gondola as it goes from the top to the bottom of its trajectory?
A. Down
B.
C. The acceleration is 0 because
the motion is uniform.

a ave 

vi


v f  vi
t

vf