Circular motion
v(t1 )
r
T - period
f =1/T - frequency


v(t2 )
-angular frequency,
or angular speed
If   const then  
l

r
v
d
dt

dl
d r 

 r
dt
dt
d
dt
- angular acceleration
a rad

 dv d vvˆ  dv
dvˆ
a


vˆ  v
dt
dt
dt
dt
a tan
v2

  2r
r
dv d


r  r
dt
dt
a  a tan  a rad
a a
2
tan
a
2
rad

dv

dr   v r
2
2

2
2
 2f .
T
 r  2 4
Circular motion and vectors


y

r

x
r  rrˆ  xiˆ  yˆj
x  r cos 
y  r sin 
   z kˆ
d
z 
;
  z
dt
d z
z 
;   z
dt

 dv d vvˆ  dv

dvˆ 
ˆ
a


vv
 a tan  a rad
dt
dt
dt
dt

dv
 dv 
 dv 
a tan 
vˆ  a tan sign  vˆ  rsign  vˆ
dt
 dt 
 dt 
 


atan    r
arad  arad rˆ
Comparison of Linear and Angular motion with Constant Acceleration
Straight-line motion
Fixed-axis rotation
v x  v0 x  a x t
 z  const
 z  0 z   z t
x x  x0  v0 x t  12 a x t 2
   0   0 z t  12  z t 2
v x  v 0 x  2a x ( x  x 0 )
 02   02z  2 z (   0 )
   0  12 ( z   0 z )t
a x  const
2
2
x  x0  12 (v x  v0 x ) t
Example: At t = 0, a grinding wheel has an angular velocity of 24.0 rad/s. It
has an constant angular acceleration of 30.0 rad/s2 until a circuit breaker trips
at t = 2.00 s. From then on, it turns through 432 rad as it coasts to a stop at
constant angular acceleration. What was its acceleration as it slowed down?
1) Angular speed at 2 s :
   0 z   z t  24.0rad / s  30.0rad / s 2 2.00s   84rad / s
2) Angular acceleration:
 02   02z  2 z (   0 )
 z2   z20 0  84rad / s 2
z 

 8.17rad / s 2
2
2432rad 
Relative motion
Galilean transformations:
relation between the description of a particle in two frames which are moving
with respect to each other with constant velocity.



rPA  rPB  rBA
P
t A  tB



v PA  v PB  v BA



a PA  a PB  a BA
y
rP,A
rB,A
A
z
If

vBA  const
rP,B



 rPA  rPB  vBAt
x
z
y
B
x
Example: Moving Sidewalk
A person walking on moving sidewalk: You can have vperson,background = 0
(not moving relative to a picture on the back wall):
Picture on the
vp,s = -v i
background
Vs,bg = +v i
bg: background
s: moving sidewalk
Clearly velocity is a reference-frame dependent quantity!
v p,bg  v p,s  v s,bg  viˆ  viˆ  0
What are some frame
independent quantities?
Mass, time, temperature…
Example: Two kids decide to race. Both kids walk with speed vw. One kid (A)
will walk on the ground while the other (B) will walk on the “moving sidewalk”
that moves with speed v0. The race is roundtrip. Which kid wins the race?
A) Kid A.
B) Kid B.
C) Tie. D) Depends on the ratio vw/v0.
E)Depends on the sign of v0
Let d = length of
d
tA  2
the moving sidewalk.
vw
Time for roundtrip, kid B: tB  tagainst SW  twith SW
Time for roundtrip, kid A:
vkid B relative to ground  vw  v 0
vkid B relative to ground  vw  v 0






d
d
2
v
d
1
1
2
v
d
w
w



tB 

 2
 tA
 2
2
2
vw  v 0 vw  v 0 vw  v 0
vw  v 0 
 v 02
1 v2 
1 v2
w 

w

1
If v 0  vw , then
 1, so tA  tB (answer A)
2
v
1  02
vw






Example: A boat can make it move at 5 m/s relative to the water and
is trying to go across a 100-m wide river to a point on the opposite
shore and right North of its starting position. The river flows due
West at 3 m/s. How long does the trip take?
N
A. 20 s
v bw
W
v bg
B. 25 s
C. 33 s
x g  100m
vbw  5 m /s
vwg  3 m /s
vbg  ?
t  ?
E
S
v wg



vbg  vbw  vwg
vbg 
t 
vbw 2  vwg 2
x g
vbg

100m

 25s
4m / s
5m / s 2  3m / s 2
 4m / s