Lecture 2: Exercise 8
Explanation
This is a straightforward calculation coupled with a visualization. The goal here is to see the relationship between position vectors, velocity
vectors, and acceleration vectors.
Hint
Recall the definitions of velocity and acceleration.
Answer
Ÿ r = Icos Ω t, eΩ t M
®
The first task is to calculate the derivative of each component with respect to time, thus giving us the velocity vector:
®
®
dr
v=
d
=
dt
We then calculate the acceleration vector
dt
Icos Ω t, eΩ t M =
d
d
cos Ω t,
dt
dt
eΩ t = I-Ω sin Ω t, Ω eΩ t M.
2
l2e8.cdf
®
®
dv
a=
d
=
dt
d
= -
ry
I-Ω sin Ω t, Ω eΩ t M
d
Ω eΩ t
Ω sin Ω t,
dt
Position Vector
dt
dt
= I-Ω2 cos Ω t, Ω2 eΩ t M.
500
400
Out[13]=
300
200
100
-1.0
-0.5
0.5
1.0
0.5
1.0
0.5
1.0
rx
Velocity Vector
vy
500
400
Out[14]=
300
200
100
-1.0
-0.5
vx
Acceleration Vector
ay
500
400
Out[15]=
300
200
100
-1.0
-0.5
ax
If you are viewing this with the CDF Player, you can manipulate the value of Ω by moving the slider—watch the vertical axis scale change as
you increase Ω.
l2e8.cdf
Ω
Position Vector
ry
500
400
Out[18]=
300
200
100
-1.0
-0.5
0.5
1.0
0.5
1.0
rx
Ω
Velocity Vector
vy
500
400
Out[19]=
300
200
100
-1.0
-0.5
vx
3
4
l2e8.cdf
Ω
Acceleration Vector
ay
500
400
Out[20]=
300
200
100
-1.0
-0.5
0.5
1.0
Ÿ r = Hcos HΩ t - ΦL, sin HΩ t - ΦLL
ax
®
The first task is to calculate the derivative of each component with respect to time, thus giving us the velocity vector:
®
dr
®
v=
d
=
@cos HΩ t - ΦL, sin HΩ t - ΦLD = B
dt dt
We use the trigonometric identities from Interlude 1:
d
dt
cos HΩ t - ΦL,
d
dt
sin HΩ t - ΦLF
sinHΑ - ΒL = sin Α cos Β - cos Α sin Β,
cosHΑ - ΒL = cos Α cos Β - sin Α sin Β.
Then we have:
v =B
d
®
dt
d
Hcos Ω t cos Φ - sin Ω t sin ΦL,
d
=
cos Ω t cos Φ dt
dt
d
sin Ω t sin Φ,
dt
Hsin Ω t cos Φ - cos Ω t sin ΦLF
d
d
sin Ω t cos Φ dt
cos Ω t sin Φ
dt
We then apply the product rule:
®
v
d
®
d
v = cos Φ
cos Ω t + cos Ω t
dt
d
cos Φ - sin Φ
dt
d
d
cos Φ
sin Ω t + sin Ω t
dt
d
sin Ω t - sin Ω t
dt
d
cos Φ - sin Φ
dt
sin Φ,
dt
d
cos Ω t - cos Ω t
dt
sin Φ
dt
Now, cos Φ is a constant,
d
®
v = cos Φ
d
cos Ω t - sin Φ
dt
d
sin Ω t , cos Φ
dt
d
sin Ω t - sin Φ
dt
cos Ω t =
dt
H-Ω cos Φ sin Ω t - Ω sin Φ cos Ω t , Ω cos Φ cos Ω t + Ω sin Φ sin Ω tL
We now reverse the trigonometric identities:
v = @-Ω sin HΩ t - ΦL, Ω cosHΩ t - ΦLD
®
We then calculate the acceleration vector
l2e8.cdf
®
dv
®
a=
@-Ω sin HΩ t - ΦL, Ω cosHΩ t - ΦLD = B-
d
=
dt dt
we then apply the trigonometric identites:
a = B-Ω
Hsin Ω t cos Φ - cos Ω t sin ΦL, Ω
d
®
dt
d
= -Ω
d
sin Ω t cos Φ + Ω
dt
d
dt
d
dt
d
cos Ω t sin Φ, Ω
dt
Ω sin HΩ t - ΦL,
d
Ω cosHΩ t - ΦLF
dt
Hcos Ω t cos Φ - sin Ω t sin ΦL,F
d
cos Ω t cos Φ - Ω
dt
sin Ω t sin Φ
dt
and then we apply the product rule
d
®
a = -Ω sin Ω t
d
d
cos Φ - Ω cos Φ
dt
sin Ω t + Ω cos Ω t
dt
d
dt
d
Ω cos Ω t
d
cos Φ + Ω cos Φ
dt
cos Ω t - Ω sin Ω t
dt
d
dt
d
dt
sin Ω t + Ω sin Φ
sin Ω t
dt
d
cos Ω t , Ω cos Φ
dt
cos Ω t ,
dt
sin Φ - Ω sin Φ
d
= -Ω cos Φ
d
sin Φ + Ω sin Φ
d
cos Ω t - Ω sin Φ
dt
sin Ω t
dt
= I-Ω2 cos Φ cos Ω t - Ω2 sin Φ sin Ω t , -Ω2 cos Φ sin Ω t - Ω2 sin Φ cos Ω t M
we reverse the trigonometric identities
a = A-Ω2 cosHΩ t - ΦL, -Ω2 sinHΩ t - ΦLE.
®
Position Vector
ry
1.0
0.5
Out[32]=
-1.0
-0.5
0.5
-0.5
-1.0
1.0
rx
5
6
l2e8.cdf
Velocity Vector
vy
1.0
0.5
Out[36]=
-1.0
-0.5
0.5
1.0
0.5
1.0
vx
-0.5
-1.0
Acceleration Vector
ay
1.0
0.5
Out[40]=
-1.0
-0.5
ax
-0.5
-1.0
If you are viewing this with the CDF Player, you can manipulate the value of Ω by moving the slider—watch the vertical axis scale change as
you increase Ω and/or Φ,
l2e8.cdf
Ω
Φ
Position Vector
ry
1.0
0.5
Out[34]=
-1.0
-0.5
0.5
-0.5
-1.0
1.0
rx
7
8
l2e8.cdf
Ω
Φ
Velocity Vector
vy
1.0
0.5
Out[38]=
-1.0
-0.5
0.5
-0.5
-1.0
1.0
vx
l2e8.cdf
Ω
Φ
Acceleration Vector
ay
1.0
0.5
Out[42]=
-1.0
-0.5
0.5
1.0
ax
-0.5
-1.0
Ÿ r = Ic cos3 t, c sin3 tM
®
The first task is to calculate the derivative of each component with respect to time, thus giving us the velocity vector:
®
®
dr
v=
Ic cos3 t, c sin3 tM =
d
=
dt
dt
d
d
c cos3 t,
dt
c sin3 t .
dt
The easiest way to do this is by the chain rule, where we choose the variables g = cos t, and h = sin t. This gives us,
d
dg
where
Ic g3 M = 3 c g2 ;
d
dh
dg
Ic h3 M = 3 c h2
dh
= -sin t;
dt
= cos t
dt
so,
®
d
v=
dg
Ic g3 M
dg
d
,
dt
dh
Ic h3 M
dh
= 3 c g2
dt
dg
, 3 c h2
dt
dt
We then calculate the acceleration vector in a similar way,
®
®
dv
a=
where,
d
=
dt
dh
dt
= I-3 c g2 sin t, 3 c h2 cos tM = I-3 c cos2 t sin t, 3 c sin2 t cos tM
I-3 c cos2 t sin t, 3 c sin2 t cos tM
9
10
l2e8.cdf
d
dg
and,
®
d
a=
dg
I-3 c g2 hM = -6 c g h;
I-3 c g2 hM
dg
d
,
dt
dh
I3 c h2 gM
d
dh
I3 c h2 gM = 6 c h g
dh
dg
= -6 c g h
dt
dh
=
, 6chg
dt
dt
H6 c g h sin t, 6 c h gcos tL = H6 c cos t sin t sin t, 6 c sin t cos t cos tL = I6 c cos t sin2 t , 6 c sin t cos2 t M
Position Vector
ry
1.0
0.5
Out[43]=
-1.0
-0.5
0.5
1.0
rx
-0.5
-1.0
Velocity Vector
vy
1.0
0.5
Out[45]=
-1.0
-0.5
0.5
-0.5
-1.0
1.0
vx
l2e8.cdf
11
Acceleration Vector
ay
2
1
Out[48]=
-2
-1
1
2
ax
-1
-2
If you are viewing this with the CDF Player, you can manipulate the value of Ω by moving the slider—watch the vertical axis scale change as
you increase c,
c
Position Vector
ry
1.0
0.5
Out[44]=
-1.0
-0.5
0.5
-0.5
-1.0
1.0
rx
12
l2e8.cdf
c
Velocity Vector
vy
1.0
0.5
Out[46]=
-1.0
-0.5
0.5
-0.5
-1.0
1.0
vx
l2e8.cdf
c
Acceleration Vector
ay
2
1
Out[49]=
-2
-1
1
ax
2
-1
-2
Ÿ r = Hc Ht - sin tL, c H1 - cos tLL
®
The first task is to calculate the derivative of each component with respect to time, thus giving us the velocity vector:
®
®
dr
v=
dt
d
d
ct-
dt
d
=
dt
d
d
c -
c sin t,
dt
@c Ht - sin tL, c H1 - cos tLD =
dt
d
c cos t = c
dt
d
dt
d
t-c
dt
c Ht - sin tL,
d
dt
d
sin t, 0 - c
dt
dt
We then calculate the acceleration vector in a similar way,
®
dv
®
a=
d
=
dt
dt
Hc - c cos t, c sin tL =
d
d
c-
dt
d
dt
d
c sin t =
c cos t,
dt
Position Vector
ry
2.0
Out[51]=
1.5
1.0
0.5
2
4
6
8
10
cos t = Hc - c cos t, c sin tL.
d
c-c
dt
rx
c H1 - cos tL =
d
cos t, c
dt
dt
sin t = Hc sin t, c cos tL.
13
14
l2e8.cdf
Velocity Vector
vy
1.0
0.5
Out[54]=
0.5
1.0
1.5
2.0
vx
-0.5
-1.0
Acceleration Vector
ay
1.0
0.5
Out[56]=
-1.0
-0.5
0.5
1.0
ax
-0.5
-1.0
If you are viewing this with the CDF Player, you can manipulate the value of Ω by moving the slider—watch the vertical axis scale change as
you increase c,
l2e8.cdf
c
Position Vector
ry
Out[52]=
2.0
1.5
1.0
0.5
2
4
6
8
rx
10
c
Velocity Vector
vy
1.0
0.5
Out[55]=
0.5
-0.5
-1.0
1.0
1.5
2.0
vx
15
16
l2e8.cdf
c
Acceleration Vector
ay
1.0
0.5
Out[58]=
-1.0
-0.5
0.5
-0.5
-1.0
1.0
ax