1
00:00:03,620 --> 00:00:06,560
In kinematics, if we
know acceleration,

2
00:00:06,560 --> 00:00:09,220
which we generally can get
from Newton's second law,

3
00:00:09,220 --> 00:00:11,160
then we'd like to
integrate the acceleration

4
00:00:11,160 --> 00:00:13,250
to get velocity and position.

5
00:00:13,250 --> 00:00:16,100
How does that work
for circular motion?

6
00:00:16,100 --> 00:00:18,810
Well, when you think
about circular motion

7
00:00:18,810 --> 00:00:23,280
and you think about the
tangential direction,

8
00:00:23,280 --> 00:00:27,730
then that, in some sense,
is a one-dimensional motion.

9
00:00:27,730 --> 00:00:30,830
And so what we'll see is that
the tangential description

10
00:00:30,830 --> 00:00:34,990
of this motion can be integrated
to get the tangential velocity

11
00:00:34,990 --> 00:00:38,190
and position, just
in a similar way

12
00:00:38,190 --> 00:00:40,400
that we did with
linear motion, where

13
00:00:40,400 --> 00:00:42,030
if we were given
the acceleration,

14
00:00:42,030 --> 00:00:44,890
we integrated
velocity and position.

15
00:00:44,890 --> 00:00:47,980
So let's look at a
particular example.

16
00:00:47,980 --> 00:00:50,790
Suppose we're given that
the tangential component

17
00:00:50,790 --> 00:00:55,210
of the acceleration is given
by some simple polynomial, A

18
00:00:55,210 --> 00:00:57,230
minus Bt.

19
00:00:57,230 --> 00:00:59,140
Now, in this case,
the acceleration

20
00:00:59,140 --> 00:01:01,970
is certainly non-constant.

21
00:01:01,970 --> 00:01:03,860
What we'd like to
do-- remember, this

22
00:01:03,860 --> 00:01:09,090
is equal to rd squared
theta dt squared.

23
00:01:09,090 --> 00:01:13,180
So if we integrate the
tangential component

24
00:01:13,180 --> 00:01:16,960
of the acceleration
with some variable

25
00:01:16,960 --> 00:01:20,730
from t prime going from
some initial value, which

26
00:01:20,730 --> 00:01:23,880
we can call 0, to
some final value,

27
00:01:23,880 --> 00:01:27,480
then that will
give us the change

28
00:01:27,480 --> 00:01:30,600
in the tangential velocity.

29
00:01:30,600 --> 00:01:34,550
So for this case,
V theta of t is

30
00:01:34,550 --> 00:01:42,660
equal to V theta t0 plus the
integral of A minus B t prime

31
00:01:42,660 --> 00:01:43,890
d t prime.

32
00:01:43,890 --> 00:01:47,950
Now, we needed an integration
variable because recall

33
00:01:47,950 --> 00:01:50,660
that our functional
dependence is the upper limit

34
00:01:50,660 --> 00:01:51,970
of the integral.

35
00:01:51,970 --> 00:01:54,060
We saw that in
one-dimensional kinematics.

36
00:01:54,060 --> 00:01:59,270
So we're going from some
t0 0 to some prime t prime.

37
00:01:59,270 --> 00:02:01,110
Well, this integral
is now-- we'll

38
00:02:01,110 --> 00:02:07,610
write this V theta 0, where,
again, t0 we're calling 0.

39
00:02:07,610 --> 00:02:11,950
And when we do this integral,
A minus B t prime and d

40
00:02:11,950 --> 00:02:17,780
t prime-- that's an integral
from 0 to t prime equals t.

41
00:02:17,780 --> 00:02:18,760
t prime equals 0.

42
00:02:18,760 --> 00:02:20,950
This integral is not hard to do.

43
00:02:20,950 --> 00:02:27,020
We get A of t minus
1/2 B of t squared.

44
00:02:27,020 --> 00:02:30,829
And there's that
constant initial term.

45
00:02:30,829 --> 00:02:36,490
And so the velocity at time
t of the tangential component

46
00:02:36,490 --> 00:02:39,930
of the velocity is given
by that expression.

47
00:02:39,930 --> 00:02:44,390
In exactly the same fashion,
the angle-- how much angle

48
00:02:44,390 --> 00:02:45,990
does this go through?

49
00:02:45,990 --> 00:02:50,640
Well, we have to
be careful there.

50
00:02:50,640 --> 00:02:55,520
So if we want to do
the arc length, s of t,

51
00:02:55,520 --> 00:02:58,890
we have to multiply the
angle by the radius.

52
00:02:58,890 --> 00:03:03,020
And that is just the integral
of the tangential component

53
00:03:03,020 --> 00:03:09,650
of the velocity from t prime
equal 0 to time t prime t.

54
00:03:09,650 --> 00:03:12,930
And so in this example, we
have a slightly complicated

55
00:03:12,930 --> 00:03:18,390
integral, V0 d t prime, 0 to t.

56
00:03:18,390 --> 00:03:20,300
I'm going to drop--
well, I'll write it

57
00:03:20,300 --> 00:03:27,290
in-- t prime plus the
integrals A t prime minus 1/2

58
00:03:27,290 --> 00:03:29,550
B t prime squared.

59
00:03:29,550 --> 00:03:32,450
I didn't need to split this up.

60
00:03:32,450 --> 00:03:33,829
But I'm splitting it up, anyway.

61
00:03:36,550 --> 00:03:41,420
And I have really one,
two, three integrals to do.

62
00:03:41,420 --> 00:03:47,300
And what I get is V
theta 0 t-- the result

63
00:03:47,300 --> 00:03:48,520
of the first integral.

64
00:03:48,520 --> 00:03:53,300
The result of the second
interval is 1/2 At squared.

65
00:03:53,300 --> 00:03:55,090
And the result of
the third integral

66
00:03:55,090 --> 00:04:00,000
is minus 1/6 B t cubed.

67
00:04:00,000 --> 00:04:02,230
And that gives me this.

68
00:04:02,230 --> 00:04:05,370
I have to be a little
careful on this side.

69
00:04:05,370 --> 00:04:14,570
It's how much has the arc length
changed as we go from time 0

70
00:04:14,570 --> 00:04:16,060
to time t.

71
00:04:16,060 --> 00:04:20,034
So that's the change
in arc length.

72
00:04:23,100 --> 00:04:25,850
And that's a typical
example of where we're

73
00:04:25,850 --> 00:04:28,390
given tangential acceleration.

74
00:04:28,390 --> 00:04:31,530
We integrate it to get
the tangential velocity.

75
00:04:31,530 --> 00:04:34,140
And then when we integrate
the tangential velocity,

76
00:04:34,140 --> 00:04:39,470
we're getting the change in
arc length around the circle.