00:04
For problem 20 -4, like problem 20 -3, we have the ladder of a fire engine, 40 foot long that is incline 30 degrees above an xy plane.
00:21
That's the rotation plane of the ladder that coincide or the rotation frame that coincides with the fixed frame and the xyz coordinates.
00:34
The latter swivels about two axes, one pointed in the positive x direction, the other in the positive z direction.
00:49
And at this particular instant, it swiveles about the z axis with an angular velocity of 0 .15 radiance per second, with an instantaneous increase of the angular velocity of 0 .2 radiance per second per second.
01:16
And likewise, about the positive x -axis, the instantaneous angular velocity is 0 .6 radiance per second.
01:26
The time rate change of that angular velocity in this rotating reference frame is 0 .4 .2.
01:35
Point four radiance per second per second.
01:38
The position of point a as will symbolize by vector r, like in problem 20 -3 -3 -2 -2 -victor i plus 20 unit vector j feet.
01:58
And this is a position vector, it has units of distance.
02:04
We also need to find the combined angular velocity vector for the composite rotations, omega 1 and omega 2.
02:16
Because they are angular velocities, that is an infinitesimally small, the limit of an infinitesimally small rotation over an infinitesimally small amount of time means that we can add two rotation operations together.
02:46
And that's how we get away with adding two angular velocities when ordinarily angular rotations are not commutative, which means they can't be added.
02:59
But that is the case for finite rotations.
03:03
Since we're working with infinitesimals and the definition of angular velocities, we can.
03:08
We combine the two by summing omega -1 and omega -2.
03:14
Which comes out to be 0 .6x component plus 0 .15z component radiance per second.
03:24
With both the position vector and the angular velocity, we can take their cross product to find the tangential velocity of point a.
03:34
Taking this cross product, which was done in depth in the solution to 20 -3, our result is negative 5 .2 unit vector i minus 12 unit vector j plus 20 .8 unit vector k.
03:54
Each of those components have linear velocity units of feet per second.
04:00
The second part of the problem asks us to find the tangential acceleration of the ladder at point a.
04:09
And this requires several steps in order to do this.
04:14
But as an overview, we'll recall that the tangential acceleration at a point given by position vector r is the sum of two cross products for a particular set of circumstances here.
04:32
That is, in the first cross product, we have angular acceleration now cross the position vector.
04:40
The point in the rotating frame, a distance r from the origin.
04:51
And for our second term, we have our angular velocity crossed with our tangential velocity.
05:06
So we'll work on the first term first here.
05:10
This angular acceleration is the sum of, of the time rate change of the angular velocity omega -1, and that's a vector, so it can change both in terms of magnitude and direction with respect to time, plus the time rate change of vector omega -2.
05:36
Both of those sum together will be our total angular acceleration.
05:40
We have to find what these are for each.
05:43
To do that, we'll use the rotation between frames, transformation equation...