00:04
For problem 2014, we're asked to find the tangential velocity and the tangential acceleration on the rim of a spinning wheel at point c.
00:17
And as you can see here, point c is along, lies on the x -axis in our coordinate system.
00:26
The wheel spins with a rate of rotation at 10 radiance per second, and that rate of rotation increases at a rate of six radians per second per second.
00:42
The spinning wheel also precesses about the z -axis at a rate of rotation of 12 radians per second, and that rate of procession increases at a rate of six radians per second per second.
01:02
In order to find the tangential velocity, rather, we will find it in terms of the cross product of the total angular velocity with the position vector from the origin of our coordinate system to point c.
01:25
That position vector is equal to 0 .15 times unit vector i meters.
01:32
And the total angular velocity is the sum of two components, the angular velocity about the axis of procession, the x -axis, plus the angular velocity of the disk about the spin axis, which runs that spin axis is along the y -axis.
01:58
So first, finding the total angular velocity, we will add 10 radiance per second in the y direction, unit vector j, plus 12 radiance per second in the k direction, that is, or the z direction rather, that's unit vector k.
02:19
Our vector has units of radiance per second and plugging in both our total angular velocity and our position vector into matrix determinant, tool that we use to find the cross product, we end up with 1 .8 times unit vector j minus 1 .5 times unit vector k meters per second.
02:48
We are also asked to find the angular acceleration, or we are, that is the tangential acceleration in this problem.
02:57
And in order to find that quantity, we first need to find the angular acceleration.
03:02
That's vector alpha.
03:04
We can express that in terms of the time rate change of the total angular velocity, that's vector omega, and breaking that down further, that vector omega dot is equal to the components of the angular velocity, each of which has undergone the time derivative operation, the vector omega sub s dot plus the vector omega of sub p.
03:35
We need to find each of those in order to find the total angular acceleration.
03:42
The first component is our vector omega -s dot.
03:47
To find this quantity, we choose a rotating reference frame for it, for which that angular velocity in the rotating frame stays constant direction.
04:03
If we choose the precessing, the frame that matches the rate of procession, we'll find that the spin axis, omega sub s, stays constant with reference to that frame.
04:16
So that is we're setting our capital omega equal to our rate of rotation of pre -session along the z axis.
04:26
That is equal to, what do we find it here? oh, yes.
04:42
Okay.
04:43
That is equal to, where is it? here it is.
04:55
There it is...