00:01
In this problem we have a thid rod that is pivoted on the left.
00:04
It's a uniform rod so its force of gravity is at its geometric center, that's where the center of mass is.
00:09
And we're asked to collect a series of questions about this configuration.
00:14
And we'll take each one at a time.
00:16
First it wants to know what's the net torque when it is released from rest in the horizontal position.
00:22
So we already are given the gravitational force acting down at the geometric center.
00:27
At the pivot we're going to have fy and fx.
00:34
These forces keep it from moving in x, moving in y, in whatever manner.
00:41
And our rotation axis is going to go through a.
00:45
That's our pivot point.
00:49
Torque, plus or minus f, are perpendicular.
00:52
F is the magnitude of the force, our perpendicular is the moment our arm is.
00:55
It's the perpendicular distance between the line of action and the rotation axis, or equivalently the shortest distance between the rotation axis and the line of action.
01:03
Plus is if the force acting alone were turned counterclockwise minus clockwise.
01:09
Now let's draw the lines of action.
01:12
So here would be the line of action through fx.
01:17
Here would be a line of action through fy.
01:21
It's just an infinite line.
01:23
Just an infinite line that goes through the force.
01:25
And here is the line of action through the weight.
01:29
Now, so let's write the three forces in here.
01:33
Fx, fy, fg.
01:36
If you're standing on the rotation axis, what's the shortest distance to get to the orange line of action for fx? you're standing on it.
01:47
Zero.
01:50
So that does not contribute rotationally around a.
01:52
What about fy and this line of action? you're standing on it also.
02:02
Zero.
02:06
So fx and fy contribute nothing to rotation around a.
02:10
So that's why we don't worry about them.
02:12
But fg does.
02:14
Now what will be the moment arm? well, it's the shortest distance, perpendicular distance.
02:24
There's the perpendicular, but it's also the shortest.
02:26
Any other line you draw would be longer than that.
02:31
So this is our perpendicular for g, which is l over 2.
02:33
And that will turn it clockwise, negative.
02:45
So we have everything.
02:49
So net torque minus mg over 2 minus 0 .5 kilograms, 9 .8 meters per second squared, 1 meter over 2 minus 2 .45 newton meter.
03:09
My sign just indicates it wants to turn it clockwise.
03:12
That torque wants to turn it clockwise.
03:17
B wants to know the angular acceleration the instant it's released.
03:23
Now we need the moment of inertia about a.
03:26
Now maybe your book has it, maybe you've seen it.
03:29
But i'm assuming you don't, because a lot of times the books don't have this for the end of a rod.
03:34
So to get it, you have to use what is called the parallel axis theorem.
03:37
You've got an axis coming out towards you through the center of mass.
03:40
Another one coming out towards you through a.
03:42
Those are parallel axes.
03:44
Parallel axis theorem.
03:46
So the way you get for the axis through a, you take the moment of inertia through the center of mass axis, plus the mass times the shortest distance, the perpendicular distance between the two parallel axes, which in this case is l over 2.
04:01
So 1 12th ml squared, you get that usually from the book table, plus ml over 2 squared.
04:10
And this is 1 12th ml squared, plus 1 4th ml squared.
04:18
But 1 4th is the same as 3 12ths.
04:20
So 3 plus 1 4th 12th, 1 3rd.
04:25
There is our moment of inertia for the axis through a.
04:30
And now to get the angular acceleration, we use newton's second law in rotational form.
04:36
The net torque is equal to the moment of inertia times the angular acceleration.
04:40
And we have our formulas.
04:46
Minus mg over 2, 1 3rd ml squared alpha.
04:53
M's go away, this power 2 goes away.
04:59
So this gives me that alpha is equal to minus 3g over 2l.
05:05
Gotta multiply both sides by 3 and divide by l.
05:13
So putting our numbers minus 3.
05:15
Again, don't be bothered by a minus sign.
05:16
It just indicates that it wants to, from rest, make it move clockwise, picking up speed in a clockwise direction.
05:25
So minus 3 times 9 .8 meters per second squared over 2 times 1 meter.
05:31
Minus 14 .7 radians per second squared.
05:37
So that's the angular acceleration.
05:40
Now part c wants the tangential acceleration, the magnitude of tangential acceleration at the right end.
05:48
And that's the formula for any point where r is the distance from the rotation axis to get your tangential acceleration.
05:57
In our case, we want it for the right end.
05:59
I'll just put a capital r on it for right.
06:02
And it's going to be alpha l.
06:06
So this is kind of easy.
06:07
14 .7 radians per second squared times 1 .0 meters.
06:18
So 14 .7 meters per second squared.
06:22
Well, it really happened.
06:24
We changed units because it's only 1.
06:26
So that's it.
06:27
That's all there is to it.
06:29
That's the magnitude.
06:30
These are all positive quantities.
06:31
It's the magnitude.
06:32
It's obviously directed straight down from a vector standpoint, if that's...
06:36
But usually you just want the magnitude.
06:41
Now part d wants to know the net torque when you are at an angle of 45 degrees.
06:47
So theta is 45 degrees.
06:50
So the gravitational force always acts l over 2 from the end.
06:56
Straight down.
06:58
So let's put in our line of action.
07:02
So our shortest distance, our perpendicular distance, will be this.
07:07
This will be r perpendicular for g.
07:12
So we got to use a little trigonometry here.
07:15
Cosine theta, jason over hypotenuse.
07:18
R perpendicular g over l over 2.
07:22
So this gives me, at the moment of arm now, l over 2 cosine theta.
07:29
So notice it changed.
07:31
The moment of arm changes.
07:32
That means torque changes.
07:34
The net torque changes, i mean.
07:36
And that means alpha changes...