00:01
Hi, let us look at another problem.
00:03
So now here we have a fixed axis at point a and we have a rod ab and from that rod ab we have another rod hanging which is rod bd and then from point d we have another rod d.
00:29
We have another rod de and now de is also fixed.
00:36
So this is also fixed to rotate so this is also fixed now what are the various lengths so a b is 175 millimeters bd is 200 millimeters then from point d to e the vertical distance is 75 millimeters and then from a to e the distance is 200 millimeters, sorry 100 millimeters only that means the total distance from d to e is 100 plus 175 that is 275 millimeters okay so we are given these values and then we are given that omega ab is four radiance per second clockwise so it's moving like this at a given instant.
01:46
Then we are asked what is the velocity of angular velocity of rod bd and angular velocity of rod d.
01:55
Okay to solve any of these kind of problems in this whole chapter we have been following this rule break the motion into translation and rotation and since rods de and ab are fixed they are going to execute perfect rotational motions so they will not have any translational component but rod b d is going to to execute a rigid body motion.
02:18
So it will have both translation and rotation.
02:20
So point b, it's part of a circle fixed axis at a and since ab is moving with a clockwise velocity of four radiance per second, the point b at this instant is going to have velocity pointing perpendicular to this radius vector ab and it's going to point up because of this omega and the value will omega ab times the length ab so so this will be four radiance per second times 175 millimeters so how much will this be so four fives are 20 four sevens are 28 plus 230 then four one is four plus seven plus three is seven so 700 millimeters per second so this is 700 millimeters per second and then if this is the translational velocity at point b, then at all points, even at point d, so let's call this vt.
03:30
At all points, it's going to have the same velocity.
03:33
So, same translational velocity.
03:37
So vt at d also will be the same throughout this rod.
03:42
That's the definition of rigid body.
03:44
Translational velocity is the same at all the points.
03:47
Then it will have some sort of rotational velocity as well.
03:50
Let us assume some omega and then as i always keep emphasizing that even if you assume omega which is not in the correct direction you will get a negative answer which means negative of incorrect direction equals correct direction so eventually we'll get the right answer.
04:07
So let us assume everything to be clockwise so this is clockwise so let us assume this to be clockwise and if we assume this to be clockwise then actually yeah so i always do that but let's today let me try to guess and see if i'm correct.
04:27
So a v is going to move clockwise.
04:29
So from the geometry it seems to me that d is going to move to the left and b is going to move up.
04:38
So this tells me that bd is actually going to probably move anti -clockwise and d is going to move clockwise.
04:46
So i'll assume like that and then we'll see whether we get the correct answer.
04:49
So let's assume omega bd to be like this and we assume b to be fixed here and motion is about point b.
04:57
Then what is the velocity of point b .d? if omega bd is like this, is going to point like this perpendicular to the vector bd and pointing to the right because of this omega, assumed omega.
05:11
So this velocity is going to be omega bd, lbd.
05:18
Now this is going to give you the net motion and the net motion now at point b since this is zero, this is translation of velocity, so at point b is the same, as i said, it's moving in a circle.
05:31
So it's constrained to, sorry, this circle.
05:36
So it's constrained to move perpendicular to the radius vector ab.
05:42
So it's going to point up, that's fine, and this is going to be 700 millimeters per second.
05:48
This is a net velocity.
05:50
And then at point d, since it's moving in a circle at this angle, a larger circle, let us see what d is going to do.
06:01
So if d is moving left, i think the angular velocity here is also going to be anti -clock.
06:08
Sorry, angular velocity here is going to be clockwise.
06:11
Yeah.
06:12
So if this is clockwise, then the velocity at point d is going to point perpendicular like this to the circle.
06:24
And the value will be omega -d -e times whatever the length of this rock.
06:31
So l no sorry l de this time this will be l de because we are looking at the circle of this radius omega de lde this is the net motion at point d okay now we have everything sorted out t plus r equals net motion so let's just break it down into components and equate the components so let us figure out angles so which angle is easiest to figure out so let's look in this diagram this angle is i think easiest to figure out because if this is the right angle triangle uh then if this is theta this is going to be theta so the angle from rod d angle of rod d from horizontal is going to be theta if this is theta then this is going to be perpendicular to the circle so this is going to be 90 minus theta so this is going to be theta this is 90 minus theta okay so now the vertical component this component is going to be omega de lde cosine theta and the horizontal component i'm drawing it a bit too large because i just want to have space to write it down this will be omega de lde sine theta okay so i think we have got it correct because you can see that the direction wise we have got all the correct directions so because this is pointing towards the right, the net velocity is to the right.
08:15
And vt is up and the net velocity is also up...