In the position shown, bar D E has a constant angular velocity of 10 rad/s clockwise. Knowing th

In the position shown, bar D E has a constant angular...

Key Concepts

Angular Velocity

Angular velocity is a measure of the rate of rotation of a rigid body around a fixed axis, typically expressed in radians per second (rad/s). It quantifies how quickly an angle changes in a rotational motion and is a fundamental parameter in the analysis of rotating systems.

Linear and Angular Velocity Relationship

This concept describes the relationship between the linear velocity of a point on a rotating body and its angular velocity, given by the relation v = r?, where r is the distance from the axis of rotation. This relationship is crucial for determining the speeds of various points in a rotating mechanism.

Rigid Body Kinematics

Rigid body kinematics studies the motion of systems where the geometry remains unchanged during motion. It incorporates both the rotational and translational components of movement and provides the foundational principles for analyzing mechanisms as interconnected assemblies of rigid elements.

Relative Motion Analysis

Relative motion analysis involves evaluating the movement of one part of a system in relation to another. In mechanisms with interconnected links, this method helps in determining the velocity and acceleration of various components by considering their motion relative to each other.

Instantaneous Center of Rotation

The instantaneous center of rotation (ICR) is the specific point in a rigid body or mechanism that, at a given instant, has zero velocity relative to a fixed reference frame. Finding the ICR simplifies kinematic analysis by allowing the transformation of complex motion into an equivalent pure rotation about this point.

Transcript

0:00 Hi.

00:02 So let us consider a fixed axis at point e and from point e we have a rod de attached to it and then from point d we have another rod at a different angle attached and at point b we have another rod go slightly more below d.

00:32 So it's more like this.

00:37 This axis a is also fixed.

00:40 And then the end of this rod is at point f.

00:43 Now, the various lengths, vertically, d .e is 200.

00:50 Horizontally, this is 100.

00:54 Then from d to b, it's 100.

00:58 And then from a to e, this is 100.

01:05 All these lengths are in millimeters.

01:09 Then, sorry, d2e already gave as 100.

01:18 So that's fine.

01:19 Then a to e is 300.

01:23 So point a to point e is 300.

01:33 Sorry, actually not up to all the weight up to point e.

01:36 Only from point a to point d is 300.

01:42 Then beyond that, we have 100 extra up to point e.

01:46 Then from point f to point a, it is h and in this problem, h has been given to be 500.

01:58 The next problem is based on the same geometry and there h is just left as a variable.

02:04 But anyways, here h is given to be 500.

02:07 Then we are given omega -de is 10 radiance per second clockwise.

02:14 And we are asked, what is the angular velocity of a rod, fbd and what is what is the velocity of point f yeah that's what is asked okay so now notice that a and e are fixed points so rod ab and rod d e are going to execute perfect rotational motions so the complex motion which is a combination of translation plus rotation will be performed by only rod fbd and here at point d we have this circle and the velocity is going to point at 90 degrees to this circle.

03:07 And i don't want to make it look perpendicular to the rod because it's not given.

03:13 So let me draw it slightly different.

03:17 So this is the circle.

03:24 Yeah, so most of these problems are basically geometry problems and trigonometry problems.

03:30 Physics is just breaking the motion into translation plus rotation.

03:33 So this velocity at point d is going to point perpendicular.

03:37 To the radius de and since the motion is given to be clockwise the direction is going to be like this and this will be the net velocity at point d so if we break the motion about point d like for example for rotational motion this is translational and for rotational motion if we keep the point d fixed then in the net velocity diagram d is going to have the velocity which is intended along tangent to the circle at point d, the circle with radius de.

04:21 So in that case, all the points are going to share the same translation of velocity.

04:26 So point f and point b are also going to have the same velocity, which is at point d.

04:34 Now for rotational velocity, we have to assume some sort of omega.

04:41 So since d is moving clockwise, we can assume any omega, but i think let's start with anticlockwise.

04:50 And let's see if we get a negative number then this would be clockwise but let's start with anticlockwise as an option and then d is fixed at point b let me write r here so that doesn't confuse so at point b the velocity is going to point perpendicular to the rod and at point f also it's going to point perpendicular to the rod now we need to figure out all the angles properly so to figure all the angles this is the right angle triangle in de and if this angle is theta then this angle is 90 degree minus theta sorry this angle is 90 degree so yeah so if this is point d this is horizontal this is vertical so if this angle is theta then this angle is 90 degree minus theta then this so what is this theta? so this theta, we can look at this right angle triangle.

06:11 So this is 100 and this is 200.

06:22 So basically this angle here, theta.

06:27 And this is the right angle triangle we are talking about.

06:30 So this is 200 here, this part is 100.

06:35 So in that case, the hypotenuse, this part is 100.

06:41 Of this right angle triangle let's actually draw the triangle separately so that it doesn't get all messy here so so this is our triangle d e theta this is 100 this is 200 so this will be square root of 100 plus 200 so this will be basically square root of 50 000 or you can think of it as 100 being common and then squared root of 2 plus 1, sorry, 2 squared plus 1 squared by pythagos theorem, that is 4 plus 1, that is 5.

07:18 So 100 root 5.

07:20 So, cos theta is going to be base upon hypotenuse, so that will be 2 over root 5, and sine theta will be perpendicular over hypotenuse, that will be 1 over root 5.

07:35 So these are the cost theta and sine theta for this diagram.

07:39 Now in this diagram what is the angle rod is making so we need to figure out this angle let's call it 5 so let's look at these triangles here now so i think it's easier to figure out rather than this angle this angle 5 so if this angle is 5 this angle is going to be 90 degree minus 5 and this angle is going to be 5 so if this angle is 5 is 5 then it's basically the same angle as in this triangle so let's draw that triangle so in this triangle if this is phi you can see that the perpendicular is hundred the base is 300 this is point b this is point t and then this is going to be square root of hundred squared plus 300 squared so hundred is common then three squared plus one squared is 9 plus 1 ten so here here, cosine theta is 3 over square root 10, and sine theta is 1 over square root 10, right? okay, so let us break the velocities into different components now.

09:18 So, just to make sure we get everything correctly, this is point e.

09:24 So, okay, so now we break the velocities into components, and at point b, we notice that b is moving in a circle with radius ab.

09:36 So the velocity at point b has to point perpendicular to this radius vector, ab.

09:43 So it has to be perfectly horizontal.

09:45 So at b, we have no vertical component...

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