Knowing that at the instant shown the angular velocity of rod A B is 15 nd / s clockwise, determ

Knowing that at the instant shown the angular velocity of...

Key Concepts

Linear and Angular Velocity Relationship

This concept establishes that the linear velocity of any point on a rotating rigid body is directly proportional to its radial distance from the axis of rotation and the angular velocity of the body. The formula v = r? is widely used in analyzing the motion of objects in mechanical systems.

Relative Velocity Analysis

Relative velocity analysis involves examining the speeds and directions of points on a rigid body relative to one another. This approach is particularly useful for systems with interconnected parts, allowing the determination of unknown quantities such as the angular velocity or the linear velocity of specific points through vector relationships.

Rigid Body Rotation

Rigid body rotation refers to the movement of a body in which all points remain at fixed distances relative to one another as the body rotates about an axis. This principle is essential when analyzing the motion of objects like rods, ensuring that the rotations and translations of different parts of the body are consistent with its overall rotation.

Angular Velocity

Angular velocity quantifies the rate at which an object rotates around a fixed axis. It is crucial in relating the rotational motion to the translational motion of any point on the object, where the linear speed of a point is the product of its distance from the rotation axis and the angular velocity.

Transcript

00:01 Hi, let us look at a situation where we have a fixed axis and from that fixed axis we have a small rod ab hanging vertically and there is another rod bd which is attached with these pins and then we have rod d where point e is again fixed to rotate about a fixed axis.

00:29 Now the various lengths are ab is to millimeters bd is 250 millimeters b e is 200 millimeters b e is 200 millimeters and d a horizontally 600 millimeters okay now if a b is moving clockwise with a velocity 15 angular velocity 15 radiance per second then we are asked what is the angular velocity of bd and let's say if c is a midpoint then what is the velocity of point c, where we know that bc is equal to cd.

01:17 Okay, now, so to solve this problem or any of the problem in this chapter, we want to break the motion in translation and rotation and look at the net motion.

01:29 So we know that a and e are fixed points, so the rod a, b and b are going to move in a circular motion, and that motion is fixed.

01:39 We can't do anything about it.

01:40 So the motion of point b, the motion of point b will be governed by a translational velocity.

01:47 That will be the net motion of point b.

01:50 And since it's moving in a circle, a clockwise velocity, this will be the radial velocity omega -cross -r, and it will point like this, perpendicular to the radius vector, obviously, joining a and b.

02:04 Since it's vertical, this is going to be perfectly horizontal.

02:08 And the magnitude will be omega times r which will be 15 times 15 radiance per second times 200 millimeters so this will be 3 ,000 millimeters per second or 3 meters per second now this if this is translational motion then all the points are going to have the same velocity so this will have vt if this is if we call this translation velocity vt then point c we also have this vt now this then it needs to have some rotational velocity.

02:41 And for rotational velocity, we assume b to be fixed.

02:44 Because in the net motion, we know that the net motion of b is just going to be in this circle.

02:49 So it's going to be like this.

02:51 Now, the net motion about point d, now, we know that d is going to move in a circle where the radius vector is actually pointing horizontal.

03:01 So now it has to be perfectly vertical.

03:07 So we can assume any given direction to solve the problem.

03:11 So let us assume that the omega is clockwise for all of them.

03:19 So i'll show you that obviously this is a weird assumption to assume all of them clockwise.

03:29 But what i want to show is that no matter what omega you assume, as long as you are consistent with the conditions of the problem and you follow your mathematics consistently, then if you're assumed omega is wrong, then you will get negative of the value, which means that negative of assumed omega, which means if your assumed omega was wrong, then negative assumed omega will give you the correct omega, which will be opposite.

03:55 So we'll see how this works out.

03:58 So if this is clockwise, then net velocity at point d is going to point up.

04:04 And it will have no horizontal component because it's executing a circular motion and the radius vector is horizontals and the velocity is always going to be perpendicular to the radius vector.

04:16 Now, what is going to be rotational velocity at point d for a clockwise omega like this? again, it's going to be perpendicular to the radius vector, so it's going to be perpendicular here, and the magnitude will be omega bd times l bd, where lbd is the length between b to d.

04:35 Okay.

04:36 Now, looking at this diagram, we can see that we need to break down this omega lbd into vertical and horizontal components.

04:48 So if this is angle theta, then we look at our original diagram.

04:54 If this is angle theta, then we are at a right angle triangle, b t, and this length is 250.

05:05 This is theta.

05:07 So if this is theta, this is 90 minus theta.

05:12 Then this angle is theta.

05:14 That means the vertical component is omega bd lbd cosine theta and this component is omega bd lbd sine theta okay now so horizontal components need to cancel so omega bd lbd let me just write it slightly more cleanly so omega bd lbd lbd sine theta so omega bd lbd sine theta so we know that we bt plus omega bd lbd sine theta has to be zero because at point d so this is at point d so at point d d horizontal velocity is zero or horizontal component of bd is zero it's perfectly vertical so this has to hold and we know that vt is basically minus three meters per second i it's pointing to the left plus now, omega bd, lbd, sine theta.

06:29 So since we have already given the directions, let's just put the magnitudes here...

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