For the oil pump rig shown, link A B causes the beam B C E to oscillate as the crank O A revolve

For the oil pump rig shown, link A B causes the beam B C E...

Key Concepts

Unit Conversion in Rotational Dynamics

Converting units, such as from revolutions per minute (rpm) to radians per second, is crucial in rotational dynamics. This ensures that all parameters are expressed in consistent units, which is necessary for correctly applying the fundamental relationships and formulas in kinematic equations.

Relative Motion Analysis

Relative motion analysis is the method of determining the motion of a point on a moving body by considering its motion relative to another point whose motion is already known. This technique, which involves vector addition and the concept of the instantaneous center of rotation, is essential for decomposing the overall motion of complex mechanisms into simpler, more manageable components.

Rotational Kinematics

Rotational kinematics involves the study of rotational motion and describes quantities like angular displacement, angular velocity, and angular acceleration. It relates the linear (tangential) velocity and acceleration of points on a rotating body to their distance from the axis of rotation, using formulas such as v = ?r and a = ?²r for centripetal acceleration.

Planar Mechanism Analysis

Planar mechanism analysis focuses on systems of interconnected rigid bodies that move in a two-dimensional plane. This branch of kinematics uses geometric relationships and vector algebra to determine the positions, velocities, and accelerations of various points on the mechanism, providing a structured approach to solving problems involving linkages and moving parts.

Transcript

00:01 For this problem on the topic of kinematics of rigid bodies, we are shown an oil pump rig in the diagram, and we told that the link ab causes the beam bce to oscillate as the crank revolves.

00:17 Now, if we know the radius of the crank oa, and we know its constant clockwise angular velocity, we want to determine the velocity and acceleration of point d at the same instant.

00:31 Now for this problem, we'll use the following unit vectors.

00:40 We'll use unit vector i to be one unit to the right, unit vector j to be one unit upward, and unit vector k to be one unit anti -clockwise.

01:00 So let's look first at the crank oa.

01:05 Now for the crank oa, we have r.

01:10 Oa to be 0 .6 meters.

01:15 The angular velocity omega -o -a is 20 revs per minute clockwise, which we can write as 2 .094 radiance a second clockwise.

01:41 The linear speed with a linear velocity of a -va is equal to omega -o -a -r -o -a, and this is 2 .0 -944 times 0 .6, which gives us 1 .25664 meters per second that's vertically upward.

02:19 The angular acceleration of the crank alpha -o -a is equal to zero, and so the tangential component of its acceleration, a t, is equal to zero.

02:36 The normal component of its acceleration, a, a, subscript n, is equal to omega -squared of the crank, the angular speed squared times r or a.

02:54 And we can calculate this to be 2 .094 squared times 0 .6, which gives us the normal component of the acceleration of a to be 2 .6319 meters per square second.

03:20 And this is to be right.

03:27 Now, let's concern ourselves with the rod ab.

03:36 So the first thing we will do for rod ab is find the velocity vb, and vb is simply the speed va and the direction is vertically upward.

03:56 Now vb and va are parallel.

03:58 So va is equal to vb and the angular speed omega ab is equal to 0 since we have straight line motion.

04:20 Now vb is equal to 1 .25664 meters per second vertically upward.

04:34 And the acceleration of b, ab is equal to the acceleration of a -aa plus the acceleration of b relative to a.

04:49 So this is the acceleration of a plus the angular acceleration of ab times k cross with r -b relative to a minus omega -squared ab times r of b relative to a.

05:16 Sorry, if we substitute the known values into this, we get 2 .6319i plus alpha abk cross with 0 .6i plus 2j minus 0 .5 .0...

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