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DE2-EA 2.1: M4DE Dr Connor Myant
4. Moving Systems
Comments and corrections to [email protected]
Lecture resources may be found on Blackboard and at http://connormyant.com
Dr Connor Myant DE2-EA2.1 M4DE 2
Contents Moving Systems ...................................................................................................................................... 3
Sliding Contacts ................................................................................................................................... 3
Moving Reference Frames .................................................................................................................. 9
Dr Connor Myant DE2-EA2.1 M4DE 3
Moving Systems In this chapter we will expand on the basic principle of rigid body kinematics to include more complex
systems. The following problems require a different method of solution to the previous sections, but
only slightly!
We will consider two types of advanced kinematics systems; sliding contacts and moving reference
frames. The important difference from the previous chapter is that the point of interest cannot be
assumed to be a point on the same rigid body as the reference point.
We will also introduce you to the concepts of primary reference frame, body fixed reference frame
and earth fixed reference frame.
Sliding Contacts
Lets consider the linkage system in Figure 4.1. Suppose that we know the angular velocity and
acceleration of the bar 𝐴𝐵, and we want to determine the angular velocity and angular acceleration
of bar 𝐴𝐶. We cannot use Equation (2.7) to express the velocity of point 𝐴 in terms of the angular
velocity of bar 𝐴𝐵, because we derived it under the assumption that points 𝐴 and 𝐵 are points of
the same rigid body. Point 𝐴 is not a part of the bar 𝐴𝐵, but moves relative to it as the pin slides
along the slot. This is an example of a sliding contact between rigid bodies. To solve such problems,
we must re-derive equations (2.7) and (3.4) without making the assumption that 𝐴 is a point of the
rigid body.
Design Engineering Example: Sliding contacts are a typical engineering problem. They involve a part, component, or machine element moving relative to another. The component assemble is not solid, or held together by pin joints. Instead, slots, sliders or extending arms form the linkages. Classic examples would be hydraulic actuators on a car boot door, or cherry picker.
Dr Connor Myant DE2-EA2.1 M4DE 4
Figure 4.1. Linkage with a sliding contact.
To describe the motion of a point that moves relative to a given rigid body, it is convenient to use a
reference frame that moves with the rigid body. We call this a body-fixed reference frame. We can
place a body fixed reference frame 𝑥𝑦𝑧 with its origin at point 𝐵, into a primary reference frame with
origin at point 𝑂, Figure 4.2. The primary reference frame is the reference frame relative to which we
are describing the motion of the rigid body.
Now lets consider point 𝐴; we do not assume that 𝐴 is a point of the rigid body. The position of 𝐴
relative to 𝑂 is;
𝑟𝐴 = 𝑟𝐵 + 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 (4.1)
Where,
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 = 𝑟𝐴/𝐵
𝑥, 𝑦 and 𝑧 are the cordiniates of 𝐴 in terms of the body-fixed coordinate system. Our next step is to
take the time derivative of this expression to obtain an equation for the velocity of 𝐴. In doing so,
we recognise that the unit vectors 𝑖, 𝑗 and 𝑘 are not constant, because they rotate with the body-
fixed coordinate system;
𝑣𝐴 = 𝑣𝐵 +𝑑𝑥
𝑑𝑡𝑖 + 𝑥
𝑑𝑖
𝑑𝑡+
𝑑𝑦
𝑑𝑡𝑗 + 𝑦
𝑑𝑗
𝑑𝑡+
𝑑𝑧
𝑑𝑡𝑘 + 𝑧
𝑑𝑘
𝑑𝑡 (4.2)
Dr Connor Myant DE2-EA2.1 M4DE 5
Figure 4.2 A point 𝐵 of a rigid body, a body-fixed coordinate system and an arbitrary point 𝐴.
What are the time derivatives of the unit vectors? If 𝑟𝑃/𝐵 is the position vector of a point 𝑃 of a rigid
body relative to another point B of the same rigid body, then;
𝑑𝑟𝑃/𝐵
𝑑𝑡= 𝑣𝑃/𝐵 = �⃑⃑� × 𝑟𝑃/𝐵
Since we can regard the unit vector 𝑖 as the position vector of a point 𝑃 of the rigid body, its time
derivative is 𝑑𝑖
𝑑𝑡= �⃑⃑� × 𝑖. Applying the same argument to the unit vectors 𝑗 and 𝑘, we obtain;
𝑑𝑖
𝑑𝑡= �⃑⃑� × 𝑖,
𝑑𝑗
𝑑𝑡= �⃑⃑� × 𝑗,
𝑑𝑘
𝑑𝑡= �⃑⃑� × 𝑘
Using these expressions, we can write the velocity of point A as;
𝑣𝐴 = 𝑣𝐵 + 𝑣𝐴 𝑟𝑒𝑙 + �⃑⃑� × 𝑟𝐴/𝐵 (4.3)
Where,
𝑣𝐴 𝑟𝑒𝑙 + �⃑⃑� × 𝑟𝐴/𝐵 = 𝑣𝐴/𝐵
And,
𝑣𝐴 𝑟𝑒𝑙 =𝑑𝑥
𝑑𝑡𝑖 +
𝑑𝑦
𝑑𝑡𝑗 +
𝑑𝑧
𝑑𝑡𝑘 (4.4)
Dr Connor Myant DE2-EA2.1 M4DE 6
is the velocity of 𝐴 relative to the body-fixed coordinate system. That is, 𝑣𝐴 𝑟𝑒𝑙 is the velocity of 𝐴
relative to the rigid body.
Equation (4.3) expresses the velocity of a point 𝐴 as the sum of three terms, shown in Figure 4.3;
the velocity of a point 𝐵 of the rigid body, the velocity �⃑⃑� × 𝑟𝐴/𝐵 of 𝐴 relative to 𝐵 due to the rotation
of the rigid body, and the velocity 𝑣𝐴 𝑟𝑒𝑙 of 𝐴 relative to the rigid body.
Figure 4.3. Expressing the velocity of 𝐴 in terms of the velocity of a point 𝐵 of the rigid body.
To obtain an equation for the acceleration of point A, we take the time derivative of Equation (4.3)
and use Equation (4.4). The result is;
𝑎𝐴 = 𝑎𝐵 + 𝑎𝐴 𝑟𝑒𝑙 + 2�⃑⃑� × 𝑣𝐴 𝑟𝑒𝑙 + 𝛼 × 𝑟𝐴/𝐵 + �⃑⃑� × (�⃑⃑� × 𝑟𝐴/𝐵) (4.5)
Where;
𝑎𝐴 𝑟𝑒𝑙 + 2�⃑⃑� × 𝑣𝐴 𝑟𝑒𝑙 + 𝛼 × 𝑟𝐴/𝐵 + �⃑⃑� × (�⃑⃑� × 𝑟𝐴/𝐵) = 𝑎𝐴/𝐵
And,
𝑎𝐴 𝑟𝑒𝑙 =𝑑2𝑥
𝑑𝑡2𝑖 +
𝑑2𝑦
𝑑𝑡2𝑗 +
𝑑2𝑧
𝑑𝑡2𝑘
Is the acceleration of 𝐴 relative to the body-fixed coordinate system. That is, 𝑎𝐴 is the acceleration
of 𝐴 relative to the primary reference frame, and 𝑎𝐴 𝑟𝑒𝑙 is the acceleration of 𝐴 relative to the rigid
body.
In the case of planar motion, we can express Equation (4.5) in a simpler form;
𝑎𝐴 = 𝑎𝐵 + 𝑎𝐴 𝑟𝑒𝑙 + 2�⃑⃑� × 𝑣𝐴 𝑟𝑒𝑙 + 𝛼 × 𝑟𝐴/𝐵 − 𝜔2𝑟𝐴/𝐵
Dr Connor Myant DE2-EA2.1 M4DE 7
The summary 𝑣𝐴 and 𝑎𝐴 are the velocity and acceleration of point 𝐴 relative to a non-rotating
coordinate system that is stationary relative to point 𝑂; to the primary reference frame – the
reference frame relative to which the rigid body’s motion is being described. The terms 𝑣𝐴 𝑟𝑒𝑙 and
𝑎𝐴 𝑟𝑒𝑙 are the velocity and acceleration of point 𝐴 measured by an observer moving with the rigid
body. If 𝐴 is a point of the rigid body, 𝑣𝐴 𝑟𝑒𝑙 and 𝑎𝐴 𝑟𝑒𝑙 are zero, and equations (4.3) and (4.5) are
identical to equations (2.7) and (3.4).
We can illustrate these concepts with a simple example. Figure 4.4A shows a point 𝐴 moving with
velocity 𝑣 parallel to the axis of a bar. Suppose that at the same time, the bar is rotating about a
fixed point 𝐵 with constant angular velocity 𝜔 relative to an earth-fixed reference frame (Figure
4.4B). You can use Equation (4.1) to determine the velocity of 𝐴 relative to the earth fixed reference
frame.
Figure 4.4. A) A point moving along a bar. B) The bar rotating. C) A body fixed reference frame. D)
Components of 𝑣𝐴.
Let the coordinate system in Figure 4.4C be fixed with respect to the bar, and let 𝑥 be the present
position of 𝐴. In terms of this body fixed reference frame, the angular velocity vector of the bar (and
the reference frame) relative to the primary earth fixed reference frame is �⃑⃑⃑� = 𝜔𝒌. Relative to the
body fixed reference frame, point 𝐴 moves along the 𝑥 axis with velocity 𝑣, so 𝑣𝐴 𝑟𝑒𝑙 = 𝑣𝒊. From
Equation (4.1), the velocity of 𝐴 relative to the earth-fixed reference frame is;
Dr Connor Myant DE2-EA2.1 M4DE 8
𝑣𝐴 = 𝑣𝐵 + 𝑣𝐴 𝑟𝑒𝑙 + �⃑⃑⃑� × 𝑟𝐴/𝐵
𝑣𝐴 = 0 + 𝑣𝒊 + 𝜔𝒌 × (𝑥𝒊)
𝑣𝐴 = 𝑣𝒊 + 𝜔𝑥𝒋
Relative to the earth fixed reference frame, 𝐴 has a component of velocity parallel to the bar and
also a perpendicular component due to the bar’s rotation (Figure 4.4D). Although 𝑣𝐴 is the velocity
of 𝐴 relative to the earth fixed reference frame, notice that is expressed in components that are in
terms of the body fixed reference frame.
Design Engineering Example:
You have been asked to design a stage for a Pop concert. The band have asked for the stage to rotate. They will move from the centre of the stage and onto a fixed walkway out into the crowd. Their movement is pre-planned in their dance sequence, but you need to define the rotational speed of the stage so that the band line up with the walkway.
In order to achieve this you will need to use a primary earth fixed reference frame and a body fixed reference frame for the rotating stage along which the band are moving. Because the band are moving relative to the rotating stage this can be considered as a sliding contact and solved using Equation (3.3);
Dr Connor Myant DE2-EA2.1 M4DE 9
Moving Reference Frames
In many situations it is convenient to describe the motion of a point by using a secondary reference
frame that moves relative to some fixed reference frame. You could envisage this for many mobile
phone mapping applications, or when we want to measure the speed of an object relative to a
moving vehicle.
We will show how the velocity and acceleration of a point relative to a primary reference frame are
related to their values relative to a moving secondary reference frame.
Equations (4.3) and (4.5) give the velocity and acceleration of an arbitrary point 𝐴 relative to a point
𝐵 of a rigid body in terms of a body-fixed secondary reference frame. But these equations do not
require us to assume that the secondary reference frame is connected to some rigid body.
They apply to any reference frame having a moving origin 𝐵 and rotating with angular velocity �⃑⃑�
and angular acceleration 𝛼 relative to a primary reference frame (Figure 4.4). The terms 𝑣𝐴 and 𝑎𝐴
are the velocity and acceleration of 𝐴 relative to the primary reference frame. The terms 𝑣𝐴 𝑟𝑒𝑙 and
𝑎𝐴 𝑟𝑒𝑙 are the velocity and acceleration of 𝐴 relative to the secondary reference frame. That is, they
are the velocity and acceleration measured by an observer moving with the secondary reference
frame (imagine you are sat on a train looking out the window at another train moving alongside
you).
Let the coordinate system be fixed with respect to the stage, and let 𝑥 be the present position of the Band, 𝐴. Relative to the primary earth fixed reference frame the angular velocity vector of the stage is �⃑⃑� = 𝜔𝑘. Relative to the body fixed reference frame, the band moves along the 𝑥-axis with velocity v, so 𝑣𝐴 𝑟𝑒𝑙 = 𝑣𝑖. Therefore;
𝑣𝐴 = 𝑣𝐵 + 𝑣𝐴 𝑟𝑒𝑙 + �⃑⃑� × 𝑟𝐴/𝐵
= 0 + 𝑣𝑖 + (𝜔𝑘) × (𝑥𝑖) = 𝑣𝑖 + 𝜔𝑥𝑗
Dr Connor Myant DE2-EA2.1 M4DE 10
Figure 4.4. A secondary reference frame with origin 𝐵 and an arbitrary point 𝐴.
When the velocity and acceleration of a point 𝐴 relative to a moving secondary reference frame are
known, we can use Equations (4.3) and (4.5) to determine the velocity and acceleration of 𝐴 relative
to the primary reference frame. There will also be situations in which the velocity and acceleration of
𝐴 relative to the primary reference frame will be known and we will want to use Equations (4.3) and
(4.5) to determine the velocity and acceleration of 𝐴 relative to a moving secondary reference frame.
Figure 4.5.
Dr Connor Myant DE2-EA2.1 M4DE 11
Lets consider the satellite shown in Figure 4.5, it is in a circular polar orbit. Relative to a nonrotating
primary reference frame with its origin at the centre of the earth, the satellite moves in a circular path
of radius 𝑅 with a velocity of constant magnitude 𝑣𝐴. At the present instant, the satellite is above the
equator. The secondary earth fixed reference frame shown is oriented with the 𝑦 axis in the direction
of the north pole and the 𝑥 axis in the direction of the satellite.
To find the satellite’s velocity and acceleration relative to the non-rotating primary reference frame
with its origin at the centre of the earth we can begin by stating, that at the present instant 𝑣𝐴 = 𝑣𝐴𝒋
and 𝑎𝐴 = −(𝑣𝐴
2
𝑅) 𝒊. The angular velocity vector of the earth points north (Right hand rule), so the
angular velocity of the earth fixed reference frame is �⃑⃑⃑� = 𝜔𝐸𝒋, where 𝜔𝐸 is the angular velocity of
the earth. From Equation (4.3);
𝑣𝐴 = 𝑣𝐵 + 𝑣𝐴 𝑟𝑒𝑙 + �⃑⃑� × 𝑟𝐴/𝐵
𝑣𝐴𝒋 = 0 + 𝑣𝐴 𝑟𝑒𝑙 + |𝑖 𝑗 𝑘0 𝜔𝐸 0𝑅 0 0
|
Solving for 𝑣𝐴 𝑟𝑒𝑙, we find that the satellite’s velocity relative to the earth fixed reference frame is
𝑣𝐴 𝑟𝑒𝑙 = 𝑣𝐴𝒋 + 𝑅𝜔𝐸𝒌
The second term on the right hand side of this equation is the satellite’s velocity towards the west
relative to the rotating earth fixed reference frame.
From Equation (4.5);
𝑎𝐴 = 𝑎𝐵 + 𝑎𝐴 𝑟𝑒𝑙 + 2�⃑⃑� × 𝑣𝐴 𝑟𝑒𝑙 + 𝛼 × 𝑟𝐴/𝐵 + �⃑⃑� × (�⃑⃑� × 𝑟𝐴/𝐵)
−𝑣𝐴
2
𝑅𝒊 = 0 + 𝑎𝐴 𝑟𝑒𝑙 + 2 |
𝑖 𝑗 𝑘0 𝜔𝐸 00 𝑣𝐴 𝑅𝜔𝐸
| + 0 + |𝑖 𝑗 𝑘0 𝜔𝐸 00 0 −𝑅𝜔𝐸
|
Solving for 𝑎𝐴 𝑟𝑒𝑙, we find the satellite’s acceleration relative to the earth fixed reference frame to
be;
𝑎𝐴 𝑟𝑒𝑙 = −(𝑣𝐴
2
𝑅+ 𝜔𝐸
2𝑅)𝒊
