DE2-EA 2.1: M4DE Dr Connor Myant

3. Planar Kinematics - Acceleration

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Contents Accelerations ........................................................................................................................................... 3

Recap: Rotation around a fixed axis ....................................................................................................... 3

General Motion: Accelerations ............................................................................................................... 4

Reminder: Cross Product ........................................................................................................................ 8

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Accelerations In a later chapter we will be concerned with determining the motion of a rigid body when we know

all the external forces and couples acting on it. This is known as Dynamics. The governing equations

of Dynamics are expressed in terms of the acceleration of the center of mass of the rigid body and

its angular acceleration. To solve such problems, we need the relationship between the accelerations

of points of a rigid body and its angular acceleration. In this chapter, we extend the methods we have

employed to calculate velocities of points of rigid bodies to accelerations.

Recap: Rotation around a fixed axis Figure 3.1 shows a rigid body rotating about a fixed axis with two lines perpendicular to the axis. The

angle, 𝜃, between the reference line and the body-fixed line describes the position, or orientation,

of the rigid body about the fixed axis. The rigid body’s angular velocity, 𝜔 (rate of rotation), and its

angular acceleration, 𝛼, are;

𝜔 =𝑑𝜃

𝑑𝑡, 𝛼 =

𝑑2𝜃

𝑑𝑡2 =𝑑𝜔

𝑑𝑡 (3.1)

Figure 3.1. Specifying the orientation of an object rotating about a fixed axis

Each point of the object not on the fixed axis moves in a circular path about the axis. Using our

knowledge of the motion of a point in a circular path, we can relate the velocity and acceleration of

a point to the object’s angular velocity and angular acceleration. Let’s view the object in Figure 3.1

down the axis of rotation, shown in Figure 3.2. The velocity of a point at a distance, 𝑟, from the fixed

axis is tangent to the point’s circular path (LHS) and is given in terms of the angular velocity of the

object by;

𝑣 = 𝑟𝜔 (3.2)

The point has components of acceleration tangential and normal to its circular path (RHS). In terms

of the angular acceleration of the object, the components of acceleration are;

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𝑎𝑡 = 𝑟𝛼, 𝑎𝑛 =𝑣2

𝑟= 𝑟𝜔2 (3.3)

Using these equations we can analyse problems involving objects rotating about a fixed axis.

Figure 3.2. (a) Velocity and (b) acceleration of a point of a rigid body rotating about a fixed axis

General Motion: Accelerations

Consider points 𝐴 and 𝐵 of a rigid body in planar motion relative to a given reference frame (Figure

3.3). Their velocities are related by Equation (2.5);

𝑣𝐴 = 𝑣𝐵 + 𝑣𝐴/𝐵

Taking the derivative of Equation. (2.5) with respect to time we obtain;

𝑎𝐴 = 𝑎𝐵 + 𝑎𝐴/𝐵 (3.4)

Where 𝑎𝐴 and 𝑎𝐵 are the accelerations of 𝐴 and 𝐵 relative to the reference frame and 𝑎𝐴/𝐵 is the

acceleration of 𝐴 relative to 𝐵. Consider the example in Figure 3.3; 𝐴 moves in a circular path relative

to 𝐵 as the rigid body rotates, 𝑎𝐴/𝐵 has normal and tangential components. The tangential

component equals the product of the distance 𝑟𝐴/𝐵 = |𝑟𝐴/𝐵| and the angular acceleration 𝛼 of the

rigid body. The normal component points toward the centre of the circular path, and its magnitude

is |𝑣𝐴/𝐵|2

𝑟𝐴/𝐵⁄ = 𝜔2𝑟𝐴/𝐵. The acceleration of 𝐴 equals the sum of the acceleration of 𝐵 and the

acceleration of 𝐴 relative to 𝐵.

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Figure 3.3 a) Point 𝐴 and 𝐵 of a rigid body in planer motion and the position vector of 𝐴 relative to

B, b) Components of the acceleration of 𝐴 relative to 𝐵, c) The acceleration of 𝐴.

Now let us consider a circular disk of radius, 𝑅, rolling on a stationary plane surface. This disk has

counter-clockwise angular velocity 𝜔 and counter-clockwise angular acceleration 𝛼 (Figure 3.4A).

The disk’s center 𝐵 is moving in a straight line with velocity 𝑅𝜔, towards the left if 𝜔 is positive.

Therefore, the acceleration of B is 𝑑

𝑑𝑡(𝑅𝜔) = 𝑅𝛼 and is towards the left if 𝛼 is positive (Figure 3.4B).

In other words, the magnitude of the acceleration of the center of a round object rolling on

stationary plane surface is the product of the radius and the angular acceleration.

Now that we know the acceleration of the disk’s center, let us determine the acceleration of the

point 𝐶 that is in contact with the surface. Relative to 𝐵, 𝐶 moves in a circular path of radius 𝑅 and

angular velocity 𝜔 and angular acceleration 𝛼. The tangential and normal components of the

acceleration of 𝐶 relative to 𝐵 are shown in Figure 3.4C. The acceleration of 𝐶 is the sum of the

acceleration of 𝐵 and the acceleration of 𝐶 relative to 𝐵 (Figure 3.4D). In terms of the coordinate

system shown;

𝑎𝐶 = 𝑎𝐵 + 𝑎𝐶/𝐵 = −𝑅𝛼𝒊 + 𝑅𝛼𝒊 + 𝑅𝜔2𝒋 = 𝑅𝜔2𝒋

The acceleration of point 𝐶 parallel to the surface is zero, but 𝐶 does have an acceleration normal

to the surface.

Expressing the acceleration of point 𝐴 relative to point 𝐵 in terms of 𝐴’s circular path about B as we

have done is useful for visualising and understanding the relative acceleration. However, just as we

did in the case of the relative velocity, we can obtain 𝑎𝐴/𝐵 in a form more convenient for applications

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by using the angular velocity vector �⃑⃑⃑� . The velocity of 𝐴 relative to 𝐵 is given in terms of �⃑⃑� by

Equation (2.6).

𝑣𝐴/𝐵 = �⃑⃑⃑� × 𝑟𝐴/𝐵

Taking the derivative of this equation with respect to time, we obtain;

𝑎𝐴/𝐵 =𝑑�⃑⃑⃑�

𝑑𝑡× 𝑟𝐴/𝐵 + �⃑⃑� × 𝑣𝐴/𝐵 =

𝑑�⃑⃑⃑�

𝑑𝑡× 𝑟𝐴/𝐵 + �⃑⃑� × (�⃑⃑� × 𝑟𝐴/𝐵) (3.5)

So we can define the angular acceleration vector �⃑⃑� to be the rate of change of the angular velocity

vector;

𝛼 =𝑑�⃑⃑�

𝑑𝑡

Then the acceleration of 𝐴 relative to 𝐵 is;

𝑎𝐴/𝐵 = 𝛼 × 𝑟𝐴/𝐵 + �⃑⃑� × (�⃑⃑� × 𝑟𝐴/𝐵) (3.6)

Using this expression, we can write equations relating the velocities and accelerations of two points

of a rigid body in terms of its angular velocity and angular acceleration;

𝑣𝐴 = 𝑣𝐵 + �⃑⃑� × 𝑟𝐴/𝐵

𝑎𝐴 = 𝑎𝐵 + 𝛼 × 𝑟𝐴/𝐵 + �⃑⃑� × (�⃑⃑� × 𝑟𝐴/𝐵) (3.7)

In the case of planar motion, the term 𝛼 × 𝑟𝐴/𝐵 is the tangential component of the acceleration of

𝐴 relative to 𝐵, and �⃑⃑� × (�⃑⃑� × 𝑟 𝐴/𝐵) is the normal component. Therefore, for planar motion, we can

write the above equation is a simpler form of;

𝑎𝐴 = 𝑎𝐵 + 𝛼 × 𝑟𝐴/𝐵 − 𝜔2𝑟𝐴/𝐵

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Figure 3.4. A) Disk rolling with angular velocity 𝜔 and angular acceleration 𝛼. B) The acceleration of

center 𝐵. C) Components of the acceleration of 𝐶 relative to 𝐵. D) The acceleration of 𝐶.

Design Engineering Example: Sports biomechanics is a quantitative based study and analysis of professional athletes and sports' activities in general. It can simply be described as the Physics of Sports. In this subfield of biomechanics the laws of mechanics are applied in order to gain a greater understanding of athletic performance through mathematical modelling, computer simulation and measurement.

Techniques such as state of the art 3D motion capture systems are employed to analyse complex human movement. This information can be feed back to design engineers, enabling them to optimise sport products and equipment.

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Reminder: Cross Product

The cross product of two vectors 𝑎 and 𝑏 is defined only in three-dimensional space and is denoted by 𝑎 × 𝑏.

Figure 3.5. The cross-product in respect to a right-handed coordinate system.

The cross product 𝑎 × 𝑏 is defined as a vector 𝑐 that is perpendicular to both 𝑎 and 𝑏, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.

Figure 3.6. Finding the direction of the cross product by the right-hand rule.

The cross product is defined by the formula;

𝑎 × 𝑏 = ‖𝑎‖‖𝑏‖ sin(𝜃)𝑛

where 𝜃 is the angle between 𝑎 and 𝑏 in the plane containing them (hence, it is between 0° and 180°), 𝑎 and 𝑏 are the magnitudes of vectors 𝑎 and 𝑏, and 𝑛 is a unit vector perpendicular to the plane containing 𝑎 and 𝑏 in the direction given by the right-hand rule (illustrated). If the vectors 𝑎 and 𝑏 are parallel (i.e., the angle 𝜃 between them is either 0° or 180°), by the above formula, the cross product of 𝑎 and 𝑏 is the zero vector 0.

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For our purposes we will most likely use a matrix notation for the cross product. Lets start by expressing a cross product as:

𝑢 × 𝑣 = |𝑖 𝑗 𝑘𝑢1 𝑢2 𝑢3

𝑣1 𝑣2 𝑣3

|

This determinant can be computed using Sarrus' rule or cofactor expansion. Using Sarrus' rule, it expands to;

𝑢 × 𝑣 = (𝑢2𝑣3𝑖 + 𝑢3𝑣1𝑗 + 𝑢1𝑣2𝑘) − (𝑢3𝑣2𝑖 + 𝑢1𝑣3𝑗 + 𝑢2𝑣1𝑘)

𝑢 × 𝑣 = (𝑢2𝑣3 − 𝑢3𝑣2)𝑖 + (𝑢3𝑣1 − 𝑢1𝑣3)𝑗 + (𝑢1𝑣2 − 𝑢2𝑣1)𝑘

Using cofactor expansion along the first row instead, it expands to

𝑢 × 𝑣 = |𝑢2 𝑢3

𝑣2 𝑣3| 𝑖 − |

𝑢1 𝑢3

𝑣1 𝑣3| 𝑗 + |

𝑢1 𝑢2

𝑣1 𝑣2| 𝑘

which gives the components of the resulting vector directly.

Figure 3.7. Use of Sarrus' rule to find the cross product of u and v