Planar Rigid Body Kinematics
Planar rigid body kinematics focuses on analyzing the motion of rigid bodies confined to a two-dimensional plane without considering the forces that cause the motion. It encompasses both translational and rotational movements, examining parameters such as the position and orientation of the body, linear and angular velocity, and linear and angular acceleration.
Planar motion can be classified into three types. Translation occurs when all points of a rigid body move along parallel paths. If these paths are straight lines, the motion is rectilinear translation, while if the paths are curved, it is curvilinear translation. In both cases, there is no rotation, and the body maintains its orientation relative to its original position.
Position of Two Points on a Body
Velocity of Two Points on a Body
Acceleration of Two Points on a Body
Velocity of a Body wrt Two Reference Frames
Acceleration of a Body wrt Two Reference Frames
Wheel Rolling on a Fixed Horizontal Surface
Wheel Rolling on a Curved Surface
Rotation about a fixed axis describes a motion where every point in the rigid body moves in a circular path around a single, stationary axis. The axis remains fixed in space, and all points on the body have the same angular velocity and angular acceleration but different linear velocities depending on their distance from the axis.
General planar motion is a combination of translation and rotation. The body experiences both a change in position and a change in orientation simultaneously. This type of motion is common in many practical applications, where the motion cannot be classified purely as translation or rotation.
Lets consider a rigid body, and identify two points \mathbf P and \mathbf Q which have a position vector in relation to a fix reference frame; in kinematics this reference frame doesn’t necessarily need to be inertial, as there is no application of Newton or Euler equations, but we are considering only the geometry.
Position of two points on a body
We can express the position of a point Q in terms of vector addition of position vectors. Let \mathbf{r}_{OQ} be the position vector from the origin O to point Q. This can be written as:
\mathbf{r}_{OQ} = \mathbf{r}_{OP} + \mathbf{r}_{PQ}
Here, \mathbf{r}_{OP} is the position vector from O to another point P, and \mathbf{r}_{PQ} is the position vector from P to Q.
Velocity of two points on a body
To obtain the velocities, we take the time derivatives of these position vectors:
\mathbf{v}_Q = \frac{\mathrm d\mathbf{r}_{OQ}}{\mathrm dt} = \frac{\mathrm d\mathbf{r}_{OP}}{\mathrm dt} + \frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt}
This simplifies to:
\mathbf{v}_Q = \mathbf{v}_P + \mathbf{v}_{QP}
where:
- \mathbf{v}_Q is the absolute velocity of point Q with respect to the reference frame.
- \mathbf{v}_P is the absolute velocity of point P with respect to the reference frame.
- \mathbf{v}_{QP} is the relative velocity of point Q with respect to point P.
It’s important to note that \mathbf{v}_Q and \mathbf{v}_P are absolute velocities measured in the fixed reference frame. The relative velocity \mathbf{v}_{QP} accounts for the motion of Q relative to P, which both points can have with respect to the reference frame.
Next, let’s consider the relative velocity \mathbf{v}_{QP}. Consider the position vector \mathbf{r}_{PQ} expressed in terms of its magnitude and direction. Let \theta be the angle that \mathbf{r}_{PQ} makes with the i-direction (the horizontal axis). We can express \mathbf{r}_{PQ} as:
\mathbf{r}_{PQ} = r_{PQ} \cos\theta \, \mathbf{i} + r_{PQ} \sin\theta \, \mathbf{j} = r_{PQ} \mathbf{u}
where \mathbf{u} = \cos\theta \, \mathbf{i} + \sin\theta \, \mathbf{j} is the unit vector in the direction of \mathbf{r}_{PQ}.
Taking the derivative of \mathbf{r}_{PQ} with respect to time to find the relative velocity \mathbf{v}_{QP}:
\mathbf{v}_{QP} = \frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt} \left( r_{PQ} \cos\theta \, \mathbf{i} + r_{PQ} \sin\theta \, \mathbf{j} \right)
Applying the product rule, we get:
\mathbf{v}_{QP} = \dot{r}_{PQ} \cos\theta \, \mathbf{i} + r_{PQ} (-\sin\theta \, \dot{\theta}) \, \mathbf{i} + \dot{r}_{PQ} \sin\theta \, \mathbf{j} + r_{PQ} \cos\theta \, \dot{\theta} \, \mathbf{j}
Simplifying, we can factor out \dot{\theta}:
\mathbf{v}_{QP} = \dot{r}_{PQ} \mathbf{u} + r_{PQ} \dot{\theta} (-\sin\theta \, \mathbf{i} + \cos\theta \, \mathbf{j})
In fixed body kinematics the magnitude of the distance between two points is fixed:
\dot{r}_{PQ} = 0
so:
\mathbf{v}_{QP} = r_{PQ} \dot{\theta} (-\sin\theta \, \mathbf{i} + \cos\theta \, \mathbf{j})
Notice that -\sin\theta \, \mathbf{i} + \cos\theta \, \mathbf{j} is equivalent to \mathbf{k} \times \mathbf{u}, where \mathbf{k} is the unit vector perpendicular to the plane (out of the plane):
\mathbf{v}_{QP} = r_{PQ} \dot{\theta} \, (\mathbf{k} \times \mathbf{u}) = \dot{\theta} \, (\mathbf{k} \times r_{PQ} \mathbf{u}) = \dot{\theta} \, (\mathbf{k} \times \mathbf{r}_{PQ})
Therefore we have:
\mathbf{v}_Q = \mathbf{v}_P + \dot{\theta} \, (\mathbf{k} \times \mathbf{r}_{PQ}) = \mathbf{v}_P + \omega \, (\mathbf{k} \times \mathbf{r}_{PQ})
with \omega = \dot{\theta} is the angular velocity.
Instant center of rotation
The instant center of rotation is a specific point on a body around which the entire body appears to be rotating at any given moment, much like a snapshot in time. This point itself has zero velocity, and there is only one such instant center for a body at each instant. The instant center can either lie on the body or somewhere outside it, in which case we refer to the “extended body”.
There are three primary methods to locate the instant center of rotation:
- Identifying a point with zero velocity: Find a point on the body that has no velocity at that instant.
- Using perpendiculars to velocity vectors: Draw perpendicular lines to the velocity vectors of different points on the body; the intersection of these lines is the instantaneous center.
- Applying geometry and similar triangles: Utilize geometric relationships and similar triangles to determine the position of the instant center.
Consider the instant center of rotation \mathbf P and another point \mathbf Q on the body. The velocity of any point on the body can be described by the angular velocity of the body multiplied by the position vector from the instantaneous center to that point Q:
\mathbf{v}_Q = \mathbf{v}_P + \omega \, (\mathbf{k} \times \mathbf{r}_{PQ})
Since the instant center of rotation has zero velocity by definition, its contribution to the equation vanishes. Therefore, the velocity of point \mathbf Q is simply the angular velocity of the body crossed with the position vector from \mathbf P to \mathbf Q.
Let’s consider an example: a wheel that rolls without slipping. Imagine a wheel in contact with the ground. At the exact moment the wheel touches the ground, the contact point has zero velocity relative to the ground. This contact point is the instant center of rotation of the wheel at that instant.
A second method for determining the instant center of rotation involves examining the velocity vectors of various points on the moving body. To apply this technique, begin by identifying the velocity vectors of at least two distinct points on the body. For each of these velocity vectors, we can draw a line that is perpendicular to the direction of the velocity at that specific point.
Since the instant center of rotation is the point about which the entire body is momentarily rotating, it must lie along these perpendicular lines. The point where these perpendicular lines intersect is the instant center of rotation. This intersection occurs because, at the instant center, the velocity of that point is zero, and all other velocities in the body are purely rotational around it.
By using this geometric approach, you can accurately locate the instant center of rotation based on the directional properties of the velocity vectors of different points on the body.
The bar AB and the bar CD are rotating around a fix axes, and therefore their velocity can only go up and down or left or right.
It is possible to draw a perpendicular from the velocity vectors and, where that crosses, that defines the instant center of rotation.
A third method for determining the instant center of rotation involves applying geometric relationships and the principles of similar triangles. This approach begins by identifying the paths or lines along which different points on the body are moving.
By analyzing the motion, it is possible to construct triangles that represent the geometric relationships between these points and the desired instant center. These triangles are arranged in such a way that they are similar, meaning their corresponding angles are equal and their sides are in proportion.
By setting up these similar triangles, the proportions between corresponding sides can be used to solve for the unknown position of the instant center. The point where the corresponding sides intersect, based on the similarity criteria, indicates the location of the instant center of rotation. This geometric method leverages the inherent properties of similar triangles to accurately determine the position of the instant center, ensuring that the rotational relationships within the moving body are properly accounted for.
Acceleration of two points on a body
Consider a rigid body with two points, P and Q, where the position vector from P to Q is denoted by \mathbf{r}_{PQ}. Let \theta(t) represent the angular displacement of the body as a function of time. The angular velocity is given by \omega = \frac{\mathrm d\theta}{\mathrm dt}, and the angular acceleration is \alpha = \frac{\mathrm d\omega}{\mathrm dt} = \frac{\mathrm d^2\theta}{\mathrm dt^2}. The angular acceleration is symbolized by {\alpha}.
The absolute velocity of point Q, \mathbf{v}_Q, can be expressed in terms of the absolute velocity of point P, \mathbf{v}_P, and the angular velocity {\omega} as follows:
\mathbf{v}_Q = \mathbf{v}_P + \omega \mathbf{k} \times \mathbf{r}_{PQ}
Differentiating both sides with respect to time, we have:
\frac{\mathrm d\mathbf{v}_Q}{\mathrm dt} = \frac{\mathrm d\mathbf{v}_P}{\mathrm dt} + \frac{\mathrm d}{\mathrm dt} \left( \omega \mathbf{k} \times \mathbf{r}_{PQ} \right)
This gives:
\mathbf{a}_Q = \mathbf{a}_P + \frac{\mathrm d}{\mathrm dt} (\omega \mathbf{k}) \times \mathbf{r}_{PQ} + \omega \mathbf{k} \times \frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt}
The term \frac{\mathrm d}{\mathrm dt} (\omega \mathbf{k}) results in the angular acceleration \alpha \mathbf{k}, so the equation becomes:
\mathbf{a}_Q = \mathbf{a}_P + (\alpha \mathbf{k}) \times \mathbf{r}_{PQ} + \omega \mathbf{k} \times \frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt}
Next, we evaluate \frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt}. From the relative velocity equation, we have:
\frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt} = \omega \mathbf{k} \times \mathbf{r}_{PQ}
Substituting this into the acceleration equation, we get:
\mathbf{a}_Q = \mathbf{a}_P + (\alpha \mathbf{k}) \times \mathbf{r}_{PQ} + \omega \mathbf{k} \times (\omega \mathbf{k} \times \mathbf{r}_{PQ})
Now, simplify the double cross product using the vector identity:
\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\mathbf{A} \cdot \mathbf{C}) - \mathbf{C} (\mathbf{A} \cdot \mathbf{B})
Here, \mathbf{A} = \omega \mathbf{k}, \mathbf{B} = \omega \mathbf{k}, and \mathbf{C} = \mathbf{r}_{PQ}. Applying the identity:
\omega \mathbf{k} \times (\omega \mathbf{k} \times \mathbf{r}_{PQ}) = (\omega \mathbf{k}) (\omega \mathbf{k} \cdot \mathbf{r}_{PQ}) - \mathbf{r}_{PQ} (\omega \mathbf{k} \cdot \omega \mathbf{k})
Since \mathbf{k} \cdot \mathbf{r}_{PQ} = 0 (as \mathbf{r}_{PQ} lies in the xy-plane and \mathbf{k} is along the z-axis), the first term vanishes. The second term simplifies to:
\omega \mathbf{k} \times (\omega \mathbf{k} \times \mathbf{r}_{PQ}) = -\omega^2 \mathbf{r}_{PQ}
Substituting this back into the equation for \mathbf{a}_Q, we obtain:
\mathbf{a}_Q = \mathbf{a}_P + (\alpha \mathbf{k}) \times \mathbf{r}_{PQ} - \omega^2 \mathbf{r}_{PQ}
Here, \mathbf{a}_P is the absolute acceleration of point P, \alpha \mathbf{k} \times \mathbf{r}_{PQ} is the tangential acceleration due to angular acceleration, and -\omega^2 \mathbf{r}_{PQ} is the normal (centripetal) acceleration due to angular velocity.
Reference frame change
Let’s consider a generic situation involving the velocities and acceleration of a point relative to two distinct frames or bodies. Take a point P, observed in two frames: the motion of P can be described relative to the body relative to either frames.
We will referring to one frame as the Fixed Frame, and the other as the Moving Frame. In this case as well, the term “Fixed Frame” doesn’t necessarily mean it is fixed in an inertial reference frame. It would only need to be fixed in an inertial reference frame if we were analyzing kinetics and applying Newton’s or Euler’s laws. For now, we’re focused on kinematics, dealing solely with the geometry of motion. So these frames simply represent two distinct frames.
In cases where multiple bodies are connected, we can analyze their kinematics by linking two frames at a time, transferring vectors between them. This approach becomes useful when studying the motion of a single point in multiple frames or bodies.
Unit vectors
The unit vectors of Frame 2 can be expressed in terms of Frame 1’s unit vectors using trigonometric relationships:
\begin{align*} \mathbf{i}_2 &= \cos(\theta) \, \mathbf{i}_1 + \sin(\theta) \, \mathbf{j}_1 \\ \mathbf{j}_2 &= -\sin(\theta) \, \mathbf{i}_1 + \cos(\theta) \, \mathbf{j}_1 \end{align*}
Since frame 2 is rotating relative to frame 1, the unit vectors \mathbf{i}_2 and \mathbf{j}_2 are functions of time. We’ll differentiate these vectors with respect to time t in frame 1.
The derivative of \mathbf{i}_2 with respect to t is:
\begin{aligned} \frac{\mathrm d\mathbf{i}_2}{\mathrm dt} & = \frac{\mathrm d}{\mathrm dt} \left( \cos(\theta) \, \mathbf{i}_1 + \sin(\theta) \, \mathbf{j}_1 \right) \\ & = -\sin(\theta) \, \dot{\theta} \, \mathbf{i}_1 + \cos(\theta) \, \dot{\theta} \, \mathbf{j}_1 \\ & = \dot{\theta} \left( -\sin(\theta) \, \mathbf{i}_1 + \cos(\theta) \, \mathbf{j}_1 \right) \\ & = \dot{\theta} \, \mathbf{j}_2 \end{aligned}
The derivative of \mathbf{j}_2 with respect to t is:
\begin{aligned} \frac{\mathrm d\mathbf{j}_2}{\mathrm dt} & = \frac{\mathrm d}{\mathrm dt} \left( -\sin(\theta) \, \mathbf{i}_1 + \cos(\theta) \, \mathbf{j}_1 \right) \\ & = -\cos(\theta) \, \dot{\theta} \, \mathbf{i}_1 - \sin(\theta) \, \dot{\theta} \, \mathbf{j}_1 \\ & = \dot{\theta} \left( -\cos(\theta) \, \mathbf{i}_1 - \sin(\theta) \, \mathbf{j}_1 \right) \\ & = -\dot{\theta} \, \mathbf{i}_2 \end{aligned}
If multiple bodies are connected and each has its own reference frame, it is possible to analyze their kinematics by linking two frames at a time. By systematically transferring vectors between adjacent frames, it is possible to describe the motion of a single point across all frames involved.
Vector derivative formula
Let’s now consider a vector \mathbf{u} expressed in two different reference frames.
In the context of kinematics, \mathbf{u} can represent various vector quantities such as position, velocity, or acceleration. We’ll express \mathbf{u} in Frame 2, which is our moving frame.
Let’s denote the unit vectors of Frame 2 as \mathbf{i}_2 and \mathbf{j}_2, and the coordinates as x_2 and y_2. Then, the vector \mathbf{u} can be expressed in Frame 2 as:
\mathbf{u} = u_{y_2} \, \mathbf{i}_2 + u_{y_2} \, \mathbf{j}_2
The objective is is to differentiate \mathbf{u} with respect to time t as observed from Frame 1. Since Frame 2 is moving relative to Frame 1, both the components u_{y_2} and u_{y_2}, as well as the unit vectors \mathbf{i}_2 and \mathbf{j}_2, may change with time.
Using the product rule for differentiation, the derivative of \mathbf{u} with respect to Frame 1 is:
\begin{aligned} \frac{\mathrm d\mathbf{u}}{\mathrm dt} \bigg|_1 & = \frac{\mathrm du_{y_2}}{\mathrm dt} \, \mathbf{i}_2 + u_{y_2} \, \frac{\mathrm d\mathbf{i}_2}{\mathrm dt} + \frac{\mathrm du_{y_2}}{\mathrm dt} \, \mathbf{j}_2 + u_{y_2} \, \frac{\mathrm d\mathbf{j}_2}{\mathrm dt} \\ & = \left( \frac{\mathrm du_{y_2}}{\mathrm dt} \, \mathbf{i}_2 + \frac{\mathrm du_{y_2}}{\mathrm dt} \, \mathbf{j}_2 \right) + \left( u_{y_2} \, \frac{\mathrm d\mathbf{i}_2}{\mathrm dt} + u_{y_2} \, \frac{\mathrm d\mathbf{j}_2}{\mathrm dt} \right) \end{aligned}
The first term presents the derivative of \mathbf{u} in frame 2 and the second term accounts for the rotation of frame 2 relative to frame 1.
From the previous derivation, we have the time derivatives of the unit vectors in frame 2 with respect to frame 1 and substituting these into the second term:
\begin{aligned} u_{y_2} \, \frac{\mathrm d\mathbf{i}_2}{\mathrm dt} + u_{y_2} \, \frac{\mathrm d\mathbf{j}_2}{\mathrm dt} & = u_{y_2} \, \dot{\theta} \, \mathbf{j}_2 - u_{y_2} \, \dot{\theta} \, \mathbf{i}_2 \\ & = \dot{\theta} \left( u_{y_2} \, \mathbf{j}_2 - u_{y_2} \, \mathbf{i}_2 \right) \\ & = \dot{\theta} \, \mathbf{k} \times \mathbf{u} \end{aligned}
As the expression can be represented as the cross product with the unit vector \mathbf{k} (perpendicular to the plane of motion).
Combining the two terms, the derivative of \mathbf{u} with respect to Frame 1 is:
\frac{\mathrm d\mathbf{u}}{\mathrm dt} \bigg|_1 = \frac{\mathrm d\mathbf{u}}{\mathrm dt} \bigg|_2 + \dot{\theta} \, \mathbf{k} \times \mathbf{u}
This is the derivative formula, which is used in transforming the derivative of a vector from a moving frame to a fixed frame.
Velocity of a body wrt two reference frames
Let’s begin by expressing the position vector. We’ll consider the position from point O_1 (the origin of frame 1) to point P as follows:
\mathbf{r}_{O_1P} \big|_1 = \mathbf{r}_{O_1O_2} \big|_1 + \mathbf{r}_{O_2P} \big|_2
To determine the velocity, we need to differentiate the position vector with respect to time. Applying differentiation, we obtain:
\frac{\mathrm{d}\mathbf{r}_{O_1P}}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{r}_{O_1O_2}}{\mathrm{d}t} \bigg|_1 + \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_1
The left side represents the velocity of point \mathbf P with respect to frame 1; the first term on the right side represents the velocity of O_2 (the origin of frame 2) with respect to frame 1 and the second term on the right represents the derivative of \mathbf{r}_{O_2P}, which is a vector expressed in the moving frame 2. We need to differentiate this vector with respect to frame 1.
Differentiating \mathbf{r}_{O_2P} \big|_2 with respect to frame 1 requires attention because frame 2 is rotating relative to frame 1. To handle this, we use the derivative formula derived previously:
\frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} \bigg|_2 + \boldsymbol{\omega} \times \mathbf{u}
Applying this formula to \mathbf{r}_{O_2P} \big|_2:
\frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_2 + \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2
Substituting this back into our original velocity equation:
\mathbf{v}_P \big|_1 = \mathbf{v}_{O_2} \big|_1 + \mathbf{v}_P \big|_2 + \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2
where:
- \mathbf{v}_P \big|_1: velocity of point \mathbf P with respect to frame 1,
- \mathbf{v}_{O_2} \big|_1: velocity of the origin of frame 2 (O_2) with respect to frame 1,
- \mathbf{v}_P \big|_2: velocity of point \mathbf P with respect to frame 2,
- \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2: term accounting for the rotation of frame 2 relative to frame 1.
Rearranging the term and dropping some indexes this formula can be written as:
\mathbf{v}_P = \mathbf{v}_{O_2} + \mathbf{v}_{\text{rel}} + \boldsymbol{\omega} \times \mathbf{r}
where:
- \mathbf{v}_P: absolute velocity of point \mathbf P with respect to frame 1,
- \mathbf{v}_{O_2}: absolute velocity of the origin of frame 2 (O_2) with respect to frame 1,
- \mathbf{v}_{\text{rel}}: relative velocity of point \mathbf P with respect to frame 2,
- \boldsymbol{\omega} \times \mathbf{r}: angular velocity of frame 2 relative to frame 1 crossed with the position of point \mathbf P in the moving frame.
Assuming planar (two-dimensional) motion, the angular velocity vector \boldsymbol{\omega} can be expressed as \omega \, \mathbf{k} (or \dot \theta \, \mathbf k), where \mathbf{k} is the unit vector perpendicular to the plane of motion, the equation becomes:
\begin{aligned} \mathbf{v}_P & = \mathbf{v}_{O_2} + \mathbf{v}_{\text{rel}} + \omega \, \mathbf k \times \mathbf{r} \\ & = \mathbf{v}_{O_2} + \mathbf{v}_{\text{rel}} + \dot \theta \, \mathbf k \times \mathbf{r} \end{aligned}
Acceleration of a body wrt two reference frames
To find the acceleration of point \mathbf P relative to the two frames, we’ll differentiate the velocity equation with respect to time in frame 1:
\begin{aligned} & \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{v}_{O_2}}{\mathrm{d}t} \bigg|_1 + \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \bigg|_1 + \frac{\mathrm{d}}{\mathrm{d}t} \left( \boldsymbol{\omega} \times \mathbf{r}_{O_2P} \right) \bigg|_1 \\ & \mathbf{a}_P \big|_1 = \mathbf{a}_{O_2} \big|_1 + \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \bigg|_1 + \frac{\mathrm{d}}{\mathrm{d}t} \left( \boldsymbol{\omega} \times \mathbf{r}_{O_2P} \right) \bigg|_1 \end{aligned}
In this formula:
- \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \big|_1 represents the acceleration of point P with respect to frame 1 (\mathbf{a}_P \big|_1),
- \frac{\mathrm{d}\mathbf{v}_{O_2}}{\mathrm{d}t} \big|_1 represents the acceleration of the origin of frame 2 (\mathbf{a}_{O_2} \big|_1),
- \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \big|_1 requires careful differentiation since \mathbf{u}_P \big|_2 is expressed in the moving frame 2,
- \frac{\mathrm{d}}{\mathrm{d}t} \left( \boldsymbol{\omega} \times \mathbf{r}_{O_2P} \right) \big|_1 is the differentiation of the rotational term involving angular velocity and position vector.
we use the derivative formula derived previously:
\frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} \bigg|_2 + \boldsymbol{\omega} \times \mathbf{u}
Applying this formula to \mathbf{v}_P \big|_2:
\frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \bigg|_2 + \boldsymbol{\omega} \times \mathbf{v}_P\big|_2 = \mathbf{a}_P \big|_2 + \boldsymbol{\omega} \times \mathbf{v}_P\big|_2
Differentiating \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2 and using \boldsymbol{\alpha} for the angular acceleration of frame 2 relative to frame 1:
\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \left( \boldsymbol{\omega} \times \mathbf{r}_{O_2P} \right)\bigg|_1 & = \frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} \bigg|_1 \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_1 \\ & = \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_1 \end{aligned}
For \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \big|_1 we use again the derivative formula and a result previously derived:
\frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_2 + \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2 = \mathbf{v}_P \big|_2 + \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2
Substituting:
\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \left( \boldsymbol{\omega} \times \mathbf{r}_{O_2P} \right) \bigg|_1 & = \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \left( \mathbf{v}_P \big|_2 + \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2 \right) \\ & = \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \mathbf{v}_P \big|_2 + \boldsymbol{\omega} \times \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2 \\ & = \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \mathbf{v}_P \big|_2 - \omega^2 \mathbf{r}_{O_2P}\big|_2 \end{aligned}
Putting all together:
\begin{aligned} \mathbf{a}_P \big|_1 & = \mathbf{a}_{O_2} \big|_1 + \mathbf{a}_P \big|_2 + \boldsymbol{\omega} \times \mathbf{v}_P\big|_2 + \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \mathbf{v}_P \big|_2 - \omega^2 \mathbf{r}_{O_2P}\big|_2 \\ & = \mathbf{a}_{O_2} \big|_1 + \mathbf{a}_P \big|_2 + \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 - \omega^2 \mathbf{r}_{O_2P}\big|_2 + 2\boldsymbol{\omega} \times \mathbf{v}_P \big|_2 \end{aligned}
where:
- \mathbf{a}_P \big|_1: acceleration of point \mathbf P with respect to frame 1,
- \mathbf{a}_{O_2} \big|_1: acceleration of the origin of frame 2 (O_2) with respect to frame 1,
- \mathbf{a}_P \big|_2: acceleration of point \mathbf P with respect to frame 2,
- \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1: tangential acceleration,
- - \omega^2 \mathbf{r}_{O_2P}\big|_2: normal or radial acceleration,
- 2\boldsymbol{\omega} \times \mathbf{v}_{O_2P}\big|_2: new term (the Coriolis acceleration) which is a new term that arise when there is angular velocity between frame 2 relative to frame 1.
Rearranging the term and dropping some indexes this formula can be written as:
\mathbf{a}_P = \mathbf{a}_{O_2} + \mathbf{a}_{\text{rel}} + \boldsymbol{\alpha} \times \mathbf{r} - \omega^2 \mathbf{r} + 2\boldsymbol{\omega} \times \mathbf{v}_{\text{rel}}
where:
- \mathbf{a}_P: absolute acceleration of point \mathbf P with respect to frame 1,
- \mathbf{a}_{O_2}: absolute acceleration of the origin of frame 2 (O_2) with respect to frame 1,
- \mathbf{a}_{\text{rel}}: relative acceleration of point \mathbf P with respect to frame 2,
- \boldsymbol{\alpha} \times \mathbf{r}: tangential acceleration of point \mathbf P with respect to frame 2,
- \omega^2 \mathbf{r}: normal acceleration of point \mathbf P with respect to frame 2,
- 2\boldsymbol{\omega} \times \mathbf{v}_{\text{rel}}: Coriolis acceleration of point \mathbf P with respect to frame 2.
Assuming planar (two-dimensional) motion, the angular acceleration vector \boldsymbol{\alpha} can be expressed as \alpha \, \mathbf{k} (or \ddot \theta \, \mathbf k) and the angular velocity \boldsymbol{\omega} can be expressed as \omega \, \mathbf{k} (or \dot \theta \, \mathbf k), where \mathbf{k} is the unit vector perpendicular to the plane of motion, the equation becomes:
\begin{aligned} \mathbf{a}_P & = \mathbf{a}_{O_2} + \mathbf{a}_{\text{rel}} + \alpha \, \mathbf k \times \mathbf{r} - \omega^2 \mathbf{r} + 2\omega \, \mathbf k \times \mathbf{v}_{\text{rel}} \\ & = \mathbf{v}_{O_2} + \mathbf{v}_{\text{rel}} + \ddot \theta \, \mathbf k \times \mathbf{r} - \omega^2 \mathbf{r} + 2\dot \theta \, \mathbf k \times \mathbf{v}_{\text{rel}} \end{aligned}
Piston motion
As an example of planar rigid body motion, we can analyze the dynamics of a piston mechanism, which is commonly used in engines. In this system, the piston moves linearly, while it is connected to a rotating crankshaft and connecting rod. Such systems exemplify how rotation and translation interact in mechanical devices. By studying the angular velocities, angular accelerations, and lengths of the links, we can determine the motion of all components, providing insights into the kinematics of rigid bodies in a plane.
- Link OA rotates with an angular velocity \omega_{OA} = -10 \, \text{rad/s} and an angular acceleration \alpha_{OA} = 5 \, \text{rad/s}^2;
- link AB rotates with an angular velocity \omega_{AB} = -2.5 \, \text{rad/s};
- the length of OA is 0.12 \, \text{m};
- the length of AB is 0.35 \, \text{m};
- the projections of the crank’s position are 4\, \text{cm} vertically and 3\, \text {cm} horizontally.
Velocity
Given that O is fixed and using the velocity formula for a rigid body:
\mathbf{v}_A = \mathbf{v}_O + \omega_{OA} \, \mathbf{k} \times \mathbf{r}_{OA}
where:
- \mathbf{v}_O = 0 as the point is fixed;
- \omega_{OA} = -10 \, \text{rad/s} (clockwise angular velocity of OA);
- \mathbf{r}_{OA} = 0.12 \, \text{m} \cdot \left( -\frac{3}{5} \, \mathbf{i} + \frac{4}{5} \, \mathbf{j} \right) = (-0.072 \, \mathbf{i} + 0.096 \, \mathbf{j}) \, \text{m}.
The velocity is:
\begin{aligned} \mathbf{v}_A & = (-10) \, \mathbf{k} \times (-0.072 \, \mathbf{i} - 0.096 \, \mathbf{j}) \, \tfrac{\text{m}}{\text{s}} \\ & = -10 \cdot (-0.096 \, \mathbf{i} - 0.072 \, \mathbf{j}) \, \tfrac{\text{m}}{\text{s}}\\ & = 0.96 \, \mathbf{i} + 0.72 \, \mathbf{j} \, \tfrac{\text{m}}{\text{s}} \end{aligned}
\mathbf{r}_{AB} can be computed using Pythagoras’s theorem. The length of AB (hypotenuse) is given as 0.33 \, \text{m}. The vertical component (height) is:
R_{ABv} = -0.12 \cdot \frac{4}{5} \mathbf j = -0.096 \, \mathbf j \, \text{m}
The horizontal component (base) is:
R_{ABh} = \sqrt{AB^2 - R_{ABv}^2} = \sqrt{0.33^2 - 0.072^2} = \sqrt{0.1089 - 0.009216} = \sqrt{0.099484} \approx 0.316 \, \mathbf i \, \text{m}
Therefore:
\mathbf{r}_{AB} = 0.316 \, \mathbf i - 0.096 \, \mathbf j \, \text{m}
The velocity of B is:
\mathbf{v}_B = \mathbf{v}_A + \omega_{AB} \mathbf{k} \times \mathbf{r}_{AB} \, \tfrac{\text{m}}{\text{s}}
Substituting:
\begin{aligned} \mathbf{v}_B & = (0.96 \, \mathbf{i} + 0.72 \, \mathbf{j}) + \omega_{AB} \, \mathbf{k} \times (0.316 \, \mathbf{i} - 0.096 \, \mathbf{j}) \, \tfrac{\text{m}}{\text{s}} \\ & = (0.96 \, \mathbf{i} + 0.72 \, \mathbf{j}) + \omega_{AB} (0.096 \, \mathbf{i} + 0.316 \, \mathbf{j}) \, \tfrac{\text{m}}{\text{s}} \\ & = (0.96 + 0.096 \, \omega_{AB}) \, \mathbf{i} + (0.72 + 0.316 \, \omega_{AB}) \, \mathbf{j} \, \tfrac{\text{m}}{\text{s}} \end{aligned}
It is possible to apply the constraints: since \mathbf{v}_B can only move horizontally, its \mathbf{j}-component must be zero:
0.72 + 0.316 \, \omega_{AB} = 0
Solving for \omega_{AB}:
\omega_{AB} = -\frac{0.72}{0.316} = -2.27 \, \tfrac{\text{rad}}{\text{s}}
Substitute \omega_{AB} = -2.27 into the \mathbf{i}-component of \mathbf{v}_B:
v_B = 0.96 + 0.096 \cdot (-2.27) \, \tfrac{\text{m}}{\text{s}} = 0.742\, \tfrac{\text{m}}{\text{s}}
The direction is known and therefore:
\begin{aligned} \omega_{AB} & = -2.27 \, \tfrac{\text{rad}}{\text{s}}\\ \mathbf{v}_B & = 0.742 \, \mathbf{i} \, \tfrac{\text{m}}{\text{s}} \end{aligned}
Acceleration
To compute the acceleration of A, we use the formula for the acceleration of a rigid body:
\mathbf{a}_A = \mathbf{a}_O + \mathbf{\alpha}_{OA} \mathbf k \times \mathbf{r}_{OA} - \omega_{OA}^2 \mathbf{r}_{OA}
where:
- \mathbf{a}_O = 0 as the point is fixed;
- \omega_{OA} = -10 \, \text{rad/s} (clockwise angular velocity of OA);
- \mathbf{\alpha}_{OA} = 5 \, \tfrac{\text{rad}}{\text{s}^2} \, \mathbf{k};
- \omega_{OA} = -10 \, \tfrac{\text{rad}}{\text{s}} \, \mathbf{k};
- \mathbf{r}_{OA} = -0.072 \, \mathbf{i} + 0.096 \, \mathbf{j} \, \text{m}.
Using these values:
\begin{aligned} \mathbf{a}_A & = \mathbf{a}_O + \mathbf{\alpha}_{OA} \mathbf k \times \mathbf{r}_{OA} - \omega_{OA}^2 \mathbf{r}_{OA} \\ & = 0 + 5\, \mathbf k \times \left( -0.072 \, \mathbf{i} + 0.096 \, \mathbf{j} \right)\tfrac{\text{m}}{\text{s}^2} - \omega_{OA}^2 \left(-0.072 \, \mathbf{i} + 0.096 \, \mathbf{j}\right)\tfrac{\text{m}}{\text{s}^2} \\ & = \left(-0.48 \, \mathbf{i} - 0.36 \, \mathbf{j} \right) \tfrac{\text{m}}{\text{s}^2} - 100 \left(-0.072 \, \mathbf{i} + 0.096 \, \mathbf{j}\right)\tfrac{\text{m}}{\text{s}^2} \\ & = \left(-0.48 \, \mathbf{i} - 0.36 \, \mathbf{j} \right) \tfrac{\text{m}}{\text{s}^2} + \left(7.2 \, \mathbf{i} - 9.6 \, \mathbf{j}\right)\tfrac{\text{m}}{\text{s}^2} \\ & = 6.72 \, \mathbf{i} - 9.96 \, \mathbf{j} \tfrac{\text{m}}{\text{s}^2} \end{aligned}
The acceleration of B is computed using the relative acceleration formula:
\mathbf{a}_B = \mathbf{a}_A + \alpha_{AB} \mathbf{k} \times \mathbf{r}_{AB} - \omega_{AB}^2 \mathbf{r}_{AB}.
Using:
- \mathbf{a}_A = 6.72 \, \mathbf{i} - 9.96 \, \mathbf{j} \, \tfrac{\text{m}}{\text{s}^2};
- \mathbf{r}_{AB} = 0.316 \, \mathbf{i} - 0.096 \, \mathbf{j} \, \text{m};
- \omega_{AB} = -2.27 \, \tfrac{\text{rad}}{\text{s}}.
\begin{aligned} \mathbf{a}_B & = \mathbf{a}_A + \mathbf{\alpha}_{AB} \mathbf k \times \mathbf{r}_{AB} - \omega_{AB}^2 \mathbf{r}_{AB} \\ & = \left( 6.72 \, \mathbf{i} - 9.96 \, \mathbf{j} \right)\tfrac{\text{m}}{\text{s}^2} + \mathbf{\alpha}_{AB} \mathbf k \times \left(0.316 \, \mathbf{i} - 0.096 \, \mathbf{j} \right)\tfrac{\text{m}}{\text{s}^2} - \left(-2.27\right)^2 \left( 0.316 \, \mathbf{i} - 0.096 \, \mathbf{j}\right)\tfrac{\text{m}}{\text{s}^2} \\ & = \left( 6.72 \, \mathbf{i} - 9.96 \, \mathbf{j} \right)\tfrac{\text{m}}{\text{s}^2} + \alpha_{AB} (0.096 \, \mathbf{i} + 0.316 \, \mathbf{j}) + \left(-1.629 \, \mathbf{i} + 0.4947 \, \mathbf{j}\right)\tfrac{\text{m}}{\text{s}^2} \end{aligned}
Since \mathbf{a}_B is only in the \mathbf{i}-direction, the \mathbf{j}-component of \mathbf{a}_B must equal 0. We can use this to solve for \alpha_{AB} \mathbf{k}, then substitute to find a_B. The \mathbf{j}-component of \mathbf{a}_B must be zero:
-9.96 + 0.4947 + 0.316 \alpha_{AB} = 0
Solving for \alpha_{AB}:
\alpha_{AB} = \frac{9.96 - 0.4947}{0.316} \approx 29.95 \, \tfrac{\text{rad}}{\text{s}^2}
And finally:
a_B = 6.72 - 1.629 + 0.096 \cdot 29.95 \approx 7.966 \, \tfrac{\text{m}}{\text{s}^2}
The direction is known and therefore:
\begin{aligned} \alpha_{AB} & = 29.95 \, \tfrac{\text{rad}}{\text{s}^2}\\ \mathbf{a}_B & = 0.7966 \, \mathbf{i} \, \tfrac{\text{m}}{\text{s}^2} \end{aligned}
Wheel rolling on a fixed horizontal surface
Let’s consider a wheel rolling (without slipping) on a fixed horizontal surface, and \omega = \dot{\theta} is the angular velocity and \alpha = \ddot{\theta} is the angular acceleration.
In the coordinate system \mathbf{i} (x-direction) points to the right, \mathbf{j} (y-direction) points downward, and \mathbf{k} (z-direction) points into the plane of motion.
The instantaneous center of rotation for a rolling wheel without slipping is the point of contact P with the ground. At any instant, this point has zero velocity.
Velocity
To find the velocity of the geometric center C of the wheel, we use the relative velocity equation:
\mathbf{v}_C = \mathbf{v}_P + \mathbf{\omega} \times \mathbf{r}_{CP}
Given that \mathbf{v}_P = \mathbf{0} (since P is the instantaneous center), the equation simplifies to:
\mathbf{v}_C = \mathbf{\omega} \times \mathbf{r}_{CP}
Here, \mathbf{r}_{CP} is the position vector from P to C. Since C is directly above P at a distance r, we have:
\mathbf{r}_{PC} = -r \mathbf{j}
The angular velocity vector is:
\mathbf{\omega} = \omega \mathbf{k}
Calculating the cross product:
\mathbf{\omega} \times \mathbf{r}_{CP} = \omega \mathbf{k} \times (-r \mathbf{j}) = \omega r (\mathbf{k} \times \mathbf{j}) = \omega r (-\mathbf{i}) = -\omega r \mathbf{i}
Considering the direction of rolling (positive in the clockwise direction), the velocity of the center C becomes:
\mathbf{v}_C = \omega r \mathbf{i}
Acceleration
To determine the acceleration \mathbf{a}_C of the geometric center, we differentiate the velocity \mathbf{v}_C with respect to time:
\mathbf{a}_C = \frac{d\mathbf{v}_C}{dt} = \frac{d}{dt} (\omega r \mathbf{i}) = r \frac{d\omega}{dt} \mathbf{i} = r \alpha \mathbf{i}
Thus, the acceleration of the geometric center C is:
\mathbf{a}_C = r \alpha \mathbf{i}
This acceleration is directed solely along the x-axis.
Next, we determine the acceleration \mathbf{a}_P of the instantaneous center of rotation P. Using the acceleration relationship for a rigid body:
\begin{aligned} \mathbf{a}_P & = \mathbf{a}_C + \alpha \mathbf{k} \times \mathbf{r}_{PC} - \omega^2 \mathbf{r}_{PC} \\ & = r \alpha \mathbf{i} + \alpha \mathbf{k} \times r \mathbf{j} - \omega^2 r \mathbf{j} \\ & = r \alpha \mathbf{i} - r \alpha \mathbf{i} - r\omega^2 \mathbf{j} \\ & = - r\omega^2 \mathbf{j} \end{aligned}
While the instantaneous center of rotation P has zero velocity at the instant considered, it possesses a non-zero acceleration directed upward.
This upward acceleration is a consequence of the rotational motion of the wheel, specifically due to the centripetal acceleration arising from the wheel’s angular velocity \omega.
Wheel rolling on a curved surface
Let’s consider a wheel rolling (without slipping) on a curved surface, and \omega = \dot{\theta} is the angular velocity and \alpha = \ddot{\theta} is the angular acceleration.
In the coordinate system \mathbf{i} (x-direction) points to the right, \mathbf{j} (y-direction) points downward, and \mathbf{k} (z-direction) points into the plane of motion.
We’ll use normal and tangential coordinates, as these are well-suited for the wheel’s motion along a curved path.
Velocity
The relative velocity equation, will be used to determine the velocity and acceleration at various points of interest. We start with the velocity of the center of the wheel, using the point of contact as our reference point.
The velocity of the center of the wheel is expressed as follows:
\mathbf{v}_C = \mathbf{v}_P + \omega \mathbf{k} \times \mathbf{r}_{PC}
Substituting the known terms, the velocity at the instantaneous center is zero (\mathbf{v}_P = 0), and the vector \mathbf{r}_{CP} is the radius r in the \mathbf{e}_n direction:
\mathbf{v}_C = 0 + \omega \mathbf{k} \times (r \mathbf{e}_n)
Crossing \mathbf{k} with \mathbf{e}_n results in the tangential direction \mathbf{e}_t:
\mathbf{v}_C = r \omega \mathbf{e}_t
The velocity is tangential to the path, and its magnitude, r \omega, represents the speed of the wheel’s center.
Acceleration
The acceleration of the center of the wheel can be found by taking the time derivative of \mathbf{v}_C. The expression becomes:
\mathbf{a}_C = \frac{\mathrm d}{\mathrm dt} (r \omega \mathbf{e}_t)
Since r is constant, the derivative expands as:
\mathbf{a}_C = r \alpha \mathbf{e}_t + r \omega \frac{\mathrm d\mathbf{e}_t}{\mathrm dt}
The derivative \frac{\mathrm d\mathbf{e}_t}{\mathrm dt} is related to the curvature of the path:
\frac{\mathrm d\mathbf{e}_t}{\mathrm dt} = \frac{\dot{s}}{\rho} \mathbf{e}_n
where \dot{s}, the magnitude of velocity, equals r \dot{\theta} = r \omega, and \rho is the radius of curvature. Substituting, the acceleration becomes:
\mathbf{a}_C = r \alpha \mathbf{e}_t + r \omega \left( \frac{r \omega}{\rho} \mathbf{e}_n \right)
Simplifying:
\mathbf{a}_C = r \alpha \mathbf{e}_t + \frac{r \omega^2}{\rho} \mathbf{e}_n
This result reveals that the center of the wheel experiences a tangential acceleration r \alpha = r \ddot{\theta} and a normal (radial) acceleration \frac{r \omega^2}{\rho} = \frac{r \dot{\theta}^2}{\rho}, directed inward.
Finally, we determine the acceleration at the point of contact \mathbf{P}. Using the relative acceleration equation:
\mathbf{a}_P = \mathbf{a}_C + \alpha \mathbf{k} \times \mathbf{r}_{CP} - \omega^2 \mathbf{r}_{CP}
The vector \mathbf{r}_{CP} is -r \mathbf{e}_n, so:
\mathbf{a}_P = \mathbf{a}_C + \alpha \mathbf{k} \times (-r \mathbf{e}_n) - \omega^2 (-r \mathbf{e}_n)
Substituting \mathbf{a}_C:
\mathbf{a}_P = \left( r \alpha \mathbf{e}_t + \frac{r \omega^2}{\rho} \mathbf{e}_n \right) - r \alpha \mathbf{e}_t + r \omega^2 \mathbf{e}_n
Simplifying, the tangential terms cancel, and the result becomes:
\mathbf{a}_P = \left( 1 + \frac{r}{\rho} \right) r \omega^2 \mathbf{e}_n = \left( 1 + \frac{r}{\rho} \right) r \dot{\theta}^2 \mathbf{e}_n
This expression describes the acceleration at the point of contact for a wheel rolling on a curved path. It accounts for the curvature of the path and combines the effects of tangential and normal accelerations.
It worth noting that as \rho \to \infty, the fixed plane curve relationship approach the fixed straight line conditions:
\begin{aligned} \mathbf{a}_C & = r \alpha \mathbf{e}_t + \frac{r \omega^2}{\rho} \mathbf{e}_n \underset{\rho \to \infty}{=} r \alpha \mathbf{e}_t = r \alpha \, \mathbf{i} \\[5pt] \mathbf{a}_P & = \left( 1 + \frac{r}{\rho} \right) r \omega^2 \mathbf{e}_n \underset{\rho \to \infty}{=} r \omega^2 \mathbf{e}_n = -r \omega^2 \, \mathbf{j} \end{aligned}
Sample problem
Let’s make an example and solve an engineering problem related to determining the velocity of a point relative to two different reference frames in planar two-dimensional motion.
We will analyze the velocity relationships between two reference frames and a specific point within a mechanical system.
The problem involves a fixed reference frame, and a moving frame 2, centered in \text{O}_2, which is welded to a vertical bar (\text{bar 2}). The common point of interest, \mathbf P, moves relative to these two frames. In this scenario, there is a collar \text C pinned to \text{bar 1} and sliding along \text{bar 2}. The angular velocity of \text{bar 1} is given as \omega_1 = -0.2 \, \frac{\text{rad}}{\text{s}}, and the acceleration is \alpha_1 = -0.25 \, \frac{\text{rad}}{\text{s}^2}.
Velocity
The goal is to determine the angular velocity of \text{bar 2} at that instant and the relative velocity of \mathbf P with respect to the moving frame 2. To achieve this, we will use the velocity relationship:
\mathbf{v}_P = \mathbf{v}_{O_2} + \mathbf{v}_{\text{rel}} + \omega_1 \, \mathbf{k} \times \mathbf{r}
where:
- \mathbf{v}_P is the velocity of point \mathbf P,
- \mathbf{v}_{O_2} is the velocity of the origin O_2 of the moving frame,
- \mathbf{v}_{\text{rel}} is the relative velocity of \mathbf P in the moving frame,
- \omega_1 is the angular velocity of the moving frame relative to the fixed frame,
- \mathbf{r} is the position vector of \mathbf P relative to O_2.
Here, \mathbf{k} points out of the plane, and \mathbf{r} is expressed in terms of the moving frame coordinates x_2, y_2.
The fixed frame 1 has its origin at O_1 with basis vectors \mathbf{i}_1, \mathbf{j}_1. The moving frame is welded to \text{bar 2}, with its origin at O_2 and basis vectors \mathbf{i}_2, \mathbf{j}_2. The position vector of \mathbf P in the moving frame is:
\mathbf{r}_{O_2P} = r_x \mathbf{i}_2 + r_y \mathbf{j}_2
In the moving frame, the relative velocity \mathbf{v}_{\text{rel}} represents the motion of \mathbf P observed from O_2. Since \mathbf P slides along \text{bar 2}, its motion is constrained to the y_2-axis:
\mathbf{v}_{\text{rel}} = v_{\text{rel}} \, \mathbf{j}_2
where v_{\text{rel}} is the magnitude of the relative velocity along \mathbf{j}_2.
From the velocity equation, we have:
\mathbf{v}_P = \mathbf{v}_{O_2} + v_{\text{rel}} \mathbf{j}_2 + \omega_2 \, \mathbf{k} \times \mathbf{r}_{O_2P}
The angular velocity \omega_2 of \text{bar 2} is unknown and must be determined. The position vector \mathbf{r}_{O_2P} is known from geometry:
\mathbf{r}_{O_2P} = -\sin\left(30 ^\circ\right) + \cos\left(30 ^\circ\right) = -0.5 \mathbf{i}_2 + 0.866 \mathbf{j}_2
The cross product yields:
\mathbf{k} \times \mathbf{r}_{O_2 P} = -0.866 \mathbf{i}_2 - 0.5 \mathbf{j}_2
Substituting into the velocity equation:
\mathbf{v}_P = \mathbf{v}_{O_2} + v_{\text{rel}} \mathbf{j}_2 - 0.866 \omega_2 \mathbf{i}_2 - 0.5 \omega_2 \mathbf{j}_2
Since O_2 is stationary in the fixed frame, \mathbf{v}_{O_2} = 0. Therefore:
\mathbf{v}_P = v_{\text{rel}} \mathbf{j}_2 - 0.866 \dot \theta_2 \mathbf{i}_2 - 0.5 \dot \theta_2 \mathbf{j}_2
From the perspective of \text{bar 1}, the velocity of point \mathbf P can be computed from the velocity of two points on the same body:
\mathbf{v}_P = \omega_1 \, \mathbf{k} \times \mathbf{r}_{O_1P} = -0.2 \, \mathbf{k} \times \mathbf{r}_{O_1P}
With \mathbf{r}_{O_1P} = -0.5 \mathbf{i}_1 + 0.866 \mathbf{j}_1, the cross product gives:
\mathbf{v}_P = 0.1732 \mathbf{i}_1 + 0.1 \mathbf{j}_1
Equating \mathbf{i}_1 and \mathbf{j}_1 components in the fixed frame gives two equations:
\begin{aligned} & -0.866 \omega_2 = 0.1732 \\ & v_{\text{rel}} - 0.5 \omega_2 = 0.1 \end{aligned}
Solving for \omega_2:
\omega_2 = -0.2 \, \frac{\text{rad}}{\text{s}}
Substitute into the second equation to find v_{\text{rel}}:
v_{\text{rel}} = 0.1 \, \frac{\text{m}}{\text{s}}
The angular velocity of \text{bar 2} is \omega_2 = -0.2 \,\frac{\text{rad}}{\text{s}}, and the relative velocity of \mathbf P in the moving frame is \mathbf{v}_{\text{rel}} = 0.1 \, \mathbf{j}_2 \, \frac{\text{m}}{\text{s}}.
Acceleration
To extend the previous problem to accelerations, we now incorporate the Coriolis term along with tangential and radial acceleration terms. The relationship for the acceleration of a point \mathbf P, relative to two different reference frames, is given as:
\mathbf{a}_P = \mathbf{a}_{O_2} + \mathbf{a}_{\text{rel}} + \alpha \, \mathbf{k} \times \mathbf{r} - \omega^2 \mathbf{r} + 2\omega \, \mathbf{k} \times \mathbf{v}_{\text{rel}}
where:
- \mathbf{a}_P is the acceleration of \mathbf P in the fixed frame,
- \mathbf{a}_{O_2} is the acceleration of the moving frame’s origin, O_2, relative to the fixed frame. In this case, \mathbf{a}_{O_2} = 0,
- \mathbf{a}_{\text{rel}} is the relative acceleration of \mathbf P in the moving frame,
- \alpha_2 is the angular acceleration of the moving frame (\text{bar 2}) relative to the fixed frame,
- \omega_2 is the angular velocity of the moving frame,
- \mathbf{r} is the position vector of \mathbf P relative to O_2,
- \mathbf{v}_{\text{rel}} is the relative velocity of \mathbf P in the moving frame.
The angular acceleration \alpha_2 and relative acceleration \mathbf{a}_{\text{rel}} of \mathbf P in the moving frame are unknown.
Since \mathbf{a}_{O_2} = 0, the equation simplifies to:
\mathbf{a}_P = \mathbf{a}_{\text{rel}} + \alpha_2 \, \mathbf{k} \times \mathbf{r} - \omega_2^2 \mathbf{r} + 2\omega_2 \, \mathbf{k} \times \mathbf{v}_{\text{rel}}
The position vector of \mathbf P relative to O_2 is:
\mathbf{r} = -0.5 \, \mathbf{i}_2 + 0.866 \, \mathbf{j}_2
It is possible to compute all the terms; the relative acceleration is to be determined:
\mathbf{a}_{\text{rel}} = a_{\text{rel}} \, \mathbf{j}_2
The tangential acceleration is:
\alpha_2 \, \mathbf{k} \times \mathbf{r} = \alpha_2 \, \mathbf{k} \times (-0.5 \, \mathbf{i}_2 + 0.866 \, \mathbf{j}_2) = 0.866 \alpha_2 \, \mathbf{i}_2 + 0.5 \alpha_2 \, \mathbf{j}_2
The centripetal acceleration is:
-\omega_2^2 \mathbf{r} = -(-0.2)^2 (-0.5 \, \mathbf{i}_2 + 0.866 \, \mathbf{j}_2) = -0.02 \, \mathbf{i}_2 + 0.03464 \, \mathbf{j}_2
Finally the Coriolis acceleration is:
2\omega_2 \, \mathbf{k} \times \mathbf{v}_{\text{rel}} = 2(-0.2) \, \mathbf{k} \times (0.1 \, \mathbf{j}_2) = -0.04 \, \mathbf{i}_2
Substituting these terms into the acceleration equation:
\begin{aligned} \mathbf{a}_P & = a_{\text{rel}} \, \mathbf{j}_2 + (0.866 \alpha_2 - 0.02 - 0.04) \, \mathbf{i}_2 + (0.5 \alpha_B + 0.03464) \, \mathbf{j}_2 \\ & = (0.866 \alpha_2 - 0.06) \, \mathbf{i}_2 + (a_{\text{rel}} + 0.5 \alpha_2 + 0.03464) \, \mathbf{j}_2 \end{aligned}
From the perspective of \text{bar 1} the acceleration of point \mathbf P can be computed from the acceleration of two points on the same body:
\mathbf{a}_P = \alpha_1 \, \mathbf{k} \times \mathbf{r} - \omega_1^2 \mathbf{r}
Substitute \alpha_1 = 0.25 \, \mathbf{k}, \omega_1 = -0.2 \, \text{rad/s}, and \mathbf{r} = -0.5 \, \mathbf{i}_1 + 0.866 \, \mathbf{j}_1:
\begin{aligned} \mathbf{a}_P & = (0.25 \, \mathbf{k}) \times (-0.5 \, \mathbf{i}_1 + 0.866 \, \mathbf{j}_1) - (-0.2)^2 (-0.5 \, \mathbf{i}_1 + 0.866 \, \mathbf{j}_1) \\ & = (0.2165 \, \mathbf{i}_1 + 0.125 \, \mathbf{j}_1) - (-0.02 \, \mathbf{i}_1 + 0.03464 \, \mathbf{j}_1)\\ & = 0.1965 \, \mathbf{i}_1 + 0.15964 \, \mathbf{j}_1 \end{aligned}
Equating \mathbf{i} and \mathbf{j} components, solve for \alpha_2 and a_{\text{rel}}:
\begin{aligned} & 0.866 \alpha_2 - 0.06 = 0.1965 \\ & a_{\text{rel}} + 0.5 \alpha_2 + 0.03464 = 0.15964 \end{aligned}
Solving:
\begin{aligned} & \alpha_2 = 2.73 \, \tfrac{\text{rad}}{\text{s}^2} \\ & a_{\text{rel}} = -0.125 \, \tfrac{\text{m}}{\text{s}^2} \end{aligned}
The angular acceleration of \text{bar 2} is \alpha_2 = 2.73 \, \frac{\text{rad}}{\text{s}^2} (counterclockwise). The relative acceleration of \mathbf P in the moving frame is \mathbf{a}_{\text{rel}} = -0.125 \, \mathbf{j}_2 \,\frac{\text{m}}{\text{s}^2}.